- https://www.desmos.com
- https://www.geogebra.org
- You can check your calculations of your own weight in Newtons, your mass in kilograms, or your weight on the Moon and on other planets using the following online calculators:
- Pre-Calculus and Trigonometry
- Linear and Angular Speed
- Great Circle Navigation
- The Vis-Viva Lab
- Delta-v for a
- Hohmann Transfer
- A Trip to the Moon
- The Rocket Equation
- Kepler’s Third Law of
- Planetary Motion, Orbital Inclination
- Single Variable Calculus
- Maximum Aerodynamic Pressure
- The Acceleration and g-Force Lab: Part 2 – SpaceX
- Falcon 9 Launch
- Differential in Velocity
- Solving Kepler’s Equation with Newton’s Method
- Distance at Engine Cut-Off
- Calculus with Parametric and Polar Equations
- Energy and Escape Velocity
- Parametric and Polar Equations in Orbit
- Tangent to Flight
- Path and Pitch
- Multivariable and Vector Calculus
- Work Done by Gravity
- Great Circle Distance and Bearing
- Optimal Staging with Lagrange Multipliers
- Differential Equations
- Solving the Rocket Equation
- Solutions
- Algebra
- Gravity and Acceleration
- Thrust-to-Weight Ratio
- The Acceleration and g-Force Lab: Part 1 – The Thumper Launch in KSP
- Orbital Speed
- Orbital Period
- Computing the Mass of a Planet or Moon
- Kinetic and Potential Energy in a Circular Orbit
- is any change in motion, including a change in direction, or an increase or decrease in speed.
- Isaac Newton’s
- First law of motion states: a force is required to cause an object to accelerate.
- Second law of motion, described by the equation F = ma, states that the magnitude of force, F, that causes a change in the motion of an object is equal to the mass, m, of the object, times the acceleration, a, of that object.
- Newton’s law of gravitation, expressed by the equation F = GMm/r2, describes how the force of gravity is related to the mass of two objects and the distance between them.
- In this equation, G, the constant of proportionality, is called the universal gravitational constant.
- The values M and m are the masses of two objects experiencing the force of gravity and r is the distance between the center of mass of each of the objects.
- From the law of gravitation, we say the force of gravity is proportional to the product of the masses and inversely proportional to the square of the distance between them.
As we have seen, acceleration due to gravity can be computed with the equation π = πΊπ / r^2
This means acceleration due to gravity changes with the values of the mass M and the distance r.
Define weight as the force due to gravity
Compute weight using the equation
πΉ = ππ = π (πΊπ / r^2 ) = πΊππ /π2
As stated previously, the value G is a constant of proportionality and is determined by the choice of units. If we measure mass in kilograms, distance in meters, and time in seconds
G = 6.67 x 10-11 m3/(kg*s2) and with these units, weight is measured in Newtons.
If weight is measured in pounds, this is also a measure of the force of gravity.
Stating an object’s weight in pounds is not directly a measure of its mass because weight changes with changes in the gravity field, that is, it depends on the values of M and r.
To convert units
1 pound = 4.45 Newtons of force
(anywhere in the universe including anywhere in the Kerbal system).
Near the surface of the Earth and Kerbin, any object with a mass of 1 kilogram has a weight of 2.2 pounds.
Note this means, at the surface of the Earth or Kerbin,
- a 1 kilogram mass has a weight of F = 2.2 pounds x 4.45 Newtons/pound = 9.8 Newtons.
We feel weight on the surface of Earth when the ground below our feet provides an equal and opposite force keeping us from accelerating toward the center of the Earth.
We also feel weight when we lift something and our muscles provide a force that opposes gravity. Imagine if you and a briefcase were both falling together: You would no longer feel the weight of the briefcase.
Objects in orbit around a planet or moon are also in free-fall, but moving so fast horizontally to the surface, without an atmosphere to slow them down, that gravity only bends their path.
Astronauts appear weightless because they are in free-fall and the only force acting on them is gravity. They appear weightless as long as there are no other forces (such as a rocket’s thrust) acting to oppose gravity and their weight is apparent only in the way gravity affects their trajectory.
As we have seen, acceleration due to gravity at the surface of the planet Kerbin and at the surface of Earth is exactly the same.
For Kerbin, with a mass M = 5.29 x 1022 kg and radius r = 600,000 meters,
π = πΊπ / π2 = (6.67×10−11)(5.29×1022) / 6000002 = 9.8 m/s2
and for Earth, with a mass M = 5.97 x 1024 kg and radius r = 6,378,000 meters,
π = πΊπ / r2= (6.67×10−11)(5.97×1024) / 600,000^2 = 9.8 m/s2
Because acceleration due to gravity at the surface of Kerbin is the same as it is at the surface of the Earth, the weight of an object on the surface of Kerbin will be the same as the weight on the surface of Earth.
However, because Kerbin is smaller than Earth, the acceleration due to gravity at the same altitude above each planet is not equal.
Thus, the weight of an object in space at equal altitudes above each planet also will not be the same.
Example
- the weight of a mass of m = 50 kg, at an altitude of 100 kilometers above each planet is computed as follows:
For Kerbin, at an orbital radius of r = 600,000 + 100,000 = 700,000 meters
πΉ = πΊππ / π2 = (6.67×10−11)(5.29×1022)(50) / 7000002 ≈ 360 Newtons
For Earth, at an orbital radius of r = 6,378,000 + 100,000 = 6,478,000 meters
πΉ = πΊππ / π2 = (6.67×10−11)(5.97×1024)(50) / 64780002 ≈ 474 Newtons
Use the previously introduced unit conversions and the equation πΉ = πΊππ /π2 to complete the
following:
A mass of 1 kilogram has a weight of 9.8 Newtons on the surface of Earth and Kerbin.
Find the weight (in Newtons) of a 3000 kilogram satellite on the surface of either planet.
A mass of 1 kilogram weighs about 2.2 pounds on the surface of Earth and Kerbin,
What is the weight (in pounds) of a 3000 kg satellite on the surface of Earth and Kerbin?
10. Given that a mass of 1 kg weighs about 2.2 lbs at the surface of Earth, find the mass, in kilograms, of a person that weighs 160lbs at the surface of Earth.
11. What is the weight, in Newtons, of a 160 lb person on the surface of the Earth?
12. What is the weight, in Newtons, of a 160 lb person in orbit 100km above Earth?
Remember Earth’s radius is about 6378 km.
The Kerbal Math & Physics Lab / Chapter 1
The Kerbal Math & Physics Lab / Chapter 1
In the formula πΉ = πΊππ, when we use the value G = 6.67 x 10-11 m3/(kg*s2), we assume distance is measured in meters, and mass is measured in kilograms.
We could combine the values GMm into one single constant and write πΉ = πΎ and use any units for F and r.
Again, with this equation, we can say the force of gravity is inversely proportional to the square of the distance.
Given any corresponding values for F and r, we can solve for K and, remaining consistent in our choice of units, then use that to answer questions such as the following:
13. Suppose an astronaut weighs 160 pounds on the surface of the Earth, which has a radius of about r = 3960 miles.
What is the astronaut’s weight in orbit at an altitude of 100 miles?
14. Suppose Jebidiah Kerman, a Kerbal astronaut, weighs 80 lbs on the surface of Kerbin which has a radius of 600 km.
What is Jeb’s weight at an altitude of 100 km above Kerbin?
15.
The International Space Station (ISS) orbits the Earth at an altitude of about 250 miles above the surface. Earth’s radius is about 3960 miles.
Calculate your own weight, in pounds, if you were onboard the ISS.
Derive a formula that you could use to find anyone’s weight, in pounds, if they were onboard the ISS.
The Kerbal Math & Physics Lab / Chapter 1 16.
Is there a point in space where an astronaut has no weight? Why or Why not?
17. Why does an astronaut in orbit feel weightless?
18. The mass of Earth’s Moon is M = 7.35 x 1022 kg and its radius is r = 1738 km.
Calculate the acceleration due to gravity at the surface of the Moon.
19. In KSP, the planet Kerbin’s nearest moon is called the Mun.
The mass of Kerbin’s moon, the Mun, is M = 9.76 x 1020 kg and its radius is r = 200 km.
Calculate the acceleration due to gravity at the surface of the Mun.
20. Calculate your own weight on Earth’s Moon, in pounds.
Derive a formula to convert any weight in pounds (on the surface of Earth) to the corresponding weight, in pounds, on Earth’s Moon.
Would your weight be the same on Kerbin’s moon as it is on Earth’s Moon?
You can check your calculations of your own weight in Newtons, your mass in kilograms, or your weight on the Moon and on other planets using the following online calculators:
https://www.thecalculatorsite.com/conversions/massandweight.php
https://www.exploratorium.edu/ronh/weight/
http://www.learningaboutelectronics.com/Articles/Weight-on-the-moon-conversion-
calculator.php
Thrust-to-Weight Ratio
Suppose the astronaut Jebidiah Kerman weighs 80lbs on the surface of Kerbin. We can calculate Jeb’s mass in kilograms as about 80 πππ × 1 ππ = 36.4 kilograms. On the surface of
1 2.2 πππ
Kerbin, with a gravitational acceleration of about 9.8 m/s2, this means Jeb has a weight of approximately F = ma = (36.4 kg)(9.8 m/s2) = 357 Newtons. This weight is the force due to gravity and is directed downwards toward the center of Kerbin. If Jeb is standing on the surface of Kerbin, an equal and opposite force is provided by the surface of Kerbin, keeping Jeb in place.
Jeb could jump off the ground if his legs provide greater than 357 Newtons of force, but without a jetpack or any other propulsive force to provide continued thrust, gravity will bring Jeb back to the ground.
It is Newton’s third law of motion that expresses the principle that a rocket engine expelling exhaust in one direction creates a force we call thrust in the opposite direction.
Jeb can achieve and maintain flight with the thrust provided by an airplane or rocket engine. We define the thrust-to-weight ratio (TWR) as a way to measure when an engine has sufficient thrust to counteract its own weight, as follows:
πππ = π‘hππ’π π‘ π€πππhπ‘
In this definition both thrust and weight are forces and should be measured in the same units (for example, Newtons or pounds) and will therefore cancel, meaning that TWR is a dimensionless value.
A rocket must have a TWR greater than 1 to lift off from the surface of a planet or moon.
As an example, the entire Saturn V rocket that launched the Apollo missions to the moon weighed about 6.5 million pounds at launch and its five first stage F-1 engines combined to provide a thrust of about 7.9 million pounds. Therefore, for a Saturn V on the launch pad, we have
TWR = π‘hππ’π π‘ = 7.9×106πππ = 1.2 π€πππhπ‘ 6.5×106 πππ
The Kerbal Math & Physics Lab / Chapter 1
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A rocket with a TWR equal to 1 would have enough thrust to support its own weight but not enough to lift that weight, while a rocket with a TWR less than 1 would not be able launch. TWR values less than or equal to 1 are important during a rocket’s landing phase. For example, the Apollo Lunar Module needed to support its own weight as it descended to the surface of the Moon and the SpaceX Falcon 9 booster rockets use TWR values less than 1 to softly touch down at their landing sites.
We can compute TWR for rocket engines as a useful comparison between different engines. Be careful to write both thrust and weight in the same units (either pounds or Newtons) and when
using metric, convert mass in kilograms into weight in Newtons. 1000 Newtons, and 1 Mega Newton = 1,000,000 Newtons.
Note that thrust-to-weight ratio is not a fixed and constant value. For rockets, weight will change as fuel is burnt, as stages are spent, and as the rocket moves through a changing gravity field. Also the amount of thrust an engine provides is usually changing during ascent and descent phases of a flight (during take-off and landing). Finally, we can calculate TWR for jet engines as well. Note that jet engines do not generally require TWR values greater than 1 to fly because the speed of the air over the wings of a jet provide lift.
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Use the definition, and appropriate units, to compute thrust-to-weight ratios for each of the following:
1.
In Kerbal Space Program, the Jumping Flea, a solid fuel rocket outfitted with a command pod and parachute, has a total mass of 2440 kilograms and a thrust of 162.9 KiloNewtons at sea level. Calculate the TWR for the Jumping Flea at launch.
2.
The Soviet R-7 rocket that launched the Sputnik satellite had a total mass at liftoff of about 267 metric tons (267,000 kilograms). The four first stage boosters of the R-7 combined to provide a total of 3.89 MegaNewtons of thrust. Calculate TWR for the Soviet R-7 at launch.
3.
The KerbalX, a rocket which can land a Kerbal on the Mun, has a total mass of 128,700 kilograms on the launch pad. At launch, the KerbalX is equipped with six LV-T45 liquid fuel engines that each provide 168 KiloNewtons of thrust and a single Rockomax Mainsail engine that provides 1379 KiloNewtons of thrust. Calculate TWR for the KerbalX at liftoff.
The Kerbal Math & Physics Lab / Chapter 1
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4. The Saturn V, which launched American astronauts to the Moon, had a total mass of approximately 2.97 million kilograms. It lifted off the launch pad with the combined thrust of 5 Rocketdyne F-1 engines that each provided about 7.02 million Newtons of thrust. Calculate the TWR for the Saturn V at liftoff.
5. NASA’s Space Shuttle weighed about 4,470,000 lbs at lift-off. The Shuttle launched with two solid rocket boosters, each providing about 2,800,000 lbs of thrust, and a main engine that provided about 1,180,000 lbs of additional thrust. Calculate the TWR for the Shuttle at lift-off.
6. In February 2018, a test flight of the SpaceX Falcon Heavy launched a red Tesla Roadster into an orbit about the Sun which reaches a distance further than Mars. At liftoff, the Falcon Heavy had a mass of about 1.42 million kilograms and 27 Merlin engines each of which provided about 845,000 Newtons of thrust. Calculate the Falcon Heavy’s TWR at liftoff.
The Kerbal Math & Physics Lab / Chapter 1
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7. In KSP, the Mun Lander is included with the stock KerbalX rocket and is capable of landing on the Mun, the nearest moon to Kerbin. Based on the Mun’s mass and radius, acceleration due to gravity at the surface of the Mun is approximately a = 1.63 m/s2. Suppose that the Mun Lander has a mass of about 15 metric tons (15,000 kilograms). What thrust (in Newtons) is required to launch the Mun Lander from the surface of the Mun?
8. The TWR of a rocket changes during flight as its mass (and thus weight) and also its thrust change. Consider the following flight data for a solid fuel rocket, based on the BACC “Thumper” Solid Fuel Booster with the Mk1 Command Pod, an Inline Stabilizer and a parachute, in KSP.
Initial mass: 8.69 metric tons
Initial thrust: 250 KiloNewtons, at sea-level (ASL)
Final mass: 2.54 metric tons (at engine cut-off)
Final thrust: 300 KiloNewtons, in a vacuum (VAC)
At launch, acceleration due to gravity on the rocket is 9.8 meters/sec2 and at engine cut- off, at an altitude of about 30,000 meters above Kerbin, acceleration due to gravity is approximately 9 meters/sec2.
Compute the rocket’s TWR at liftoff and right before engine cut-off.
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A scenario similar to that given previously in question 8 is explored by Mike Aben on YouTube in his Let’s Do the Math series, available at https://www.youtube.com/watch?v=VS1XACh4upc and also at https://sites.google.com/kspmath.
In Kerbal Space Program, create a single stage rocket and record mass, weight, thrust, and thrust to weight ratio (TWR) at launch and before engine cut-off as done by Mike Aben and in the previous section of this workbook. Check the TWR values you calculate against what is displayed in game. What is the max TWR encountered by your rocket? How does this value relate to g-forces experienced by the rocket?
Can you build a rocket with a TWR just above 1 that can hover above the launch pad, like a SpaceX Falcon 9, Star Hopper or StarShip? Describe your approach, perhaps you can find an already built ship at KerbalX.com. Describe the challenges of maintaining a TWR at or just above 1.
The Acceleration and g-Force Lab, Part 1 – The Thumper Launch in KSP
In this lab exercise we’ll examine the thrust-to-weight ratio, delta-v, acceleration and g-force
experienced during a rocket launch.
