Consider a spacecraft in an elliptical orbit around the earth. At the low point,
ID: 1323363 • Letter: C
Question
Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is 200km above the earth's surface; at the high point, or apogee, it is 2000km above the earth's surface.
1 What is the period of the spacecraft's orbit?
2 Using conservation of angular momentum, find the ratio of the spacecraft's speed at perigee to its speed at apogee.
3 Using conservation of energy, find the speed at perigee and the speed at apogee.
4 It is necessary to have the spacecraft escape from the earth completely. If the spacecraft's rockets are fired at perigee, by how much would the speed have to be increased to achieve this?
5 What if the rockets were fired at apogee?
6 Which point in the orbit is more efficient to use?
Explanation / Answer
1) for finding the period of space craft orbit
use keplers third law.....
Kepler's 3rd Law: T=(2*pi*a3/2)/ sqrt(GME)
2)Using conservation of angular momentum, find the ratio of the spacecraft's speed at perigee to its speed at apogee
ANS ---->>>
we know that semimajor axis is a = (1/2)[(Re + hp)+(Re +ha)] = 9.13*106 m Re = radius of earth = 6380 km = 6380000 m hp = height of perigee = 200 km = 200000 m ha = height of apogee = 2000 km = 2000000 m M = mass of the earth = 5.98*1024 kg G = 6.67*10-11 Nm2/kg2 we know that time period of revolution of the spacecraft is
T= 2?a3/2/?GM Then put the values in equation ....u will get the value of T
We know that according to conservation of angularmomentum mvprp =mvara vp = spacecraft's speed at perigee va=spacecraft's speed at apogee Then vp /va =ra/rp = (Re+ha)/(Re +hp) = 6380+2000/6380+200 =1.27
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