Consider a spacecraft in an elliptical orbit around the earth. At the low point,

Question

Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is 350 km above the earth's surface; at the high point, or apogee, it is 2000 km above the earth's surface.

A) What is the period of the spacecraft's orbit?

B) Using conservation of angular momentum, find the ratio of the spacecraft's speed at perigee to its speed at apogee. 1.25

C) Using conservation of energy, find the speed at perigee and the speed at apogee. (I keep getting 17400,13900 but its wrong)

D) 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?

E) What if the rockets were fired at apogee?

F) Which point in the orbit is more efficient to use?

I cant get past part B. Please help.

Explanation / Answer

Apply Time period T^2 = 4pi^2 R^3/GM

here

semi major axis R = Ra+ Rp/2

R = (350+ 2000 )/2 = 1175 km

so

T^2 = 4*3.14* 3.14 * 1175000*1175000*1175000/(6.67e-11 * 5.97 e24)

T = 40103,89 secs

-----------------------------

speed V ^2 = GM/Rp

V^2 at perigee = 6.67 e-11 * 5.97 e24/(2000000)

Vp^2 = 1.9 e 8

Vp = 14.1 k m/s

---------------------

V at Apogee = Va^2 = GM/Ra

Va^2 = 6.67 e-11 * 5.97 e24/(350000)

Va = 33.72 km/s

so

ratio is 33.72/14.1

= 2.39

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