To illustrate, let’s consider a rocket built in KSP, the “Thumper Test Rocket” made with the following components: a MK1 command pod, an Advanced Inline Stabilizer, the TD-12 decoupler, the BACC Thumper solid fuel booster and 4 basic aerodynamic stabilizer fins. For this example, we’ll reduce the BACC Thumper solid fuel booster to 50% thrust and reduce the amount of fuel in the booster to 500 units.
With 500 units of fuel, the “Thumper” test rocket has a total mass of 6370 kg on the launch pad and is expected to burn approximately 3750 kg of fuel in 52 seconds, after which the total mass of the rocket will be about 2580 kg.
At sea-level with 50% thrust, the BACC Thumper engine produces about 125 kiloNewtons of thrust. Assuming constant thrust during flight, the initial TWR is about 125000/(9.8*6370) = 2 while the final TWR at engine cut-off is approximately 125000/(9.8*2620) = 4.9.
Finally, the estimated ∆π£ (delta-v or change in speed) for the Thumper rocket is about 1525 meters/second. This is the theoretical expected change in velocity provided by the rocket, between launch and engine cut-off, and does not include the effects of gravity and air resistance. Launching from the surface of Kerbin, with an atmosphere similar to Earth’s, theactual change in velocity will be less than the theoretical value of 1525 m/s.
The Kerbal Math & Physics Lab / Chapter 1
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In this lab, we will compute TWR, record speed during launch, maximum altitude reached and compute the rocket’s acceleration and the g-forces experienced by the pilot. The Thumper rocket is designed to reach space so we’ll launch vertically and then pitch slightly eastward during ascent to insure we return with a splashdown near the Kerbal Space Center.
There are several ways to record flight data in KSP. The simplest method would be to hit the escape-key periodically to pause the game and write down speed and altitude. Another method, that will also track other in-flight data, is to download and install the free game mod “MechJeb” (short for Mechanical Jeb). With Mechjeb, players have access to a flight data recorder that can download data in .csv format to analyze in a spreadsheet like Microsoft Excel or paste into Desmos.
We will use the rocket’s speed over time to estimate its acceleration and the g-forces experienced by the rocket and pilot.
Acceleration is change in speed over time, so the units are meters/second per second, or m/s2. There are two ways we can estimate acceleration based on the data collected:
Estimate acceleration (a) by dividing total change in velocity (∆π£) by the length of time of the engine burn (∆π‘): That is, π = ∆π£.
∆π‘
Estimate acceleration (a) by finding the slope of the line of best fit (a linear regression) for the velocity data.
Both of these methods of calculation provide estimates for the average acceleration over a given time interval and both can be referred to as estimates of the average rate of change in velocity. A primary topic in calculus involves finding instantaneous rates of change, for example the acceleration at a single instant in time. We can return to this lab in chapter 3 to estimate the instantaneous acceleration on a rocket.
To compute g-force (GF), consider the upward (positive) force N produced by the rocket and the downward (negative) weight mg of the rocket, where m is the rocket’s mass, and g is the acceleration due to gravity at the surface of the planet. If the sum, F, of these forces produces anacceleration(a),wehaveF=ma=N–mg. SolvingforN,wehaveN=ma+mg. Wedefine g-force as the ratio of rocket’s thrust to the weight of the rocket. In other words, g-force is mathematically equal to TWR. and we have
πΊπΉ =πππ = π =ππ+ππ=ππ+ππ=π+1 ππ ππ ππ ππ π
Because the factor m drops out in the equation above, we can refer to a rocket’s thrust-to- weight ratio, or the g-force experienced by the rocket and pilot, as the same thing.
The Kerbal Math & Physics Lab / Chapter 1
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The Kerbal Math & Physics Lab / Chapter 1 We now have the relation πππ = π + 1 which relates thrust-to-weight ratio, acceleration, and
π
surface gravity. As the rocket picks up speed, the acceleration increases as does the thrust-to- weight ratio.
Acceleration and g-Force Lab Data:
Build and launch the Thumper Test Rocket as described above (or a similar rocket), and collect flight data to complete the table below. Record data for at least 10 different points in time between launch and engine cut-off.
Rocket Name: _______________ Launch Date: ________ Launch Site: ___________ Initial Mass: __________ Engine burn time (in seconds): _______
Total Mass at engine cut-off: _____________
Engine Thrust: ________________
Initial TWR: _________ Final TWR (at engine cut-off): ______________
Theoretical delta-v: ____________ (use values provided in-game or the ∆π£ formula in Ch. 2)
* ASL means Above Sea-Level
** For Mach speed, divide speed (in meters/sec) by 343. Mach 1 (343 m/s) is the speed of sound, Mach 2 is twice the speed of sound, in the atmosphere of Earth and Kerbin.
Complete the previous table, or reproduce the table on a separate sheet of paper, and use a separate sheet of paper to complete the following:
Use Excel or Desmos to graph your collected velocity data (from launch to engine cut- off) with seconds on the horizontal axis and speed on the vertical axis.
Use Excel or Desmos to graph the height data (from launch to engine cut-off) with seconds on the horizontal axis and height (in meters) on the vertical axis.
Use Excel or Desmos to graph velocity and height data for the entire flight from launch to splash down.
Estimate the acceleration on the rocket (from launch to engine cut-off) using π = ∆π£ ∆π‘
where ∆π£ is the total change in velocity during the burn and ∆π‘ is the time interval of the burn.
Estimate the acceleration on the rocket (during engine burn) by finding the slope of the line of best fit for the velocity data over the interval of the engine burn. Use Excel, Desmos, the calculator or computer to find the line of best fit.
Use your estimates for acceleration computed above to estimate the average G-force experienced by the rocket and pilot between launch and engine cut-off.
What is the g-force experienced by the pilot on the launch pad before the engine burn?
What is the g-force experienced by the pilot at lift-off immediately after the rocket engine fires?
What is the maximum g-force experienced by the pilot (immediately before engine cut- off)?
What is the maximum g-force a human pilot can typically endure before losing consciousness?
The Kerbal Math & Physics Lab / Chapter 1
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Orbital Speed
Orbital speed for a circular orbit is given by the formula π£ = √πΊπ , where G is the universal π
gravitational constant 6.67 x 10-11 m3/(kg*s2), M is the mass of the object being orbited (like a Sun, planet or moon) and r is the radius of the orbit. Throughout, we are assuming the mass of the orbiting body, m, is much smaller than the mass, M, of the object being orbited.
In the formula above, because of the value we use for G, we measure the mass M in kilograms and the distance r in meters. Also, r represents the entire radius of the orbit, which is the total distance from the center of the orbiting body to the center of the body being orbited. This means we need to remember to add the radius of any planet or moon to the altitude of an orbiting rocket or satellite to determine the full radius of the orbit.
1. A rocket orbits at an altitude of 100 km above the surface of Kerbin which has a radius of 600 km and a mass of M = 5.29 x 1022 kg. Find the orbital radius of the rocket and then compute the orbital speed at 100 km.
You can check your answer above with the Kerbal 1 rocket (available in Sandbox mode in KSP). Check your rocket’s speed in the Navball display and in Map View.
2. What is the orbital speed at 100 km above the surface of Kerbin in miles/hour? Use the fact that 1 mile = 1609 meters.
The Kerbal Math & Physics Lab / Chapter 1
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3. What is the orbital speed at 100 km above the surface of the Earth in miles/hour and also in meters/sec? Use the facts that the mass of the Earth is M = 5.97 x 1024 kg and its radius is approximately r = 6,378,000 meters.
4. Compute the acceleration due to gravity an altitude of 100 km above Kerbin. What is the force of gravity (in Newtons) on a ship with a mass of 3000 kg at this altitude?
5. Does orbital speed in a circular orbit depend on the mass of the orbiting body? What is the orbital speed for a Kerbal astronaut on EVA outside a spacecraft orbiting in a circular orbit at 100 km above Kerbin?
The Kerbal Math & Physics Lab / Chapter 1
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The Kerbal Math & Physics Lab / Chapter 1 6. Does orbital speed depend on the size of the planet or moon being orbited? Explain.
7. The Mun is the closest moon to the planet Kerbin and orbits in a circular orbit at an altitude of 11,400 kilometers above the surface of Kerbin. Determine the orbital radius of the Mun. Compute the orbital speed of the Mun. This value can be confirmed in Map View in the game.
8. If a rocket increased its altitude to a circular orbit at 150 km above the surface of Kerbin, what is the orbital speed at this higher altitude? Is it traveling faster or slower than it was at 100 km altitude?
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9. The mass of the Mun is 9.76 x 1020 kg and its radius is 200 km. Compute the orbital speed of a rocket in a circular orbit around the Mun at an altitude of 13 km.
Build a rocket in KSP to reach the Mun or use the stock KerbalX rocket available in sandbox mode. There are many tutorials on reaching the Mun available on YouTube, including some by Mike Aben at sites.google.com/view/kspmath and detailed instructions on the KSP Wiki, on how to reach a Mun orbit in KSP.
During flight, check orbital speed in flight above Kerbin and above the Mun to compare with the previous calculations.
The Kerbal Math & Physics Lab / Chapter 1
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Orbital Period
For a circular orbit, we can determine orbital period using the equation π = 2ππ where T is the π£
period (in seconds) of the orbit, r is the radius of the orbit (in meters), and v is the orbital speed (in meters/sec). Show all work required to answer the following.
1. Find the period of a circular orbit about Kerbin at an altitude of 100km above the surface. Write the orbital period in terms of minutes and seconds.
2. The International Space Station orbits above the Earth in nearly circular orbit at an altitude of approximately 250 miles. Compute the orbital speed and period of the ISS. Write the speed in meters/sec and miles/hr and write the period in minutes and seconds.
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3. The Mun’s orbital radius about the center of Kerbin is exactly 12,000 km and the Mun’s orbital speed is 542.5 m/s. Compute the period of the Mun’s orbit about Kerbin.
The planet Kerbin orbits around a sun called Kerbol. The mass of Kerbol is M = 1.76 × 1028 kg and Kerbin orbits at a radius of approximately 1.36 x 1010 meters from its sun.
4. Use the values above to calculate Kerbin’s orbital speed about its sun.
5. Find the period of Kerbin’s orbit about the sun. This value represents the approximate length of a Kerbal year.
Check answers in game using Map View or at wiki.kerbalspaceprogram.com/wiki/Main_Page.
The Kerbal Math & Physics Lab / Chapter 1
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The previous discussion dealt only with circular orbits. We now consider the period of an elliptical orbit. Kepler’s third law of planetary motion states the square of the orbital period for any elliptical orbit is proportional to the cube of the semi-major axis of the orbit. Mathematically, this can be written as
Here T is the orbital period, a is half of the length of the larger axis of the ellipse as shown in the diagram above, G is the universal gravitational constant, and M is the mass of the primary body being orbited. In this form of the equation we assume the mass of the body being orbited is significantly larger than the orbiting body, like a planet that orbits about a sun, or a rocket orbiting a planet or moon. Also, if we use G = 6.67 x 10-11 m3/(kg*s2) then the units of T are seconds, a is measured in meters and M is measured in kilograms.
In Kepler’s third law as written above, what is the constant of proportionality?
Solve for T in the equation describing Kepler’s third law and write a formula for orbital period in an elliptical orbit.
The Kerbal Math & Physics Lab / Chapter 1
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8. In the Kerbal system, the planet Duna orbits in an elliptical orbit with a semi-major axis of a = 2.07 x 1010 meters around its sun Kerbol which has a mass of 1.758 x 1028 kg. Use Kepler’s third law to find the orbital period of the planet Duna.
9. Suppose that astronomers on the planet Kerbin discover the planet Eeloo orbiting about the sun with a period of approximately 1.57 x 108 seconds. Use Kepler’s third law to determine the semi-major axis of Eeloo’s orbit.
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Satellites in geostationary orbit remain in a fixed position above the surface of the Earth. This has useful applications for communications, weather observation and navigation. To remain in a fixed position relative to a point on Earth, as the Earth rotates on its own axis, the period of a geostationary orbit must be approximately 23 hours, 56 minutes and 4 seconds. This is the time it takes the Earth to rotate about its axis, relative to the apparent fixed background of the stars, and is defined as the length of a sidereal day.
Geosynchronous orbits, like geostationary orbits, also have an orbital period equal to the length of a sidereal day. While satellites in geostationary orbit remain at a fixed point relative to the surface of the Earth and require a circular orbit directly above the equator, geosynchronous orbits can be elliptical or inclined relative to the equator.
10. Use the equation π2 = 4π2 π3 to solve for a, the semi-major axis of an elliptical orbit, in πΊπ
terms of the period, T, and mass, M, of the primary body.
11. Determine the number of seconds in the orbital period of a geosynchronous orbit. Use this result above to determine the semi-major axis of a geosynchronous orbit.
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12. A geostationary orbit (GEO) is a circular orbit with a radius equal to the semi-major axis of a geosynchronous orbit. Calculate orbital speed for a geostationary orbit.
13. Look up the length of a sidereal day on Kerbin. Use the result to find the radius of a Kerbin-stationary (KEO) orbit.
14. What is orbital speed for a satellite in Kerbin-stationary orbit?
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Computing the Mass of a Planet or Moon
Assuming an object is in a circular orbit, we know orbital speed is given by π£ = √πΊπ , where v π
is the speed in meters/sec, M is the mass of the body being orbited, r is the orbital radius and G is the universal gravitational constant 6.67 x 10-11.
1. Use the formula for orbital speed in a circular orbit to solve for M.
2. The Mun orbits Kerbin at 542.5 m/s in a circular orbit with an orbital radius of 12,000 km. Use these values to compute the mass of Kerbin.
3. Kerbin orbits its sun, Kerbol, at a speed of 9285 m/s in a circular orbit with a radius of 13,599,840,256 meters. Use this to compute the mass of the Sun in the Kerbal system.
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Given the period of a circular orbit can be expressed as π = 2ππ where r is the orbital π£
radius and v is the linear speed, find an expression for the linear speed in a circular orbit in terms of orbital radius and period.
Given that Moon orbits about Earth at an orbital radius of about 239,000 miles with an orbital period of approximately 27.3 days, estimate the linear speed of the Moon’s orbit about Earth and use this result to estimate the mass of the Earth.
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The vis-viva equation relates the speed of an object in an elliptical orbit, with distance from the primary body, the length of the semi-major axis, and the mass of the body being orbited. One form of this equation is given below
π£2 =πΊπ(2−1) ππ
In this equation, v is speed (in meters/sec), G is the universal gravitational constant, M is the mass of the body being orbited (in kg), r is the distance (in meters) between the centers of mass of the two bodies and a is the semi-major axis of the ellipse (also in meters).
6. Use the above equation to solve for M and find a formula for the mass of a primary body based on speed, distance and semi-major axis of another object in orbit about the primary body.
7. A KerbalX rocket is in an orbit above the Mun at an altitude of 70 kilometers at a speed of about 492.2 meters/sec. Suppose the semi-major axis of the orbit is a = 271,111 meters. Recall that the gravitational constant is G = 6.67 x 10-11 and the Mun’s radius is 200 km. Compute the mass of the Mun.
Confirm the value for the mass of the Mun either by looking up the value on the Kerbal Wiki or checking the map view in the game.
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Kinetic and Potential Energy in a Circular Orbit
Kinetic energy (KE) is given by the formula πΎπΈ = 1 ππ£2 and potential energy (PE) is given by 2
the formula ππΈ = −πΊππ where G is the universal gravitational constant, r is the distance π
between the centers of mass of two bodies, M the larger mass being orbited (the sun, a planet or moon) and m, the smaller mass of the orbiting body (a planet, rocket, satellite or moon).
Recall that velocity in a circular orbit is given by the equation π£ = √πΊπ. π
1. Use algebraic substitution to show that, for a circular orbit, the kinetic energy of an orbiting body is exactly half the absolute value of its potential energy.
2. The Mun orbits Kerbin at an orbital radius of r = 1.2 x 107 meters. The Mun’s orbital speed is 543 m/s. The Mun’s mass of 9.76 x 1020 kg and Kerbin’s mass is 5.29 x 1022 kg. Compute kinetic and potential energy for the Mun. How do these values relate?
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3. Launch a rocket into a circular orbit about Kerbin. Use the MechJeb mod or a spreadsheet to determine kinetic and potential energy at different radius orbits. Complete the table below and describe the relationship between kinetic and potential energy in a circular orbit.
