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Problem set on two stage rocket - Dynamics - MIT
Typology: Exercises
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Consider the design and optimization of a 2 stage rocket. We desire to determine the optimum mass ratio between the first and second stages of a rocket. The payload is 10% of the total mass of the rocket M. The mass of the first and second stages together is then 90% of the total mass M. 90% of the mass of each stage is fuel. The mass of the first stage is R times the combined mass of the two stages, (m 1 =. 9 ∗ R ∗ M ), leaving (1 − R) times the combined mass of the two stages for the mass of the second stage (m 2 =. 9 ∗ (1 − R)M ). (R is a number less than 1, say .55 as a typical value.) Now let’s launch. Take c = 4500 and for convenience take the total mass M equal to 1 (payload plus both stages; M = 1). Ignore gravity and use the expression for the ”ideal” Δv from a rocket firing from Lecture 14, Δ(v)ideal = −cln( mmf 0 ) = −cln(μ). Calculate the velocity after the first stage is fired to comple tion and about to be thrown away. Then calculate the additional velocity from the firing of the upper stage after the first stage has been discarded.( Using (v 0 )ideal ignores the effect of gravity and gives the most optimistic prediction of final velocity.)
1-a) Now fire the first stage and fly it to completion. For a general value of R, and the specified masses of the payload, fuel and second stage, what is the final velocity of the rocket after the first firing? What is μ for this part of the calculation?
1-b Now fire the second stage, and fly it to completion. What is the final velocity of the rocket vf after this firing. Express this final velocity as a function of R, the mass ratio of the first stage m 1 to the total rocket mass m 1 + m 2 (excluding payload). What is μ for this part of the calculation?
1-c Find the optimum value of R that maximizes the final velocity. You may do this by either plotting vf inal(R) vs. R, or by taking ( (^) dRdv^ )f inal, setting it equal to zero, and solving for R.
1-d Compare this ”optimum” velocity to the final velocity for a single stage rocket. (What would R be for a single stage rocket?)
The Columbia Accident Investigation Board recommended that, for safety reasons, all future shuttle missions should be able to utilize the Space Station as a ”safe harbor” in the event of an emergency. This led to the reconsideration of all future Hubble Space Telescope repair missions. Why is it so difficult to get from the Hubble to the Space Station? To accommodate the participation of the Russians in the space station program–so that they could launch from their Baikonur launch site–the inclination of the space station orbit is 51. 6 o^ from the equator; its altitude is 400 km from the earth’s surface. The inclination of the Hubble orbit–minimizing the energy required for a launch from Kennedy Space Center in Florida– is 28. 5 o^ from the equator and it altitude is 600 km from the earth’s surface. Create an emergency-transfer orbit to transfer the shuttle and its crew from their mission in the Hubble orbit to the space station in the space station orbit. This transfer involves both an angular change and an orbital radius change. Do the angular change first: change orbit inclination without changing orbit altitude.
The two points where the orbit intersects the earth equatorial plane are called the nodes, and the line passing through the nodes is called the line of nodes. Assume that the line of nodes is the same for the Hubble and the Space Station. This means that the angular transfer can be done at the equatorial plane and that the angle of transfer is the difference between the inclination of the two orbits. Recall from the rocket equation that ΔV = −cln(mf /m 0 ) (1) 3-1 What percent of the total shuttle mass would need to consist of fuel at this point in its mission to accomplish the required orbit inclination change? Justify the form of the rocket equation used. Take c as 4500 m/sec.
3-2 Now complete the transfer with a Hohmann transfer orbit. Don’t worry about time synchroniza tion between the space station and the Hubble at this point. What ΔV and what percentage of shuttle mass are required to perform this maneuver? How does this compare to that required for the inclination change. 3-3 Now considerer time synchronization. How many opportunities per week exist for the total maneuver?
Image by MIT OpenCourseWare.
Consider two bodies of mass M = 5. 97 × 1024 kg and m = 7. 36 × 1022. Non-dimensionalize distances by the radius of the earth (Re = 6. 36 × 106 m and time by 1 day = 24 hours = 3600 ∗ 24 sec. The masses are separated by a non-dimensional distance of 60.25. They orbit about their common center of mass. Formulate and solve the ”Kepler problem” for the orbital motion of each body about their common center of mass. The universal gravitational constant is G = 6. 67 × 10 −^11 m^3 /kgsec^2.
4-1 Where is the center of mass of the system. How do the bodies move about this center of mass.
4-2 What non-dimensional velocities v 1 and v 2 will produce circular orbits for both bodies?
4-3 What is the non-dimensional radius r 1 of the orbit of M?
4-4 What is the non-dimensional radius r 2 of the orbit of m?