_ _ _
*PreCalculus & *Trigonometry
Linear and Angular Speed
Linear speed measures a difference in position over time relative to some fixed point in space, for example with units such as meters per second or miles per hour. We calculate the average
linear speed of an object as π£ = πππ π‘ππππ. π‘πππ
Angular speed measures a difference in the angular position of an object over time relative to a fixed point in space, and can be measured with units such as radians per second, degrees per hour, or revolutions per day.
Linear and angular speed can both be used to describe motion on simple straight paths and on complex changing curved paths. When an object moves at a constant speed on a circular path, there is a simple equation relating linear and angular speed: π£ = ππ where π£ is linear speed, π is the radius of the circle, and π is the angular speed. In this equation, angular speed is measured in radians per unit time.
In this section, we’ll analyze uniform circular motion, which refers to any object moving at a constant speed around a circle. It is helpful to make a distinction between velocity and speed: Speed is a scalar value; it is just a single number and describes distance travelled over time. Velocity is a vector and describes both speed and direction. Technically, it is possible to use positive and negative scalar values to indicate direction, like up and down or left and right, but in the case of circular motion additional information about direction, captured by the mathematics of vectors, is required to measure direction.
We can describe two-dimensional uniform circular motion with the position vector
π(π‘) = 〈ππππ (ππ‘), ππ ππ(ππ‘)〉
Here π(π‘) is a vector that locates the position at time t, of an object in uniform circular motion,
π
revolution about the circle, and the linear speed π£ = ππ of the object.
The Kerbal Math & Physics Lab / Chapter 2
moving about a circle with constant radius π, with constant angular speed π.
We can also use these values to describe the period π = 2π , the time it takes to complete one
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In uniform circular motion, an object moves around on a circular path with a constant linear and angular speed, but because the direction of motion is constantly changing, we say that its velocity is changing. In fact, the velocity of an object in uniform circular motion is changing at a constant rate described by the acceleration of the object. Acceleration is also a vector, giving both the magnitude and direction of the change in velocity. We measure the magnitude of the acceleration of an object in uniform circular motion with the expression π = ππ2, where again π is the constant radius of the circle, and π is the constant angular speed.
Previously, we have seen acceleration due to gravity given by the expression π = πΊπ where G is π2
the universal gravitational constant, M is the mass of a primary body, and r is the distance between the center of mass of the primary body and any other object in the gravity field of the primary body.
Suppose the acceleration due to gravity equals the acceleration required for uniform circular motion a distance r from a primary body with mass M. Find a formula for the corresponding angular and linear speed of the object and simplify each.
A satellite orbits Kerbin in a circular orbit at an altitude of 100 km. What is the linear and angular speed?
What is the period of a satellite in a circular orbit at an altitude of 100 km above Kerbin?
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4. The International Space Station (ISS) orbits above the Earth in a roughly circular orbit at an altitude of approximately 400 km (about 250 miles). Assuming the Earth’s radius is approximately 6,371 km, find the linear and angular speed of ISS and the orbital period.
5. A geostationary orbit is a circular orbit above the Earth with a period of 23 hours, 56 minutes and 4 seconds. Find the orbital radius of geostationary orbit and the corresponding linear and angular speed.
6. The wheels of the Mars Perseverance rover have a radius of approximately 10 inches. If the wheels complete 1 rotation in about 6 seconds, find the rover’s average angular speed (in radians/sec) and linear speed (in miles/hour). Use Desmos to graph the position vector of a point on the surface of the wheel as it rotates through one complete rotation about its center. In Desmos, you can graph the position vector with theform (ππππ (ππ‘),ππ ππ(ππ‘)).
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Great Circle Navigation - Introduction
A great circle path is the shortest route between any two points on a sphere and is a portion of a great circle, as shown in the figure below. A great circle is a circle on the surface of a sphere with the same center and radius as that sphere. In the figure, points A and B represent endpoints of a great circle path.
The initial bearing for the great circle path is the initial direction of the great circle path, as measured from the north pole of the sphere.
The formulas below can be used to calculate the initial direction, called the initial bearing, from any point A to any point B on any spherical planet or moon.
In these formulas (ππ΄ , ππ΄ ) represent latitude and longitude for point A, while (ππ΅ , ππ΅ ) are latitude and longitude for point B. The value ∆π is the difference in the longitudes (that is,
∆π = ππ΅ − ππ΄). Further, we use negative values for angles in southern latitudes and negatives to indicate western longitudes.
π₯ = πππ ππ΄ π ππππ΅ − π ππππ΄ πππ ππ΅πππ (∆π) π¦ = π ππ(∆π)πππ ππ΅
We compute the initial bearing using ππππ‘ππ (π¦), however we’ll need to consider the quadrant π₯
of the terminal side of the angle. Because the arctangent function is defined with a range between -900 and 900 and we’ll need to determine a bearing from 00 to 3600, the “two- argument arctangent” is useful here.
The two-argument arctangent is defined as follows.
Using the two-argument arctangent as defined above, we find an angle π½′ between -1800 and 1800 corresponding to the direction of travel relative to the positive x-axis in the standard coordinate system.
In the context of navigation, it is most common to measure π½ in degrees with 00 representing directly north, 900 due east, 1800 representing directly due south and 2700 as directly due west. To adjust the results of the two-argument arctangent to this orientation we include the following last step:
If the angle π½′ is positive, π½ = π½′ is the initial bearing.
If the angle π½′ is negative, π½ = π½′ + 360° is the initial bearing.
An explanation of why these formulas work involves spherical coordinates and vector arithmetic and can be found at youtube.com/c/csvaughen. Look for the Kerbal Guide to Navigation videos, parts 1 and 2.
Example: On Earth, the initial bearing for the shortest route from New York to Tokyo is about 3330 (which you can check online at gcmap.com) and be computed as follows:
New York: (40.70 North, 74.00 West) Tokyo: (35.60 North, 139.70 East)
∆π = 139.7 − (−74) = 213.70
y = cos(35.60)sin(213.70) = -0.4511
x = cos(40.70)sin(35.60)-sin(40.70)cos(35.60)cos(213.70) = 0.8824
Finally, arctan(-.4511/.8824) = -27.080, and since this value is negative, we’ll add it to 3600. Therefore, the initial bearing is approximately 3600 + -27.10 = 332.90 (or about 3330).
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Great Circle Navigation - Examples
1. Find the initial bearing for the shortest route from the Kerbal Space Center (the KSC) located at 0.050 South, 74.60 West to the K’Yangtze river canyon located at 22.30 North, 38.60 West.
Calculate the initial bearing and use this as a constant heading to navigate to the K’Yangtze river. The trip will take a little over 10 minutes traveling at about 700 m/s (about Mach 2). The stock jets Ravenspear Mk1 or Mk4 will work for this mission.
Once you find the K’Yangtze, try flying through the river canyon. Learn more about the River Run Air Race Challenge at forum.kerbalspaceprogram.com/index.php?/topic/176287-river-run- air-race-challenge/
You also can find the K’Yangtze river and other locations on Kerbin using an online map of the planet at https://ksp.deringenieur.net/
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2. Find the initial bearing for the shortest route from the KSC (0.050 South, 74.60 West) to the Desert Pyramids (6.50 South, 141.70 West).
Try making the trip by maintaining a constant heading set at this initial bearing. The trip takes about 15 minutes traveling at around 800 m/s. Look for the pyramids to be nestled in some mountains a little to the right of your approach. Why doesn’t a constant heading take you directly to the pyramids?
3. Explain the difference between a rhumb line path and a great circle path. In what cases are rhumb line and great circle paths the same?
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4. Find the initial bearing from the Kerbal Space Center (0.050S, 74.60W) to the location of the UFO buried in an ice sheet near the north pole of Kerbin (at 820N, 128.50W). Flying at a constant heading set to this initial bearing will not lead to the UFO. Why not? What happens instead?
The rhumb line heading to reach the UFO from the KSC is about 3400. It takes about 20 minutes flying at this constant heading to reach the UFO with a maximum speed around 1200 m/s during the trip. Use one of the stock jets in KSP and follow this constant heading to find the UFO.
5. Calculate the initial bearing for the shortest route from Philadelphia (400N, 75.20W) to Miami (25.80N, 80.20W). Confirm this result online with gcmap.com, www.movable- type.co.uk/scripts/latlong.html, wolframAlpha.com or flightaware.com.
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6. Compute the initial bearing for the great circle path from New York’s JFK airport to London’s Heathrow airport. Check with an online calculator such as WolframAlpha.com or gcmap.com. Print a map showing the route.
7. Compute the initial bearing for the great circle path on the Moon from the Apollo 11 landing site to the Apollo 12 landing site. Use the Apollo 11 landing site coordinates at (0.680 North, 23.40 East) and the Apollo 12 landing site coordinates at (3.20 South, 23.40 West). Show work using the method presented here.
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The Vis-Viva Lab
In this section we derive the very important vis-viva equation and study principles involved in that derivation. The term vis-viva is Latin for “living force” and the vis-viva equation relates directly to the principle of conservation of energy in orbital mechanics.
In Kerbal Space Program, and for highly accurate models in real life, orbital paths are determined by the principles of conservation of energy, conservation of angular momentum, and patched conic trajectories. Conservation of energy implies that throughout an orbit, in the absence additional forces, the sum of an orbiting body’s kinetic and potential energy remain constant. Conservation of angular momentum refers to a constant relationship between velocity, distance and direction of travel during orbit, and the assumption of patched conic trajectories means we assume only two masses at a time when computing orbits. That is, the Sun, the planets and moons (in the real solar system and in the Kerbal system) each have a defined sphere of influence (SOI) within which orbital paths are determined based only on the masses of a primary body and the orbiting body and the conservation of energy and angular momentum within that SOI. Travel between the SOIs of the Sun, the planets and moons results in paths that are made of conics (ellipses, circles, parabolas or hyperbolas) patched together as a spacecraft moves from one SOI to another.
It is interesting to note the parabolic arc of a projectile in free-fall, familiar to students of algebra and calculus, is derived from the assumption of a constant acceleration on the projectile, in a gravity field pointing in a constant direction downward, everywhere in a two- dimensional space. This provides a good approximation for projectile motion when the distance involved is small compared to the planet or moon on which that motion occurs. However, in a three-dimensional gravity field surrounding a spherical planet or moon, where that gravity field everywhere is pointing toward a single point in space (toward the center of mass of the planet or moon), every sub-orbital trajectory is actually a “truncated” ellipse. That is, both a powered suborbital trajectory and the path of a projectile in free-fall, are actually elliptical paths intercepting the surface of a sphere at two points (at launch and at landing).
We can determine whether an object is following an elliptical, parabolic or hyperbolic trajectory by measuring the sum of its kinetic and potential energy.
We define an object’s specific orbital energy, π, as follows π = 1 π£2 − πΊπ
2π
Here v is the speed of an orbiting body, G is the universal gravitational constant and r is the distance between the center of mass of the orbiting body and a primary body (with mass M). Note that this represents the total kinetic and potential energy of the orbiting body divided by its mass m.
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The specific orbital energy π of an orbiting body will remain constant on an entire orbital path, unless or until it is either acted upon by a force, such as an engine’s thrust or atmospheric drag, or if it moves from one sphere of influence to another, such as when exiting the SOI of a planet and entering the SOI of a moon or the Sun.
We characterize an orbit as gravitationally bound (elliptical) or unbound (parabolic or hyperbolic) based on the value of π. Circular orbits are a special case of elliptical orbits and can be identified by calculating orbital eccentricity, which we will get to shortly. Be careful not to confuse specific orbital energy, π, with orbital eccentricity, e.
If π is negative, an orbiting body is gravitationally bound to the primary body and will orbit in an elliptical (or circular) path, unless it encounters other forces such as engine thrust, or drag or impacts the surface of the primary body by landing or crashing.
If π is zero, the orbit is parabolic (and not gravitationally bound to the primary body) and has a trajectory that, unless altered later by an added force, will escape the SOI of the body being orbited.
If π is positive, the orbit is hyperbolic (and not gravitationally bound to the primary body) and again represents a trajectory that, unless altered later, will escape the SOI of the body being orbited.
We can also use the equation π = 1 π£2 − πΊπ to determine escape velocity. 2π
Ifwesetπ=0,wehave1π£2 −πΊπ =0andsolvingforv,wegetπ£2 =2πΊπ andthusπ£=√2πΊπ , 2πππ
which is tells us the velocity just sufficient to escape the gravity of a planet or moon of mass M from an initial distance r.
Because Earth has a radius of 6,371 km and mass of 5.97 x 1024 kg, escape velocity on Earth’s surface is π£ = √2(6.67×10−11)(5.97×1024) = 11,181 m/s.
6371000
Kerbin, with radius 600 km and mass 5.29 x 1022 kg, has an escape velocity at its surface equal
toπ£=√2(6.67×10−11)(5.29×1022) =3,429m/s.
The principle of conservation of angular momentum relates the mass, speed and direction of travel of an orbiting body relative to the body it orbits.
We calculate the magnitude of the specific angular momentum of an orbiting body, h, using the equation h = ππ£π πππ where r is the distance between the centers of mass of the primary and orbiting body, v is the speed of the orbiting body, and π is the angle between the radial vector and velocity vector of the orbiting body (see below). Like specific energy, the specific angular momentum is angular momentum with mass, m, of the orbiting body divided out.
These diagrams show a planet with center of mass at point A and a spacecraft in orbit at point B. The radial vector rextends from A to B while the velocity vector v indicates the direction of travel of the spacecraft at point B (called theprograde direction). If we imagine extending the radial vector in a line through point B away from A, the angle π is formed between the velocity vector and this extended line.
The flight path angle, π, of a spacecraft in orbit is the angle between its velocity vector and the horizon, and because the horizon is always perpendicular to the radial vector, if π ≥ 0 then π and π are complementary angles.
A spacecraft’s pitch is its orientation relative to the horizon. If the spacecraft can change its orientation during flight, the pitch and flight path angle may not be aligned.
In KSP, we can read a spacecraft’s pitch from
the Navball display. However, to calculate specific angular momentum, we must the align pitch with the flight path angle by orienting the spacecraft in the prograde direction (the direction of travel). Once facing prograde, pitch equals flight path angle π, and using π = 90° − π, with distance r, and speed v, we can calculate a spacecraft’s specific angular momentum.
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Angular momentum is technically itself also a vector, and is studied in detail in vector calculus, but for the purpose of deriving the vis-viva equation, we make the observation that at the periapsis and apoapsis of an elliptical orbit, when oriented prograde, flight path angle (and pitch) equal 00, that is π = 0°, and therefore π = 90° and because sin(90°) = 1, we have
h = ππ£π ππ(90°) = ππ£ (1) which simplifies the calculation of specific angular momentum at those two points.
Lastly, we can determine orbital eccentricity, e, using the equation π = √1 + 2πh2
π2
where
e = orbital eccentricity
π = specific orbital energy
h = specific angular momentum
π = GM, the gravitational parameter of the primary body
Note for any orbit, π ≥ 0 and the shape of the orbit can be identified as follows:
We can now use all of these results to classify orbits based on some basic flight data and also conclude this section by deriving the visa-viva equation, something we use extensively for other purposes, such as computing velocity and direction of orbital maneuvers.
The Vis-Viva Lab Flight Report
In KSP, launch several different rockets from Kerbin into sub-orbital, orbital and escape trajectories. Record speed, pitch, and distance from Kerbin at two different points in flight and calculate specific orbital energy and specific angular momentum at each point. Calculate eccentricity using either point or by averaging the values from each point. Remember to include Kerbin’s 600 kilometer radius in determining the orbital radius. Also, it may be easier to use Kerbin’s gravitational parameter π = 3.53 × 1012 instead of the product GM when computing specific energy.
Include units below. Attach supporting calculations or spreadsheet, confirm that orbital eccentricity matches the given trajectory.
Flight Report Analysis
Is specific energy constant during each flight? Explain why or why not.
Is specific angular momentum constant during each flight? Explain why or why not.
Use the formula for eccentricity, and speed in a circular orbit, to write an expression for specific orbital energy, in a circular orbit, in terms of only π and r.
In the absence of forces like atmospheric drag and engine thrust, and assuming a constant mass
for the orbiting body, specific energy at one point in flight, given by π = 1 π£2 − πΊπ (with 1 2 1 π1
velocity v1 and radius r1) will equal specific energy at another point in flight π = 1 π£2 − πΊπ 2 2 2 π2
(with velocity v2 and radius r2). Setting these values equal, we have 1π£2 −πΊπ =1π£2 −πΊπ
21 π1 22 π2
Rearranging and moving those terms that involve velocity, v1 and v2 to the same side, and multiplying through by 2, we arrive at
π£2 − π£2 = 2 (πΊπ − πΊπ) (2) 1 2 π1 π2
At this point, we complete the derivation assuming an elliptical trajectory, however the resulting equation will apply for any orbit whether circular, elliptical, parabolic or hyperbolic.
Recalling a previous conclusion (1), made for the case of an elliptical orbit, we observed that specific angular momentum is h = rv at both periapsis and apoapsis of the orbit. Because angular momentum is conserved, we know specific angular momentum is rv at all points in orbit. Therefore, at any two points in orbit, π π£ = π π£ .
We let r1 represent the distance from the center of mass of the primary body to the point of periapsis and similarly let r2 represent the distance to the point of apoapsis.
We solve for v2, writing π£2 = π1 π£1 and substituting this into eq. (2), we have π2
π£2 − (π1 π£ )2 = 2 (πΊπ − πΊπ) 1π21 π1π2
and with some further algebraic rearranging and factoring,
π£2 =πΊπ(2 −1) (3)Now, solving for v1, we get
where π = π1+π2 is the semi-major axis of the elliptical orbit. 2
11 22
π2 11 π£2(1−1)=2πΊπ( − )
1π2 ππ 212
1 π1π
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In equation (3), the values r1 and v1 represented distance and speed at periapsis in an elliptical orbit. In fact, the this equation holds at any point in any orbit, with distance r between the primary and orbiting body, and speed v of the orbiting body. In general, the equation holds for any orbit whether circular, elliptical, parabolic and hyperbolic. This form represents a common way to write the vis-viva equation.
π£2 =πΊπ(2−1) ππ
In the vis-viva equation, v is linear speed, r is distance between center of mass of the orbiting body and primary body, and a is the semi-major axis of the orbit. We will follow the convention that, in the vis-viva equation, if an orbit is elliptical a > 0, if parabolic then a is undefined, and if the orbit is hyperbolic, a < 0. Note also for circular orbits, a = r.
One important application of the vis-viva equation is determining the change in velocity required for orbital maneuvers such as traveling to the moon or other planets.
Delta-v for a Hohmann Transfer
A Hohmann transfer orbit is an elliptical orbit used to move from one circular orbit to another. In the figure below, we see two circular orbits, both with center at C, one with radius r1 and the other with radius r2. The point C represents the location of the primary body, which could be a planet, moon or Sun.
Suppose a rocket will maneuver from a circular orbit of radius r1 to a larger circular orbit of radius r2.
To do this, the rocket burns its engines to increase its velocity. The point on the initial circular orbit where this occurs becomes the periapsis of the transfer orbit, at point P. With added velocity, the rocket follows an elliptical transfer orbit to the apoapsis of the orbit at point A.
The apoapsis of an elliptical orbit is always the point in the orbit where an orbiting body will be moving at its slowest speed during the orbit.
At point A, the apoapsis of an elliptical orbit, the rocket is not traveling fast enough to maintain a circular orbit of radius r2 and thus, without additional thrust, would fall back toward the periapsis of the transfer orbit. Thus, to circularize the orbit at point A, the rocket would need to fire its engines a second time and increase its velocity to the speed required to maintain a circular orbit with radius r2.
Because the rocket follows an elliptical transfer orbit, from one circular orbit to another, we can relate the rocket’s velocity and position with the vis-viva equation π£2 = π (2 − 1), where π =
ππ
πΊπ is the gravitational parameter of the primary body at point C, r is the distance between therocket and the primary body, and π = π1+π2 is the semi-major axis of the transfer orbit. We also 2
know the rocket’s speed in any circular orbit is given by π£ = √π where π is the same π
gravitational parameter used for the transfer orbit and r is the radius of a circular orbit.
1. Do an internet search for Walter Hohmann. When and where did he live? What were his contributions to science?
2. Use the vis-viva equation, and the formula for speed in a circular orbit, to derive a general formula for the initial change in velocity (∆π£1) required for a Hohmann transfer from a circular orbit of radius r1 to a circular orbit of radius r2, where r2 > r1.
3. Use the vis-viva equation, and the formula for speed in a circular orbit, to derive a general formula for the change in velocity (∆π£2) required for the circularization burn to enter into an orbit of radius r2 after completing a Hohmann transfer orbit from an orbit of radius r1, where r2 > r1.
4. How do the general formulas for ∆π£1 and ∆π£2 derived above change for a transfer orbit from a larger radius orbit r2 to a smaller radius orbit r1?
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A Trip to the Moon
The Mun is Kerbin’s nearest moon and orbits at an altitude of 11,400 km in a circular orbit above the surface of Kerbin. The Mun’s radius is 200 km and its mass is 9.76 x 1020 kg.
According to Kepler’s laws, a bound orbit is always an ellipse with the body being orbited at one of the foci of the ellipse (the primary focus). A circular orbit is a special case where both foci coincide at one point. The semi-major axis of an orbit is half of the axis containing the foci, which is always the larger axis. The apoapsis of an orbit is the point in an orbit where an orbiting body is furthest from the primary focus, and the periapsis is the point where the orbiting body is closest to the primary focus. The eccentricity of an orbit is the distance from the center of the major axis to either focus divided by the length of the semi-major axis.
We can graph an orbit in two dimensions in rectangular coordinates or in polar form, although for most purposes in orbital mechanics, polar form is most convenient.
For elliptical orbits, we can graph the orbit with either of the following equations:
A rocket on a mission to reach the Mun is orbiting in a circular orbit at an altitude of 100km above Kerbin. To reach the Mun, the rocket will follow an elliptical transfer orbit with apoapsis 12,000 km and periapsis 700km, as measured from the center of Kerbin, the primary focus of the orbit. Including the radius of Kerbin, this means the rocket will reach an altitude of 11,400 km above the surface at its furthest point (apoapsis), and an altitude of 100km at its nearest point in the orbit (periapsis).
Find the semi-major axis and eccentricity of an elliptical transfer orbit (a Hohmann transfer orbit) with periapsis 100km and apoapsis 11,400 km above Kerbin.
Write an equation for the elliptical transfer orbit found above in rectangular coordinates with the primary focus of the ellipse, and center of mass of Kerbin, at the origin.
Assume the major axis of the elliptical orbit is horizontal in the xy-plane.
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3. Write the equation for the elliptical transfer orbit in polar coordinates with the center of Kerbin at the origin.
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Use Desmos.com to create a scale graph of the elliptical transfer orbit determined above. Include circles to represent the planet Kerbin and its moon, the Mun. Sketch the graph by hand below and then print the graph from Desmos, labeling the center, apoapsis and periapsis of the orbit.
Use the vis-viva equation (or the formula for ∆π£1 derived in the previous section) to compute the required delta-v for a transfer orbit to the Mun from an initial circular orbit at 100 km above Kerbin. Assume the apoapsis of the transfer orbit is the Mun’s distance from Kerbin.
6. In practice, a transfer orbit with apoapsis at the Mun’s altitude is not required to reach the Mun. Why?
Choose the stock KerbalX or build your own rocket and fly to the Mun. Do you want Kerbals to pilot the craft or will this be an unmanned satellite like the Luna or Ranger missions of the early Russian and American space race? Plan your trip by choosing an initial circular orbit about Kerbin and then record the semi-major axis and eccentricity for a transfer orbit to the Mun. Estimate the delta-v requirements for your trip. Is a return trip in your mission objectives? Compare your actual trajectory with your mission plans. Note: some of the calculations in this section are worked out in the video series A Trip to the Moon by Christopher Vaughen, available at youtube.com/c/csvaughen and at sites.google.com/view/kspmath.
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In the Apollo program of the 1960s and 1970s, the Saturn V rocket put the Apollo Command and Lunar Module (CLM) into an initial approximately circular orbit above the Earth at an altitude of approximately 190 km (120 miles) above the surface before sending the CLM to the Moon. At the initial orbital altitude, the Saturn V and CLM orbited Earth at about 7.7 km/sec (17,000 mph). To reach the Moon, the Saturn V then fired its engines again for about 6 minutes, for the Translunar Injection Burn, to increase the CLM’s speed and send it to the Moon. The Moon orbits the Earth at a nearly circular orbit of approximately 384,000 km from the center of the Earth and the Earth’s radius is approximately 6378 km. The Earth’s gravitational parameter is π = πΊπ = 3.986 × 1014 m3 /s2
7. Use the vis-viva equation, or the formula for the delta-v of Hohmann transfer, to estimate the delta-v requirements for the translunar injection burn for the Saturn V. Write the answer in km/sec, in miles/hour and also in feet/sec. Check your answer by looking up the Apollo 11 flight transcript at https://history.nasa.gov/afj/ap11fj/02earth- orbit-tli.html#0024416
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8. Calculate the orbital speed of the Moon.
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9. Calculate the orbital speed of the CLM at the apoapsis of the transfer orbit (assuming no Moon encounter).
10. What delta-v would the CLM require to match the Moon’s orbital speed?
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The Rocket Equation
Do an online search for the “ideal rocket equation”. Write that equation below. Define each of the variables, include the appropriate units. The ideal rocket equation is also referenced in chapter 6 of this workbook, in an application of differential equations.
Who was Konstantin Tsiolkovsky ? When and where did he live? What were his contributions to rocket science?
Suppose a rocket has an initial mass of 2480 kg and a final mass of 1430 kg. Suppose the rocket’s specific impulse is 140 seconds (at sea level). Use the ideal rocket equation to estimate the total ∆π£ for this rocket.
The initial and final mass above are those of a small rocket with an RT-5 solid rocket engine (also known as the “flea”) in Kerbal Space Program. Use the stock “Jumping Flea” rocket (available in Sandbox mode) or use the RT-5 engine to build your own similar rocket. Launch and compare velocity at engine cut-off with the theoretical value computed above.
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Launch an RT-5 “Flea” solid rocket from the KSC Launchpad and determine its velocity at engine cut-off. Confirm initial and final mass for the RT-5 rocket in the Vehicle Assembly Building (VAB) at the Kerbal Space Center. How does the ∆π£ recorded after the launch compare with the value predicted by the ideal rocket equation?
initial mass = _________ final mass ___________
delta-v (based on ideal rocket equation) _____________
initial speed (at launch) ___________ speed at engine cut-off ___________ delta-v during engine burn ___________Explain why maximum velocity of the RT-5 rocket launched from surface of Kerbin is different from the value predicted by the ideal rocket equation.
Kepler’s Third Law of Planetary Motion
As introduced in the previous chapter, Kepler’s third law of planetary motion states that for an object in an elliptical orbit, the square of the orbital period is proportional to the cube of the semi-major axis. This can be written as
4π2 π2 = πΊπ π3
where T is the orbital period, a is the semi-major axis, G is the universal gravitational constant and M is the mass of the body being orbited. This form assumes the body being orbited (for example a sun) is significantly more massive than the orbiting body (like a planet).
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We can confirm Kepler’s third law of planetary motion in the Kerbal system using the following data found at the Kerbal Wiki and also in map view in the game.
We will use the fact that mass of the sun Kerbol is M = 1.758 x 1028 kg while the universal gravitational constant is G = 6.67 x 10-11 m3/(kg*s2).
1. Complete the table below for each planet in the Kerbal system, where a is the semi-major axis of the orbit (in meters) and T is the orbital period (in seconds).
2. Let x = log(a) and y = log(T). Plot these values in an xy-plane. How do the values of log(a) and log(T) relate? Use extra paper and plot values by hand or use Desmos.com.
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Use the points x = log(a) and y = log(T) from the table (and graph) to determine the
slope and y-intercept for the line of best fit for this data.
How do the slope and y-intercept for the line of best fit for the log-log plot of orbital data relate to Kepler’s third law?
5. Suppose Kerbal astronomers discover a new planet, which they name Eeloo, with an orbital period of approximately 1.57 x 108 seconds. Based on this measurement, determine the semi-major axis for Eeloo’s orbit. Confirm your result on the Kerbal Wiki or in map view in the game.
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6. Compute log(a) and log(T) for Eeloo’s orbit, where a is the semi-major axis (in meters) and T is the period (in seconds). Include the results in the previous table of orbital data for the planets of the Kerbal system .
7. Graph the log(a) vs. log(T) data from the observations of the planet Eeloo and confirm the values are consistent with Kepler’s third law of planetary motion.
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Orbital Inclination
We conclude this chapter by investigating the relationship between launch site latitude, launch azimuth angle and orbital inclination. First, we’ll define these terms:
The latitude of the launch site is the angle measured at the center of a planet or moon between the launch site and equator, where positive angles measure points north of the equator and negative angles measure points south of the equator.
The launch azimuth angle is the initial the direction of ground track of the rocket, measured at the surface of the planet or moon, from the geographic north pole to the ground track (figure 1). This is the same angle as the initial heading discussed in the previous section on navigation and great circle paths, introduced at the beginning of this chapter.
The orbital inclination is an angle measured between the plane of the orbit and the plane of the equator (see figure 1).
Any object orbiting around a planet or moon will remain in the same orbital plane unless acted upon by an added force such as the thrust of a rocket engine. As the object completes an orbit, its ground-track along the surface is a great circle path. We can use the plane containing the ground-track as determining the orbital plane regardless of whether the orbit is circular, elliptical or hyperbolic.
Orbital inclination is an angle between two planes and can be measured between 00 and 1800. When the angle is between 00 and 900, we refer to the orbit as prograde. Objects in prograde orbits move in the same direction as the rotation of the planet or moon. Objects in retrograde orbits, with inclination between 900 and 1800, move in a direction opposite to the rotation of the planet or moon. A polar orbit has an inclination of 900, and equatorial orbits have of 00 inclination (prograde) or 1800 inclination (retrograde).
The relationship between orbit inclination (i), launch azimuth angle (A) and launch site latitude (π) is given by the equation
cos(π) = sin(π΄)cos(π)
A proof of this relationship requires an extended discussion of spherical trigonometry; however, the resulting equation is surprisingly simple.
In this section we use the equation above to compute orbital inclinations and find the required launch azimuth angles corresponding to different launch site latitudes and also investigate some fundamental mathematical restrictions on these values.
The calculations that follow are based only on the equation above and do not take into account a planet’s oblateness or the effects of rotation of a planet or moon in determining precise azimuth and inclination angles. Thus, results should all be taken as approximations.
1. The International Space Station (ISS) orbits the Earth at an inclination of approximately 51.60.
a) Suppose SpaceX plans to launch a resupply mission to the ISS from Kennedy Space Center in Florida, which is at a latitude of 28.60 N. Find two launch azimuth angles at Kennedy that result in an orbital inclination of 51.60.
b) Suppose NASA plans to launch a resupply mission to the ISS from the Baikonur Cosmodrome in Kazakhstan, located at 45.60 N. Find two launch azimuth angles at Baikonur that will result in the target orbital inclination of 51.60.
At Kennedy Space Center (KSC), allowable launch azimuth angles range between about 350 and 1200 due to safety considerations. Launching rockets from Kennedy at azimuth angles less than 350 or greater than 1200 risk sending errant rockets, or the dropped lower stages of a multi- stage rocket, into the east coast of the United States, or the Bahamas or other islands in the Caribbean. For similar reasons, allowable launch azimuth angles at Vandenberg Air Force Base in California range from approximately 1400 to 2010, as shown below.
The ISS orbits at an inclination of 51.60 because the first components of the station were launched from Baikonur, from a latitude of 45.60 N, and local launch site azimuth restrictions (between about -130 measured counter-clockwise from north to about 650 clockwise from north) allow an orbit inclination of 51.60.
Any mission to dock with the ISS requires an orbital inclination of 51.60 to match the inclination of the station. This is because even for two spacecraft orbiting at the same altitude and velocity relative to a planet, if their orbital inclinations are different, their relative speeds due to their different inclinations would make docking impossible.
For this reason, any mission launched from the KSC to a rendezvous with the ISS will leave the KSC with a launch azimuth angle of approximately 450 as computed above in question 1, part (a). Missions to the ISS can also be launched from Wallops Island, off the coast of Virginia, which is at a latitude of 37.90, because allowable azimuths there permit the required orbit inclination.
2. What launch azimuth angles are required from Vandenberg Air Force Base in California (at latitude 34.70 N) to reach the orbital inclination of the International Space Station? Is either angle within the ranged of allowable launch azimuths at Vandenberg?
An interesting consequence of the geometry of an orbital plane is that the minimum possible orbital inclination that can be achieved directly from launch is equal to the latitude of the launch site and occurs exactly when the launch azimuth is 90 (or 270) degrees. To visualize why, imagine a satellite in orbit around a planet at some nonzero inclination. The minimum possible inclination occurs if the satellite passes over the launch site in a direction parallel to the equator, corresponding to the point where the satellite is farthest from the equatorial plane. If the satellite passes over the launch site in a direction not parallel to the equator (either increasing or decreasing in latitude), then the orbit inclination must be greater than the launch site latitude.
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In other words, if we imagine the orbital plane as a sheet of paper that contains the launch site, the center of the planet and all points along the orbit, the minimum inclination of that orbit is when the paper intersects the launch site in a line parallel to the equator. This also explains why azimuth angles of 900 (or 2700) achieve the minimum possible inclination (see figure 3 below).
(a) A = 550, i = 450 (b) A = 900, i = 300 (c) A = 1250, i = 450
The Earth spins on its axis eastward and for a rocket to take full advantage of this spin for the added velocity to reach orbital speed, it is preferable to launch eastward into a prograde orbit from a point close to the equator. For example, launching directly eastward (A = 900) from Kennedy Space Center (at latitude 28.60) results in a 28.60 inclined orbit and this is the minimum possible orbit inclination that can be achieved directly from a launch at Kennedy. Of course, rockets launched from Kennedy can reach lower inclination orbits, but only after additional orbital maneuvers lower their inclination.
A geo-stationary orbit is particularly useful for communications because satellites in this type of orbit remain in a fixed location above the Earth. A geo-stationary orbit requires a satellite to be at certain altitude so that its orbital speed exactly matches the spin of the Earth and it also requires the satellite to be in a 00 inclination (or equatorial) orbit.
The delta-v required for an orbital inclination change depends on the speed of the orbiting body and the angle of desired inclination change. Higher orbital speeds and larger inclination changes require larger amounts of delta-v. For this reason, launch sites are chosen based on latitude, allowable launch azimuth and their corresponding orbital inclinations.
While launching eastward from sites close to the equator provides the greater added speed due to the rotation of the Earth, launching near the equator is not always desirable. For example, launching into polar (or near polar) orbits and retrograde orbits on Earth is best done from launch sites at more northern or southern latitudes, where the spin of the Earth contributes less to the launch vehicle’s initial orbital speed.
3. Show mathematically why it is not possible to launch directly into an orbit with an inclination of 28.60 from the Baikonur Cosmodrome located at 45.60 latitude by using the equation cos(π) = sin(π΄)cos(π) and solving for the azimuth angle A.
4. Is it mathematically possible to launch directly into the orbital inclination of the ISS from Russia’s Plesetsk Cosmodrome at latitude 62.90 north? Justify your conclusion using the equation cos(π) = sin(π΄)cos(π).
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5. Determine the orbit inclination of a spy satellite launched from Vandenberg Air Force Base (at latitude 34.70 N) with a launch azimuth of 1800.
6. Suppose SpaceX plans to launch Starship from Boca Chica, Texas (latitude 260 N) with a launch azimuth of 930. Calculate the resulting orbital inclination.
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7. Search online to find the latitude and longitude for the Wenchang Launch Site on the island of Hainan, China. What launch azimuth (assuming launch directed southward over the ocean) is required from Wenchang to achieve an orbital inclination of 97.4 degrees?
8. A weather satellite launched by Rocket Lab from the Mahia peninsula in New Zealand (latitude 39.20 S) has a target orbit inclination of 98 degrees. Assume azimuth restrictions at the Mahia peninsula require launches to be directed southward out over the Pacific Ocean. Find the required launch azimuth for this satellite.
9. The European Space Agency launches missions from the Guiana Space Centre in Kourou, French Guiana at latitude 5.20 N on the east coast of South America. Given launch azimuth restrictions that range from 3490 (or -110) to 900, what range of orbit inclinations can be achieved directly from launch at the Guiana Space Centre?
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10. The Making History expansion in Kerbal Space Program includes the option to launch vehicles from the Woomerang launch site at latitude 45.30 N, a site analogous to the Baikonur Cosmodrome on Earth.
a) What is the expected orbital inclination resulting from a launch directly to the east at Woomerang (azimuth 900)? Launch a vehicle from Woomerang in KSP and confirm your calculation.
b) What is the expected orbital inclination the results from a launch directly west from Woomerang? Confirm your calculations with a launch in KSP.
c) Find launch azimuth angles from Woomerang that will reach an orbital inclination of 60 degrees and confirm these values in KSP.
d) Find launch azimuth angles from Woomerang that will reach orbital inclinations of 120 degrees and confirm these values in KSP.
_ _ _
*Single Variable Calculus
Maximum Aerodynamic Pressure
Aerodynamic pressure on a rocket is based on the speed of the rocket and the density of the atmosphere it travels through and can be described with the equation π = 1 ππ£2, where Q is
2
the aerodynamic pressure, p is atmospheric density and v is a rocket’s speed.
We can see in the equation above, the value of Q, aerodynamic pressure, is zero if either v = 0 or p = 0. For example, if a rocket is on the launch pad with zero velocity (v = 0), then aerodynamic pressure is also zero. Or if the rocket is in space, where there is no atmosphere (p = 0), then again, aerodynamic pressure is zero.
We can further conclude that when launching from the surface of a planet with an atmosphere, as a rocket’s speed increases, aerodynamic pressure (Q) will increase, but as the rocket travels higher, and the atmospheric density decreases, the value of Q will decrease. Therefore, aerodynamic pressure can increase and then decrease. The point of maximum aerodynamic pressure is called max-Q.
Assume aerodynamic pressure, Q(t), on a rocket during launch is a nonnegative, continuous and differentiable function of time. Further, suppose Q(t0) = 0 at liftoff, and again Q(t1) = 0 at some later time, t1, when the rocket reaches space. What theorem of calculus, combined with the fact that Q(t) is nonnegative, guarantees the existence of a point in time when Q(t) will reach a maximum value?
The function π(π‘) = −0.2π‘3 + 2.8π‘2 + 6.1π‘ models aerodynamic pressure on an RT-5 “Jumping Flea” solid fuel rocket outfitted with a command pod and parachute during the first 15 seconds of flight. In this model, Q(t) is the aerodynamic pressure in KiloPascals and t is time in seconds. The RT-5 isn’t powerful enough to reach space on its own, however aerodynamic pressure increases as velocity increases and then decreases as the rocket slows. Find the instant of maximum aerodynamic pressure (Max-Q) based on the given model. Graph Q(t) on the interval [0,15.9] and label the point of Max-Q. Use extra paper to show work and include the graph.
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Acceleration and g-Force Lab, Part 2 – SpaceX Falcon 9 Launch
The table below lists time, speed and altitude for the first 2 minutes 40 seconds of the SpaceX Falcon 9 launch of a set of Starlink satellites on September 3rd, 2020.
Copy the table above into Excel and include additional columns for speed in meters/sec and miles/hr. Include additional columns for altitude both in meters and also in miles.
Graph the speed and altitude data with Excel or Desmos using a scatter-plot.
Find a quadratic curve of best fit for the speed of the rocket with form π£(π‘) = ππ‘2
where v(t) is measured in meters/sec and t is measured in seconds.
Graph the mathematical model v(t) along with the speed data using Desmos on the
same graph.
Find the derivative of the curve of best fit, v’(t) = a(t).
Use the mathematical model a(t) to estimate acceleration of the rocket at t = 10
seconds, at t = 1 minute 10 seconds (Max-Q) and at t = 2 minutes, 38 seconds (Main
Engine Cut-Off or MECO).
Use the formula from Chapter 1, πππ = π + 1 to compute the Thrust-to-Weight ratio
π
(TWR) for the Falcon 9 at time t = 10 seconds, at Max-Q, and at MECO. Note: TWR is the same as the g-Force experienced by the rocket.
Speed Altitude (km/hr) (km)
0 0
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Differential in Velocity
The formula π£ = √π gives orbital speed for a circular orbit at an orbital radius r about a mass π
given by the gravitational parameter π = πΊπ where G is the universal gravitational constant and M is the mass of the body that is being orbited (the primary mass).
1. Assuming π is constant and taking a derivative with respect to r, find a formula for ππ£, the differential in velocity required to change an orbital radius.
The differential in velocity can be used to estimate delta-v, the actual change in velocity required to change an orbital radius.
2. Given π = 3.53 x 1012 , the gravitational parameter of the planet Kerbin, use the differential ππ£ to estimate the delta-v requirement for a change in orbital radius from 700,000 meters to 750,000 meters above Kerbin.
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3. Use the vis-viva equation, and the principle of a Hohmann transfer orbit developed in chapter 2, to find the actual delta-v required to change the radius of a circular orbit from r = 700,000 meters to r = 750, 000 meters.
4. Use the differential formula derived above to estimate the delta-v required to increase from a 150 km to 160 km altitude circular orbit above Kerbin. Check this result with the vis-viva equation.
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Solving Kepler’s Equation with Newton’s Method
Kepler’s equation can be written as π = πΈ − ππ πππΈ, where M is the mean anomaly, E is the eccentric anomaly and e is the eccentricity of an object in an elliptical orbit. The meanings of these terms, and their application to the calculation of position in an elliptical orbit, are explored in this lab.
Kepler’s equation applies to the motion of any object in an elliptical orbit. For example, Earth or Mars in orbit around the Sun, or any moon orbiting a planet, or a rocket or satellite in orbit about any planet or moon. This equation is applies anywhere in the real universe and it is modeled in the game Kerbal Space Program.
In this lab we’ll apply Kepler’s equation to calculating the position of a rocket in an elliptical orbit as a function of time. We’ll use Kerbal Space Program to develop an understanding of Kepler’s equation and its application by placing rockets and satellites in different orbits and computing their position as a function of time and checking our answers in the game.
For circular orbits, finding position as a function time is relatively simple. In a circular orbit, the orbiting body moves at a constant speed with a constant orbital radius and therefore we can determine its position over time by multiplying the object’s angular speed by the elapsed time. If we know an angular change in position and the radius of the orbit, we could locate the body at any point in that orbit.
Computing position over time for elliptical orbits is more complicated because linear and angular speeds change during the orbit. The orbiting body speeds up and slows down as its position changes relative to the body it is orbiting.
Let’s consider a rocket in an elliptical orbit around a planet. To calculate the rocket’s position over time, we will need to understand the mean anomaly, eccentric anomaly, and the true anomaly of the rocket.
We’ll begin with mean anomaly: In orbit, a rocket will slow down as it moves away from the planet, reducing its angular speed. As the rocket approaches the planet again, its angular speed increases. To simplify, it is useful to find the average angular speed (in radians/sec) during the entire orbit. We call this average angular speed the mean motion of the rocket. The mean anomaly is the angle swept out during an elapsed period of time based on the average angular speed, or mean motion, of the rocket.
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We calculate mean anomaly by associating a circle with radius a equal to the semi-major axis of a given elliptical orbit, as shown below. As the rocket travels along the elliptical orbit, a corresponding point, the mean motion point, is traced out along the circle. The position of the mean motion point is based on the average angular speed (or mean motion). The angle M, the mean anomaly, can be visualized as the angle measured at the center of the circle with radius a, from periapsis (point of closest approach) to the mean motion point.
Above (Figure 1), we see an elliptical orbit with semi-major axis a, circumscribed by a circle with radius a. A rocket follows an elliptical orbit about a planet, as shown, with the planet located at one focus of the ellipse. As the rocket follows the elliptical path, the mean motion point follows a circular path. The movement of the mean motion point is based on the average angular speed of the rocket. The mean anomaly of the rocket is the angle drawn at the center of the circle, from periapsis to the mean motion point. In the figure, mean anomaly for the rocket at its current position as shown is M = 38.50.
The true anomaly, the angle π (theta), is also shown in the figure above. The true anomaly is the angle drawn at the planet (or body being orbited) from periapsis to the orbiting body, in this case a rocket. In the figure above, the true anomaly for the rocket is π = 117.20.
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We are going to calculate the position of a rocket in an elliptical orbit as a function of time and we’ll describe that position in terms of distance from the planet and the angle of true anomaly.
To do this we will need a way to compute true anomaly based on mean anomaly and while there are direct methods, they are beyond the scope of this lab. Instead, we follow a commonly used method entirely within the scope of introductory calculus. Derivations of the formulas we use here require only geometry and trigonometry and are all within reach of calculus students, however also including them here would greatly extend the length of this introduction. Instead, we’ll include links to those derivations as the formulas are introduced. Exploring those derivations would provide an interesting extension to this lab.
To continue here we’ll need to introduce one more term: The angle E, the eccentric anomaly of an orbiting body. We’ll examine this angle in a moment, but to get a sense of where it fits in, we step back and look at the procedure as a whole. Given mean anomaly, we will calculate eccentric anomaly, which we use to find true anomaly and then ultimately we use true anomaly to the find the position in orbit.
The full procedure can be summarized as follows:
Given: Average angular speed in an elliptical orbit with semi-major axis a, eccentricity e Input: Any elapsed time, t, in seconds
Step 1: Calculate angle M, the mean anomaly, based on elapsed time t.
Step 2: Calculate angle E, the eccentric anomaly, based on the mean anomaly M.Step 3: Calculate angle π, the true anomaly, based on the eccentric anomaly E.
Step 4: Calculate r, the distance from primary to orbiting body, based on true anomaly π, the semi-major axis a, and eccentricity, e.
Output: Location of orbiting body in terms of distance r and true anomaly π.If that seems complicated, well it is rocket science! Just remember, this will work for any object in any elliptical orbit anywhere in the universe and, of course, in Kerbal Space Program. So, once we get it here, we can use it anywhere. Second, we’ll make good use of technology to speed up the process. Imagine the work that Kepler, Newton, and many other scientists faced before the advent of digital computers.
In step 2 we use Kepler’s equation π = πΈ − ππ πππΈ to solve for E with given values for M and e. In general, it is impossible to find an exact solution to Kepler’s equation and thus some type of numerical approach must be used to find an approximate solution. Even if we use a computer, the solution is still approximate and based on a very fast numerical method. Students in calculus are typically introduced to a numerical method called Newton’s method and we’ll use that here to solve Kepler’s equation.
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The Kerbal Math & Physics Lab / Chapter 3 Returning to the task of calculating position as a function of time, we now define the angle E,
called the eccentric anomaly, as shown in the figure below (figure 2).
To find the angle E, first consider the line of apsides, the line through the center of the orbit, connecting the apoapsis and periapsis. This line contains the major axis of the ellipse. Imagine a perpendicular to the line of apsides, passing through the orbiting body at its current position in the elliptical orbit. This perpendicular line will also cross through the associated circle (with radius a) that circumscribes the ellipse, at point K as shown below. We call point K, the eccentric anomaly point. The eccentric anomaly is defined as the angle measured at the center of the ellipse (and circle), from the periapsis to the eccentric anomaly point (K). The eccentric anomaly, which is angle E, is shaded with dark stripes in the figure below.
In the figure, with the rocket at its current position in the orbit, eccentric anomaly is E = 59.67o, mean anomaly is M = 300 and true anomaly is π = 97.840. The rocket is following an elliptical orbit with semi-major axis a = 25 and eccentricity e = 0.6. Eventually, we’ll need all of these values to calculate the position of the rocket over time.
Incidentally, the word “anomaly”, which generally means some deviation or error, was used in this context because early astronomers considered the elliptical paths of the planets to be deviations from a simple circular path.
With some trigonometry, it can be shown the relation between π (true anomaly), e
(eccentricity), and E (eccentric anomaly), is given by
π‘ππ (π) = √1+π π‘ππ (πΈ) (1)2 1−π 2
A derivation of this formula is at https://en.wikipedia.org/wiki/Eccentric_anomaly. We use equation (1) above to solve for π after we have known values for e and E.
Finally, with known values for π (true anomaly), e (eccentricity) and a (semi-major axis), we can calculate the distance r to the orbiting body using the polar equation
π = π(1−π2) (2) 1+ππππ π
Equation (2) is the polar form of an ellipse with one focus at the origin of coordinate system. We use equation (2) to locate the position of the orbiting body as a function of π.
Example 1
Consider a rocket in an elliptical orbit with semi-major axis a = 25 units and eccentricity e = 0.6 as shown in the figure below. Suppose the rocket is travelling with an average angular speed of
π radians/second. Calculate the position of the rocket 10 minutes after periapsis passage. 3600
figure 3
We will calculate the distance from the planet to the rocket, and the rocket’s true anomaly, 10 minutes after it passes periapsis.
The following example illustrates the calculation of position at some elapsed time after periapsis passage. This is because periapsis is a convenient fixed point on any elliptical orbit. We could further generalize the following procedure to calculate position with any elapsed time from any arbitrary initial point in the orbit. To simplify, in this lab we’ll just calculate position given time elapsed from periapsis passage.
Given:
Input: Elapsed time t = 10 minutes = 600 seconds (after passing periapsis)
Step 1: Calculate angle M, mean anomaly, based on elapsed time t.
Mean motion or average angular speed is n = π rad/sec. 3600
A rocket in an elliptical orbit with semi-major axis a = 25, eccentricity e = 0.6, and average
angular speed π radians/second 3600
In 600 seconds, mean anomaly is M = π ∙ π‘ =
π (600) = π radians (or 30 degrees) 3600 6
Step 2: Calculate angle E, the eccentric anomaly, based on the mean anomaly M.
We solve for E in Kepler’s equation π = πΈ − ππ πππΈ
withM=π ande=0.6,wehave π =πΈ−0.6π πππΈ 66
There is no analytic method to find an exact solution for E. With a numerical method, such as Newton’s method, a computer or the calculator, we arrive at the solution E = 1.0415 radians (about 59.67 degrees). This is shown in the previous figure.
Step 3: Calculate angle π, the true anomaly, based on the eccentric anomaly E. With E = 1.0415 radians, we calculate true anomaly using equation (1)
π‘ππ (π) = √1+.6 π‘ππ (1.0415) 2 1−.6 2
ΞΈ = 2 ∙ tan−1 (√1.6 π‘ππ(0.52075)) .4
ΞΈ = 2 tan−1(1.14712) = 1.7076 radians (about 97.84 degrees).
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Step 4: Calculate r, the distance from primary to orbiting body, based on true anomaly π, the semi-major axis a, and eccentricity, e.
At true anomaly 97.840, with eccentricity e = 0.6, and semi-major axis a = 25, the radial position of the rocket is computed with equation (2) as
π = 25(1−0.62) = 17.43 units 1+0.6cos (97.84°)
Note that we can use either degrees or radians for the angle of true anomaly in this equation and the units of distance r match the units used for the semi-major axis a.
Output: Location of the orbiting body in terms of distance r and true anomaly π.
The rocket is at a distance of 17.43 units from the planet at a true anomaly of 97.840.Example 2:
A rocket is in an elliptical orbit about the planet Kerbin. Orbital semi-major axis is a = 2160 kilometers with eccentricity e = 0.65. The mass of Kerbin is 5.29 x 1022 kilograms. Calculate the position of the rocket 15 minutes after periapsis.
Solution:
Kepler’s third law states π = 2π√π3, where T is the orbital period, a is the semi-major axis and π = GM isπ
the gravitational parameter of the body being orbited. We find the average angular speed with π = 2π π
radians/second. In general, π = √ π is the average angular speed, or mean motion, of the orbiting π3
body.
Using universal gravitational constant G = 6.67 x 10-11 and mass M = 5.29 x 1022, for this example wehave π = (6.67x10-11)(5.29x1022) = 3.53 x 1012. These values assume distance in meters and time in seconds. Thus average angular speed for the rocket is π = √3.53×1012 = 5.9184 × 10−4 rad/sec.
21600003
Using this average angular speed, we calculate the position of the rocket as follows:
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Finding position as a function of time:
Step 1: Elapsed time is t = 15 minutes = 900 seconds. Thus mean anomaly is M = π ∙ π‘ =(5.9184 × 10−4 πππ) (900 π ππ) = .5327 radians (about 30.5 degrees) π ππ
Step 2: From Kepler’s equation, M = E – esinE, we solve for E. Solving 0.5327 = E – 0.65sinE, we find solution E = 1.117 radians (about 64 degrees). Note we arrive at this solution by Newton’s method or other numerical method on a computer or calculator.
Step 3: With E = 1.117 and e = 0.65, we find true anomaly π, as follows
π = 2 tan−1 (√1+.65 tan (1.117)) = 1.8712 radians (about 107.2 degrees)1−.65 2
Step 4: Calculate position.
At t = 15 minutes after periapsis, the rocket is located at π = πππππππ(π−.πππ) = π, πππ, πππ metersπ+.πππππ (πππ.π)
from the center of Kerbin at a true anomaly of 107.20.
Thus the altitude above the 600 km radius planet is 1,544,214 – 600,000 = 944,214 meters.
It’s important to note that the values of M and E in Kepler’s equation must be measured inradians, that is, they must be dimensionless as are the values of e and sinE.
A video derivation of Kepler’s Equation can be found at the channel TeachMeEngineering with the link https://www.youtube.com/watch?v=wJf8B0SUdYM. This particular video also includes a discussion of Newton’s method and an example using Newton’s method to solve Kepler’s Equation.
Example 3:
We solve Kepler’s equation M = E – esinE using Newton’s method and check the result with a computer. Suppose we are given the values M = π and e = 0.5. We have π = πΈ − 0.5π πππΈ , and
so first re-write the equation as
π − πΈ + 0.5π πππΈ = 0 6
Following Newton’s method, let π(π₯) = π − π₯ + .5π πππ₯. Thus π′(π₯) = −1 + .5πππ π₯. 6
Taking our first guess to be π₯0 = π = π, we iterate the formula 6
π₯π+1 = π₯π − π(π₯π) π′(π₯π)
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84
After 4 iterations, the values for π₯π+1 approach approximately 0.922007 which we can confirm graphically on a computer or calculator.
Listing successive iterations by Newton’s method in a table, we have:
Remember, in general, we are not able to find exact values for solutions to Kepler’s equation, even on a computer or calculator. The solution illustrated above in Desmos is a decimal approximation. Also, because Newton’s method requires an initial guess for the solution, when applied to solving Kepler’s Equation, we can take our initial guess to be the value M.
To conclude, in general, Kepler’s Equation M = E – esinE cannot be solved exactly by an algebraic method. However, solving this equation is critical to finding the position of an orbiting body over time. The primary method used to solve Kepler’s equation is the calculus based method called Newton’s method.
Use Newton’s method, and the procedure outlined previously, to calculate the position of an object in an elliptical orbit as a function of time.
Use Newton’s method to solve for x in the equation 0.532 = x – 0.65sinx. Use initial guess x0 = 0.532 and 4 iterations. Create a table listing each iteration and write the final answer accurate to 3 places after the decimal. Use extra paper and attach.
Solve 0.67195 = E – 0.6sinE for E by Newton’s method. Using a first guess E = 0.67195, find a solution accurate to 3 places after the decimal. How many iterations were required for this level of accuracy? Check the solution graphically using a computer, print the graph and label the solution in the graph. Use extra paper and attach.
A satellite in an elliptical orbit has an average angular speed 0.05 radians/second. What is the mean anomaly after 20 seconds of elapsed time?
Suppose a satellite follows an elliptical orbit with eccentricity e = 0.25. Suppose the mean anomaly of the satellite is M = 1 radian. Calculate the eccentric anomaly E. Attach graph or work using Newton’s method to support your conclusion.
A satellite in an elliptical orbit with eccentricity e = 0.25 has an eccentric anomaly of E = 70.8 degrees as measured from periapsis. Calculate true anomaly for the satellite.
6. A satellite in an elliptical orbit with eccentricity e = 0.25 and semi-major axis a = 1000 kilometers is at true anomaly π = 94°. Find distance to the satellite from the center of the planet. If the planet has a radius of 600 kilometers, calculate the satellite’s altitude above the surface of the planet.
In February 2018, SpaceX launched a red Tesla Roadster into an orbit around the Sun which passes far enough from the Sun to cross over the orbit of Mars. The Tesla Roadster, with a mannequin named Starman in the driver’s seat, orbits about the Sun with a semi-major axis of approximately 1.325 au and an eccentricity of 0.256.
Note: 1 au (an astronomical unit) is 149.6 million kilometers.
The data below is from NASA’s Jet Propulsion Laboratory and is used by the websites https://spacein3d.com/starman and https://www.whereisroadster.com/ to track the position of Starman and the Roadster. Below is the Roadster’s position data for June 24, 2018.
Use the Sun’s gravitational parameter of 1.327 x 1020 m3/s2 and the given semi-major axis to calculate the period of the Roadster’s orbit. Write answer in seconds and also in years, rounding to the nearest decimal.
Calculate the average angular speed (in radians/second) of the Roadster in its orbit about the Sun.
Given a mean anomaly of approximately 90.46 degrees, how much time would have passed since the point of periapsis in the Roadster’s orbit about the Sun?
87
10. Use the given mean anomaly and eccentricity of the orbit to calculate the true anomaly for the Roadster on June 24, 2018.
11. Plot the orbit of the Roadster with periapsis on the positive x-axis. Plot the location of the Roadster in the orbit on June 24, 2018. Use Desmos or another graphing program and attach your work.
Elliptical Orbit Lab
In Kerbal Space Program, launch a rocket or satellite into an elliptical orbit around Kerbin or another celestial body. Complete the following lab report using data collected in game.
Record spacecraft name, planet or moon being orbited, and the gravitational parameter.
Record the eccentricity and semi-major axis of the orbit.
Write the polar form for the orbital path, print and attach a graph of the orbit.
Determine periapsis and apoapsis points of the orbit and label these in the graph.
Use mean anomaly for the orbiting spacecraft to determine times at which the
spacecraft is 1⁄4, 1⁄2, and 3⁄4 of the orbital period from periapsis.
Calculate eccentric and true anomaly for the spacecraft at each of these points.
Determine points in polar form (π, π) where the spacecraft is 1⁄4, 1⁄2 and 3⁄4 of the way
through its orbital period from periapsis. Label these points on a graph of the orbit,
include units, write π in degrees.
Use the vis-viva equation to compute linear speed for the spacecraft at each of these
points and summarize all results in a table.
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Distance at Engine Cut-off
The function π£(π‘) = 2.05π‘2 + 35.12π‘ models the velocity of the Jumping Flea, a small sub- orbital rocket with a solid fuel engine, during the first 10 seconds of flight, where v is the speed in meters/sec and t is time in seconds.
What is the velocity at 10 seconds?
Assuming the rocket launches from an initial height of 74 meters, use integration to find the rocket’s height after 10 seconds.
Write a formula for the acceleration of the Jumping Flea over time.
What is the initial acceleration of the Jumping Flea? What is the acceleration of the Jumping Flea at t = 10 seconds?
Look for the Jumping Flea rocket in Sandbox mode in KSP. Launch and compare your results with the mathematical model above. How does this model compare with an actual flight? How do you think this model was created?
_ _ _
*Calculus with *Parametric & *Polar Equations
Energy and Escape Velocity
In math and physics, work measures a change in energy and can be determined in multiple
ways, one of which is by multiplying force times distance. In orbit, if a distance is large enough
where the force of gravity changes significantly over that distance, we compute work done by
gravity by integrating the force of gravity over a distance travelled as follows:
W = − ∫
GMm
r
2
r2
r1
dr
Here W is the work done by gravity, the fraction GMm
r
2
represents the force of gravity, r1 and r2
represent different radius orbits around the mass M, and the negative sign indicates that work
done is negative whenever r2 > r1.
We can also compute work done by finding a change in energy. In space, we measure kinetic
energy with the equation KE =
1
2
mv
2
and potential energy with the equation PE =
−GM
r
.
The planet Kerbin has a mass of 5.29 x 1022 kg and a radius of 600 km. Recall, the gravitational
constant G = 6.67 x 10-11 N*m2
/kg2
.
Suppose a ship, with mass m = 3000 kg, orbits above the atmosphere of the planet Kerbin.
1. Use the integral formula above to compute the work done by gravity as the rocket
maneuvers from a circular orbit at 100km to one at 150km above the surface of Kerbin.
2. Find the change in the rocket’s potential energy between orbits at 100km and 150km.
3. Use the formula v = √
GM
r
to find the speed of the rocket at circular orbits of 100km and
also 150km and then find the change in kinetic energy between these orbits.
4. How does the change in kinetic energy of the rocket relate to the work done by gravity,
and the change in potential energy, as the rocket moves between orbits at 100 km and
150 km?
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5. Use an improper integral, and the formula for work done by gravity, to find a formula
for the escape velocity at any distance r1 from any planet or moon with mass M. Check
this formula with the formula for escape velocity derived in chapter 2 (Pre-calc and Trig).
6. Compute escape velocity on the surface of Kerbin, write in m/s and miles/hr.
7. Compute escape velocity on the surface of Earth, write in m/s and miles/hr. What was
the first man-made object to reach escape velocity above Earth? Where is it now?
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8. Write the formula that relates escape velocity with orbital speed v for a circular orbit.
9. Compute the escape velocity at an altitudes of 100km and 150 km above the surface of
the planet Kerbin.
10. Find the escape velocity at an altitude of 14 km above the surface of the Mun.
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Parametric and Polar Equations in Orbit
A spacecraft is in a circular orbit around the planet Kerbin at an altitude of 100 km. The radius
of Kerbin is 600 km and Kerbin has a mass of M = 5.29 x 1022 kg. The universal gravitational
constant is G = 6.67 x 10-11
.
1. Write parametric equations for the position, velocity and acceleration vectors of the
spacecraft in its two-dimensional orbital plane.
2. What is the angular speed, linear speed, period, radial distance, and acceleration of the
rocket?
3. Using Desmos.com, graph a circle for Kerbin with radius 600 km. Graph the circular
orbit of the rocket at a 100 km altitude using parametric, polar and rectangular
equations. Graph all forms of the equations of the orbit together to confirm they
coincide as the same path around the planet.
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Tangent to Flight Path and Pitch
An elliptical orbit can be written in the polar form r =
a(1−e
2)
1+ecosΞΈ , where a is the semi-major axis
of the ellipse, e is the eccentricity, ΞΈ is the true anomaly and r is the radial distance from the
primary to the orbiting body. The true anomaly, ΞΈ, is the angle between the line from the
primary focus to the periapsis (Pe) and the line from the primary focus to the orbiting body (m),
as shown in the diagram below.
Assume the orbital plane is the xy-plane.
1. Use the polar form of an elliptical orbit given above to find a formula for dy
dx ⁄ , the
slope of the tangent to the flight path in the xy-plane.
2. An elliptical transfer orbit from Kerbin to the Mun has a semi-major axis of a = 6350 km
and an eccentricity of e = 0.8898. Compute the slope of the tangent to this orbital path
at a true anomaly of 90^0
3. Find the true anomaly at the instant the slope of the tangent in the xy-plane is 0.
4. What is the slope of the tangent in the xy-plane at periapsis and apoapsis?
5. In general, for any ellipse with eccentricity e, what is the slope of the tangent in the xy- plane at a true anomaly of 900?
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96
In the diagram below, the angle Ο is called the flight path angle. The flight path angle is the angle between the velocity vector v and the horizon. For a spacecraft in orbit, the horizon is equivalent to a perpendicular to the radial vector r.
We can use the relation π‘πππ = 1 ππ to determine the flight path angle in orbit. π ππ
A proof of this relation is outlined at youtube.com/c/csvaughen.
6. If π = π(1−π2) describes an elliptical orbit, find an expression for 1 ππ . 1+ππππ π π ππ
Use the relation π‘πππ = 1 ππ to find a formula for the flight path angle of an orbiting π ππ
spaceship.
8. Compute the flight path angle for a spaceship in a circular orbit.
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The Kerbal Math & Physics Lab / Chapter 4 A crew of Kerbal astronauts on a journey from Kerbin to the Mun in a KerbalX rocket follow an
elliptical transfer orbit with semi-major axis a = 6350 km and eccentricity e = 0.8898. 9. Compute flight path angle for this trajectory at a true anomaly of 900.
10. The flight path angle is also called the spacecraft’s pitch. What is the maximum pitch for the KerbalX during the transfer orbit? At what angle of true anomaly does this occur?
_ _ _
*Multivariable and *Vector Calculus
Work done by Gravity
A rocket with mass m = 3000 kg orbits around the planet Kerbin with periapsis at 100 kilometers above the surface and apoapsis at 11,400 kilometers above the surface. The planet Kerbin has a radius of 600 kilometers and a mass of 5.29 x 1022 kg.
The force of gravity on the rocket in its orbital plane is given by the 2D vector field
−πΊππ π⃗ π2 ‖π‖
where G = 6.67x10-11 is the universal gravitational constant, M is the mass of Kerbin and m is the mass of the rocket. The vector π⃗ = π₯π̂ + π¦π̂ is the position vector for the rocket.
The following example is solved in the video “Work Done by Gravity in Orbit (Calc 3)” on YouTube at https://www.youtube.com/c/csvaughen. Look in the Calc 3 playlist.
1. Use the fundamental theorem of line integrals to compute the work done by gravity as the rocket orbits from periapsis to apoapsis.
The Kerbal Math & Physics Lab / Chapter 5
⃗
πΉ ( π⃗ ) =2. Given the rocket’s speed at periapsis is 3087 meters/second and 180 meters/second at apoapsis, check the answer above by computing the change in kinetic energy of the rocket.
100
Great Circle Distance and Bearing
To find the shortest distance between points A and B on a sphere, we find the length of the arc of the great circle between those points. Given the locations of points A and B in terms of latitude and longitude, we use the following spherical to rectangular conversion formulas to
⃑ ⃑⃑
write two vectors π΄ and π΅, whose initial points are at the center of the sphere and whoseterminal points are on the surface of the sphere at locations A and B.
π₯ = ππ ππππππ π π¦ = ππ ππππ πππ π§ = ππππ π
Here, π is the radius of the sphere, π is the “co-latitude” (an angle measured from the positive z-axis down to the point) and π is the longitude. Typically, latitude and longitude are measured in degrees (between 00 and 900 for latitude and between 00 and 1800 for longitude). We follow the convention that southern latitudes and western longitudes are indicated by negatives. Be careful when finding π for points in the southern hemisphere of a planet. Because π represents the angle from the positive z-axis down to a point on the surface, if the point is was in the southern hemisphere, say at latitude -50, then π = 90° − (−5°) = 95°
To find the distance between points A and B on the surface of the sphere, we next find the ⃑ ⃑⃑
angle between vectors π΄ and π΅, in radians, as follows:
⃑ ⃑⃑πΌ = cos−1 (π΄ ∙ π΅) π 2
⃑ ⃑⃑
where πΌ is the angle between vectors π΄ and π΅, and R is the radius of the sphere. With theangle πΌ (in radians), we calculate the distance, d, between points A and B as π = π πΌ.
As an example, let’s calculate the distance from Philadelphia to Miami. Starting with Philadelphia at approximately 400 North, 750 West and Miami at about 260 North, 800 West, and taking the Earth to be a sphere with a radius of about 3960 miles, we write these points in spherical coordinates in the form (π, π, π) as follows:
A: Philadelphia: (3960, -750, 500)
B: Miami: (3960, -800, 640)
Converting to vectors in rectangular coordinates we get π΄ =< 785.14, −2930.17, 2545.44 >⃑
⃑⃑ −1 15174697.72and π΅ =< 618.05, −3505.15,1735.95 > and πΌ = cos ( 39602 ) = .25495 radians Therefore, the distance from Philadelphia to Miami is about 3960(.25495) = 1009 miles.
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Use the method just presented above to answer the following questions.
1. The Kerbal Space Center is located at (0.050 South, 74.60 West) and the K’Yangtze river canyon is located at (22.30 North, 38.60 West) on the planet Kerbin. Kerbin has a radius of 600 km. Find the distance from the KSC to the K’Yangtze river.
2. Based on an average speed of 700 m/s, estimate the time it takes to make the trip from the KSC to the K’Yangtze river.
The KSP stock Ravenspear Mk1 or Mk4 jets can average about 700 m/s. Confirm your calculations in KSP by flying to the K’Yangtze River.
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In Chapter 4, formulas to compute bearing between any two points on any spherical planet or moon were introduced. For each of the following missions in KSP, find distance and bearing and then fly the mission to confirm your calculations.
3. Calculate the distance and bearing from the KSC to the Island Runway (1.50 S, 71.90 W).
4. Calculate the distance and bearing from KSC to the Desert Pyramids (6.50 S, 141.70 W).
5. Calculate the distance and bearing from the KSC to the UFO buried near the North Pole at (820 N, 128.50 W).
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The formulas and methods presented here will work on any size spherical planet or moon. We can apply these formulas for a flight from New York to Tokyo or to navigate a rover from the Apollo 11 landing site to the Apollo 12 landing site on the Moon.
Note calculations done using the formulas for bearing in chapter 4 of this workbook produce the “true heading” which can be different from the “magnetic heading”, particularly on Earth where the magnetic north pole is not the same as the geographic north pole. To learn more about the difference between true and magnetic north pole in navigation check out the video: https://youtu.be/dXcphJ1RjDc at https://www.youtube.com/c/csvaughen
6. Find distance and initial bearing for a flight from New York to Tokyo. Look up latitude and longitude for New York and Tokyo and confirm your final answers using a website like flightaware.com, gcmap.com or wolframalpha.com.
The Kerbal Math & Physics Lab / Chapter 5
7. Compute the distance and initial bearing for the shortest path on the Moon from the Apollo 11 landing site to the Apollo 12 landing site. Use the Apollo 11 landing site coordinates at .680 North, 23.40 East and the Apollo 12 landing site coordinates at 3.20 South, 23.40 West. The Moon is nearly spherical with a radius of about 1737 km. Write distance in kilometers and in miles.
104
Optimal Staging with Lagrange Multipliers
In this lab we use the method of Lagrange multipliers to find the optimal mass of a multi-stage rocket. We compare our results with modern and historical rockets and with models built in Kerbal Space Program. The mathematical approach presented here can also be found in the books Calculus – Early Transcendentals by James Stewart, Orbital Mechanics for Engineering Students by Howard Curtis, and Space Mathematics by Bernice Kastner.
We begin our analysis with the ideal rocket equation (which also appears in chapters 4 and 6)
where
∆π£ = change in velocityπ£π = exhaust velocity relative to the rocket ππ= initial mass of rocket
ππ = final mass of rocketFor a two-stage rocket, the ideal rocket equation becomes π£π = πππ ( π1+π2+π΄ ) + πππ ( π2+π΄ )
And for a three-stage rocket, we have
π£π = πππ ( π1+π2+π3+π΄ ) + πππ ( π2+π3+π΄ ) + πππ ( π3+π΄ )π π1+π2+ π3+π΄ π π2+π3+π΄ π π3+π΄
(Equation 2)
(Equation 3)
∆π£ = π£πππ(ππ) ππ
(Equation 1)
To analyze multi-stage rockets and their potential to lift specific payloads to orbit, we will modify the ideal rocket equation and include several simplifying assumptions:
First, we’ll assume that each stage is similar and produces the same constant exhaust velocity, π£π = π, and that each stage has the same ratio of empty (“dry”) mass to fully-fueled (“wet”) mass. This ratio is called the “structural factor”, which we will represent with the value s. After developing results with the assumption of similar stages, we can extend our approach to include different exhaust velocities and structural factors for each stage of the rocket.
And to further simplify, we won’t immediately include the effects of gravity or air-resistance in the rocket equation. Instead, we adjust the total required velocity to compensate for velocity lost overcoming gravity and air-resistance during launch.
The Kerbal Math & Physics Lab / Chapter 5
π π1+π2+π΄ π π2+π΄
To understand these expressions, note each term on the right-hand-side is derived from the ideal rocket equation (Equation 1) and represents the change in velocity contributed by each stage of the rocket.
The value c represents the relative exhaust velocity produced by each stage. The values π1, π2, π3 represent initial mass for each component or individual stage (called the “step mass”), the value A represents the payload mass and π£π is the final velocity resulting from the combined delta-v from all stages.
The numerators in each argument of the natural log function (in Equations 2 and 3 above) represent total initial mass for the rocket at each stage because lower stages carry higher stages as payload, while the denominators represent the final mass for the rocket at each stage after the fuel is spent.
For example, π1 + π2 + π΄ is the combined total mass of the first stage of a two-stage rocket carrying payload A, before the fuel of the first stage is spent. The individual step mass of the first stage is π1, however the first stage also carries the second stage (with step mass π2) and the payload with mass A, as shown in the diagram to the left.
The structural factor for each stage can be expressed as
π = πππππππππ‘π¦("πππ¦")πππ π =ππ ππππ‘πππ ππ’ππππ ("π€ππ‘") πππ π π0
We can think of s as representing the fraction of each step mass that is the structural component of the rocket (such as tanks, engines, hull, equipment, etc.)
In particular, for the step mass π1, we will say it begins initially filled with fuel, with mass π1)0 and after burning all of its fuel, will have mass π1)π and thus
π = π1)π π1)0
or π1)π =π ∙π1)0
In other words, the final step 1 mass, π1)π, after completing the stage 1 burn is π ∙ π1)0We simplify this notation and write π π1 to represent the stage 1 step mass after the stage 1 burn is complete. That is, if π1 is the initial step mass and s represents the fraction of each stage that is structural, then π π1 is the individual stage 1 step mass after the fuel of stage 1 is spent. Similarly, if ππ is the step mass for stage π then π ππ is the step mass after the fuel for stage π is spent.
We conclude, after the first stage (of a two-stage rocket) completes its burn (and before stage 1 is dropped), the final mass of the rocket is π π1 + π2 + π΄. Thus, from the ideal rocket equation, the delta- v that results from the stage 1 burn will be
∆π£1 = πππ ( π1 + π2 + π΄ ) π π1 + π2 + π΄
Continuing with this example, after the stage 1 burn is complete, the first stage is dropped and the total
mass of the rocket becomes π2 + π΄, (i.e., second stage step mass π2 and payload π΄).
And as before, after the second stage completes its burn, the resulting final mass of the rocket will beπ π2 + π΄ and the resulting delta-v from the second stage burn will be ∆π£2 = πππ ( π2 + π΄ )
π π2 + π΄
Thus, the final velocity of the two-stage rocket, given constant exhaust velocity c, after the stage 1 and stage 2 burn, will be
π£π =∆π£1+∆π£2 =πππ(π1+π2+π΄)+πππ(π2+π΄) π π1 +π2 +π΄ π π2 +π΄
In this analysis we do not consider any fuel contained within the payload, so for the two-stage rocket, the second stage and payload achieve the same final velocity. Similarly for a three-stage rocket, the third stage and payload achieve the same final velocity.
Also note there is no delta-v gained directly from dropping stages alone because the rocket and the lower stage are traveling at the same speed when a stage is dropped. The delta-v gains come only from expelling mass from the rocket at high-speeds relative to the rocket.
Our objective in this lab will be to find the lowest possible total mass, and the corresponding step mass for each stage in a multi-stage rocket, with the constraint that the rocket achieve a given final velocity (for example, a final velocity sufficient to reach orbit).
We will find the theoretical optimal (minimum) value for each step mass and thus also the total minimum combined mass for a multi-stage rocket by the method of Lagrange multipliers, develop formulas for each step mass, and the combined total mass, for a given payload and target velocity, and apply these formulas to specific examples using Excel and Mathematica. We will then compare theoretical optimal values with modern and historical rockets such as the Falcon 9 and Saturn V, and with rocket models built in Kerbal Space Program.
We continue now by developing formulas for optimal staging for a two-stage rocket with similar stages (having the same structural factor and exhaust velocity). These formulas can then be extended for three-stage rockets, and for variable structural factors and variable exhaust velocities.
To apply the method of Lagrange multipliers, we must identify the objective and constraint functions: In this context, with a two-stage rocket, we wish to minimize the objective function
πΉ(π1,π2) = π0 = π1 + π2 + π΄ (Equation 4)
which represents the total mass of the rocket in terms of the step masses π1, π2 and the constantpayload π΄. Additionally, we use
πΊ(π1,π2)=π£π =πππ(π1+π2+π΄)+πππ(π2+π΄) (Equation5)π π1 +π2 +π΄ π π2 +π΄
as the constraint, where the values c , s and A each represent constants.107
Write an objective function F and constraint function G for a three-stage rocket in the form of Equations 4 and 5.
Show that, regardless of the number of stages, the minimum value of π0 (the total rocket mass) occurs at the same point as the minimum value of ππ (π0) where A is assumed to be constant
π΄
and π0 > 0.
The functions F and G given in Equations 4 and 5 above will be difficult to manipulate precisely in both the algebra and calculus of the method of Lagrange multipliers, therefore to simplify the anticipated steps, we define new variables in terms of the given variables, apply the method of Lagrange multipliers in terms of the new variables, and then solve for π1 and π2.
Therefore, continuing with the assumption of a two-stage rocket, we define two new variables: π = π1+π2+π΄ and π = π2+π΄.
1
3.
4.
5.
π π1+π2+π΄ 2 π π2+π΄
Show (1−π )π1 = π1+π2+π΄ and (1−π )π2 = π2+π΄ 1−π π1 π2+π΄ 1−π π2 π΄
The Kerbal Math & Physics Lab / Chapter 5
Show π0 = π΄
(1−π )2π1π2 (1−π π1)(1−π π2)
Showππ(π0)=2ln(1−π )+πππ +πππ −ln(1−π π )−ln(1−π π )
π΄
1212
To proceed by the method of Lagrange multipliers, we find the minimum value of the function
πΉ (π ,π )=2ln(1−π )+πππ +πππ −ln(1−π π )−ln(1−π π ) π121212
subjecttoπΊ (π ,π )=ππππ +ππππ =π£ π1212π
108
The Kerbal Math & Physics Lab / Chapter 5 One approach to the method of Lagrange multipliers is to write the “Lagrange Function”, H, which
combines the objective function and constraint and introduces the Lagrange multiplier π as follows: π»(π ,π ,π)=2ln(1−π )+πππ +πππ −ln(1−π π )−ln(1−π π )+π(π£ −ππππ −ππππ )
121212π12
Note this approach requires rewriting the constraint equation as the function π£ − ππππ − ππππ . π12
For the function H given above, find ππ» , ππ» and ππ». ππ1 ππ2 ππ
Solve the following system of equations for π, π , π and show π = π . 1212
1+π =ππ
π 1−π π π 111
1+π =ππ π2 1−π π2 π2
π£ =ππππ +ππππ π12
Note that under the assumption of similar stages (constant exhaust velocity and constant structuralfactorsineachstage),theconclusionπ =π impliesthemassratiosineachstage
are the same and the delta-v produced by each stage is the same, however it does not imply the mass of each stage is the same.
Letπ=π =π andsolveforπintermsofπ£ andπ. 12π
(a) Solve for π2 in terms of π£π, π, s, and A.
(b) Solve for π1 in terms of π2, π£π, π, s, and A.
12
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We have now developed formulas for the step masses π1 and π2 and the total mass π0 = π1 + π2 for the optimal (minimum) mass for a two-stage rocket given fixed payload A, structural factor s, relative escape velocity π£π = π, and target orbit velocity π£π. To be thorough, it remains to be shown that the values for π1 and π2 correspond to a minimum value for the total mass of the rocket subject to the final velocity constraint.
10. Show the total differential in H
2 2 π2π»
π2π» = ∑∑ π=1 π=1
ππ ππ ππ
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ispositivewhenππ(ππ−1)2 +2ππ−1>0. Notecomputingππ(ππ−1)2 +2ππ−1and confirming it is positive in a specific example is one way to confirm that the solution found is a minimum.
11. Repeat the steps outlined by questions 1 to 9 to find formulas for the theoretical minimum total mass, and the theoretical minimum mass for each stage of a 3-stage rocket given payload A. Assume the same relative exhaust velocity c, and structural factor s for each stage of the rocket.
ππ ππ ππ
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The orbital velocity at an altitude of about 200 km above Earth (at low Earth orbit or “LEO”) is about 7.8 km/s or about 17,500 miles/hour. However, for a rocket to reach this altitude, approximately 2 km/s (or 5000 mph) is spent overcoming the pull of gravity (climbing vertically to reach altitude) and traveling through the atmosphere. Thus, the total minimum delta-v required to reach orbital velocity launching from the surface of the Earth is approximately 10 km/s (or approximately 22,000 mph total).
12. A single stage rocket has a payload with mass A = 50 metric tons. Suppose the rocket generates a relative exhaust velocity of 3 km/s and suppose a total velocity of 10 km/s is required for orbit.
13.
Use another modified version of the ideal rocket equation (for a single stage rocket) to show such a rocket could not reach the orbital velocity with a structural factor of s = 0.1 (meaning 90% of the mass of the rocket is fuel).
Determine the maximum structural factor s for which a single stage rocket can reach orbital speed assuming a payload of 50 metric tons, relative exhaust velocity of 3 km/sec and a target orbital speed of 10 km/s.
Assume a single stage rocket has payload 50 metric tons, relative exhaust velocity of 3 km/sec has a structural factor equal to the value found in part (b). What is the rocket’s required theoretical minimum mass and what percent of the total mass is the payload?
Consider a two-stage rocket with a payload mass A = 50 metric tons with a target orbital velocity of 10 km/s. Compute the minimum mass and each step mass assuming a structural factor of s = 0.1 and relative exhaust velocity of 3 km/s.
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14. Suppose a three-stage rocket must lift a payload with a surface weight of 500 pounds. Suppose this rocket will have a structural factor of s = .2 at each stage and that each stage can produce a relative exhaust velocity of 6,000 miles/hour. Suppose the payload requires an Earth escape velocity of 24,700 miles/hour. Find the theoretical minimum mass for the rocket and the mass of each stage.
15. The Saturn V, which launched the Apollo missions to the Moon between 1969 and 1972, was a 3-stage rocket that carried a payload of approximately 50 metric tons to the Moon. The payload consisted of the Command, Service and Lunar Excursion Modules and provided astronauts the capability to land on the Moon and return to Earth. Assume a structural factor of 0.05 for all three stages of the Saturn V and assume a constant exhaust velocity at each stage of 3.8 km/s. Assume the total delta-v requirements to reach the Moon after launching from the surface of the Earth is about 14 km/s.
Compute the theoretical optimal mass of each stage and the total mass of the Saturn V.
Compare the theoretical results determined above with the following approximations to the actual mass of each stage of the Saturn V.
Stage 1 mass: 2,280 metric tons Stage 2 mass: 485 metric tons Stage 3 mass: 120 metric tons Payload: 50 metric tons
Total mass (including payload): 2,935 metric tons
Determine the percent of total mass of the Saturn V that is payload using both the theoretical optimal values from part (a) and using the actual values shown above.
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The SpaceX Falcon 9 is a two-stage rocket capable of launching about 23 tons to low Earth orbit.
a) Assume a constant structural factor of 0.04 for each stage, a relative exhaust velocity of 3.5 km/sec, and a target orbital velocity of 10 km/sec, and find the theoretical optimal mass for each stage of the Falcon 9 and the total mass of the rocket.
b) Look up design specifications for each stage and maximum payload capability for the SpaceX Falcon 9 “Full Thrust” Block 5 variant which began use in May 2018 and compare with the theoretical values determined above. What simplifying assumptions were made to produce the theoretical optimal values? How do these simplifying assumptions differ from actual performance requirements and operating conditions for the Falcon 9?
The Starship and Super Heavy Booster currently under development at the SpaceX site in Boca- Chica, Texas, will form a 2-stage rocket with a first orbital test flight possible in 2022. Based on information provided on the SpaceX website and Wikipedia, Starship is intended to lift 100 metric tons or more to low Earth orbit, to the Moon and to Mars.
a) Assume a payload of 150 metric tons, a structural factor of 0.06 for both Starship and Super Heavy, and a relative exhaust velocity of 3.7 km/sec for each stage. If a total delta-v of 10 km/sec is required for the initial orbit, what is the theoretical optimal mass required for each stage?
b) Based on the theoretical optimal values found above, what percentage is the payload mass of the total mass of the rocket (payload + super heavy booster + starship)?
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18. In Kerbal Space Program, a two-stage rocket will lift a payload of 500 kg and has the following specifications: Structural factor s = 0.5 and relative exhaust velocity c = 2700 m/s on both stages. Assume a target velocity of 3000 m/s for low Kerbin orbit.
(a) Findthetheoreticaloptimalminimumrequiredmassforeachstageandthetheoreticaltotal minimum mass required for this rocket.
(b) Buildatwo-stagerocketinKerbalSpaceProgramtoliftapayloadof500kgtolowKerbin orbit (approximately 100 km altitude) and compare theoretical optimal and actual mass values for the rocket and each stage.
19. The Acapello, a rocket in Kerbal Space Program included in the Making History DLC, is similar to the real Saturn V rocket with 3 stages and is designed to lift a payload consisting of a Command, Service and “Munar” Excursion Module to Kerbin’s moon, known as the Mun. Consider the following approximations for the design of the Acapello rocket:
Assume all three stages of the Acapello are similar with a structural factor s = 0.45, relative exhaust velocity c = 3200 m/s and a total delta-v requirement to reach the Mun of 5500 m/s.
Compute the theoretical total minimum required mass for the Acapello rocket with these specifications and find the minimum required mass for each stage. Compare these values with the actual values for the Acapello available in game in the Vehicle Assembly Building. Note the free downloadable mods Kerbal Engineer or MechJeb can be helpful in identifying delta-v and the individual mass for each stage of a rocket built in KSP.
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Finally, we can extend the previous derivations for 2 and 3 stage rockets with similar stages to allow for an arbitrary number of stages, K, each with different structural factors and different relative exhaust velocities by defining π π as the structural factor and ππ as the relative exhaust velocity at each stage π with π = 1,2,3 ... , πΎ.
We proceed in the same manner as previously with similar stages and define the mass ratios of each stage as follows:
π = π1+π2+⋯+ππΎ+π΄
1
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π 1π1 +π2 +⋯+ππΎ +π΄
π2= π2+⋯+ππΎ+π΄ π 2π2 +⋯+ππΎ +π΄
⋮
ππ= ππΎ+π΄ π πππΎ +π΄
Following the approach used previously for 2 or 3 similar stages leads to the following Lagrange function:
πΎπΎπΎπΎ
π»(π,π,...,π,π)=∑ln(1−π )+∑πππ−∑ln(1−π π)+π(π£ −∑ππππ) 12ππππππππ
π=1 π=1 π=1 π=1
20. Given the function H above, find ππ» and ππ». πππ ππ
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21. Solving the resulting system of equations by the method of Lagrange multipliers, assuming an arbitrary number of stages, each with different structural factors π π and exhaust velocities ππ, leads to the following single equation which must be solved for π.
πΎπΎπΎ
∑ππln(πππ −1)−πππ∑ππ −∑ππln(πππ π)=π£π π=1 π=1 π=1
Suppose K = 3, π1 = 3.924, π2 = 3.434, π3 = 2.943, π 1 = 0.1,π 2 = 0.15,π 3 = 0.2, and π£π = 10.
Use a calculator or computer to find a numerical approximation for π rounding to 4 decimal places.
22. A three-stage rocket has the following design specifications:
Payload: 5000 kilograms (5 metric tons), target orbital velocity: 10 km/sec.
Compute the theoretical optimal mass for each stage and the theoretical minimum total mass for the rocket. Compute propellant and dry mass for each stage.
Note relative exhaust velocity π£π is computed from engine πΌπ π with the definition π£π = πΌπ π π0 where π0 is the Earth surface gravity 9.81 m/s2 or 32 ft/s2.
_ _ _
*Differential Equations Solving the Rocket Equation
The ideal rocket equation, introduced in chapter 2, is written either in the form ∆π£ = π£πππ(π0)
where
ππ
or
∆π£ = πΌπ ππ0ππ(π0) ππ
∆π£ = change in velocity (delta-v)
π£π = velocity of exhaust gases expelled relative to the rocket
π0 = initial mass of rocket
ππ = final mass of rocket (after engine burn)
πΌπ π = specific impulse (a measure of rocket efficiency)
π0 = acceleration due to gravity at the surface of Earth (or Kerbin)The equation π£π = πΌπ ππ0 relates relative exhaust velocity, π£π, specific impulse, πΌπ π and the acceleration due to gravity, π0, at the surface of Earth (and Kerbin). Specific impulse is a convenient way to measure a rocket engine’s efficiency regardless of whether metric or customary units are used. Thus, because πΌπ π = π£π , the units of πΌπ π are always seconds.
π0
The ideal rocket equation relates initial and final mass of a rocket to a resulting change in velocity and does not take into account the effects of gravity. This statement might be confusing because the factor π0 appears in the equation. To clarify, π0 is always used, whether the engine burn takes place on the surface of the Earth or Kerbin, or at the Moon, or on Mars or in deep space, either in real life or in Kerbal Space Program. In every case, π0 is simply a constant (either 9.8 m/s2 in metric or 32.2 ft/s2 in customary units) that allows for a conversion of the relative exhaust velocity, whether in metric or customary units, to something independent of either unit of measure.
For example, suppose a rocket engine’s thrust expels exhaust at π£π = 1372 m/s at sea level. Because 1 meter is approximately 3.281 feet, this is equivalent to approximately
π£π = 1372 meters × 3.28 feet = 4500.2 ft/s (at sea level). 1 second 1 meter
Thus πΌ = π£π = 1372 π/π = 140 seconds and similarly, πΌ = π£π = 4500.2 ππ‘/π ≈ 140 seconds π π π0 9.8 π/π 2 π π π0 32.2 ππ‘/π 2
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That is, πΌπ π is the same regardless of whether metric or customary units are used and the factor π0 here is really just a conversion factor. Further, notice the ratio π0 in the ideal rocket
ππ
equation is dimensionless and therefore the units of ∆π£ depend entirely on the choice of unitsfor π0.
A rocket engine’s performance changes depending on whether it is at sea-level or in space because the velocity of the exhaust changes with the surrounding atmospheric pressure. Therefore, rocket engine specific impulse is generally stated for different environments, such as at sea-level or in a vacuum.
1. Suppose a rocket has an initial mass of 2480 kg and a final mass of 1430 kg. Suppose the rocket’s specific impulse is 140 seconds (at sea level). Use the ideal rocket equation to estimate the total delta-v for this rocket.
2. If a rocket is launched from the surface of a planet with an atmosphere, what forces would lead to a final velocity lower than that computed by the ideal rocket equation?
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The Kerbal Math & Physics Lab / Chapter 6 With the ideal rocket equation, ∆π£ = π£πππ (π0) or ∆π£ = πΌπ ππ0ππ (π0), we calculate a change
in velocity that would result from a change in mass with some given constant relative velocity for expelled exhaust gasses (or given the corresponding specific impulse). Written in either of these forms, these represent an “ideal” because neither equation takes into account the rocket’s weight (the force of gravity) or atmospheric drag. Thus the ideal rocket equation is most applicable in a case where a rocket is in deep space, where any force due to gravity is small and doesn’t change during the engine’s thrust and where there is no force due to atmospheric drag.
In this chapter we develop, and then solve, a differential equation that represents a version of the rocket equation that will take into account both the weight of a rocket and the force of atmospheric drag during ascent through an atmosphere.
In this model, let’s assume forces directed upward are positive and those directed downward are negative. Recall that the total force on a rocket can also be expressed as the time rate of
change of its momentum. That is, πΉ = π (ππ£) where m is the rocket’s mass and v is its ππ‘
velocity. We will also take the following simplifying assumptions: the rocket provides a constant upward thrust (such as a solid fuel rocket), and that during the first few seconds of flight the coefficient of drag, k, is constant and finally, acceleration due to gravity, g, will be assumed constant while the rocket is still near the surface of the planet.
3. Write a version of the rocket equation that includes gravity and atmospheric drag proportional to the square of velocity with the assumptions described above. Define each of the variables. Write the order of equation and classify whether it is linear or nonlinear.
ππ ππ
4. Who is Leonard Euler? When did Euler develop the method we now call “Euler’s method” for solving differential equations?
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A rocket with an initial mass of 2480 kg is launched vertically from an initial height of 74 meters.
The rocket engine will burn for approximately 8.5 seconds with a constant thrust of 163,000
Newtons and a fuel consumption rate of 131.25 kg/sec. Assume a drag coefficient for the term
kv2 of k = 0.3 and a constant acceleration due to gravity near the surface of the planet of g = 9.8 m/s2.
The corresponding differential equation cannot be solved exactly, thus a numerical approach like Euler’s method is required to find velocity over time.
5. To apply Euler’s method, solve the equation for dv/dt and write a recursive formula for vn, the velocity of the rocket at each time increment.
6. Use Euler’s method to solve the rocket equation from t = 0 seconds to t = 9 seconds, using a step size of 0.5 seconds. Use an online application or Excel to execute Euler’s method. Print a table of values and a graph for the velocity during the engine burn. Restrict the graph to the interval 0 < t < 9. Use an appropriate vertical scale. Be sure to label the values in the table and label the axes and curve in the graph.
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7. Use a computer or calculator to find a curve of best fit for the velocity data. Write the resulting velocity function.
According to the mathematical model, what is the velocity at 8.5 seconds? How does this compare with the prediction from the ideal rocket equation (in question 1 above)?
Integrate the velocity function to find a height function. Write the height function and graph the height function. Label the axes and the curve. Restrict the graph to the interval 0 < t < 9, and use an appropriate vertical scale.
10. Create a table indicating the height of the rocket at each second during the engine burn. Label the values in the table. According to the model, what is the predicated height at 8.5 seconds?
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The Euler-1B test rocket has an initial mass of π0 = 3380 kg, and a final mass of ππ = 2480 kg after all fuel is spent. The rocket will be launched from an initial height of 832 m above sea level where the rocket’s engine has a specific impulse (Isp) of 179.1 seconds. The Euler-1B engine provides a constant thrust of 153.51 kiloNewtons for 10.3 seconds. Assume a constant acceleration due to gravity of g = 9.8 m/s2 during the flight and a drag coefficient of k = 0.38 kg/m in the rocket equation.
11. Calculate the theoretical ∆π£ for the Euler-1B based on the ideal rocket equation.
12. Write the rocket equation (in a form that includes gravity and air-resistance) with the values given above and solve for ππ£⁄ . Assume the mass of the rocket drops at a constant rate given
ππ‘ by π, the fuel burn-rate.
Write an expression for velocity in the form π£π+1 = π£π + h ∙ π(π‘π, π£π) which can be used in Euler’s method.
Using the initial conditions π‘0 = 0, π£0 = 0, and Euler’s method with step size 0.5 seconds, solve for the rocket’s speed at 10.5 seconds.
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Write a linear regression model (based on the data from Euler’s method) for the speed of the rocket during the first 11 seconds of flight.
How does the velocity at t = 10 or t =11 seconds in the model compare with the change in velocity predicted by the ideal rocket equation? Explain any differences in these values.
Use the linear regression model to determine the rocket’s acceleration during the first 11 seconds of flight.
Based on the acceleration of the rocket, calculate the g-force experienced by an astronaut on board the Euler-1B.
Use the linear regression model for rocket’s velocity to find a formula for the height of the rocket during the first 11 seconds of flight.
What height does the model predict for the rocket at t = 10 and t = 11 seconds?
Check your work on the Euler-1B in Kerbal Space Program by building a similar rocket as shown in this video clip: https://drive.google.com/file/d/1AN3TD7w-tOlQcTvqFCiH_zT9gJaEJUsc/view?usp=sharing
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