Question

A satellite is injected into an elliptical orbit with a semi-major axis equal to 4 DU. When it is precisely at the end of the semi-minor axis it recevies an impulsive velocity change just sufficient to place it into an escape trajectory. What was the magnitude of the velocity change?

Answer 1

1 DU = 6378.1 km

4 DU = 25512.4 km = 2.55 x 10^7 m

Vesc = sqrt(2 x 3.987 x 10^14) / 2.55 x 10^7

Vesc = 5.592 x 10^3

In order to determine the speed before the launch, we must use the energy equation.

E = 1/2mv^2 - GM/r ..............(1)

E = -GM/2a

E = (3.987 x 10^14) / (2 x 2.55 x 10^7)

E = -7.186 x 10^6

Put in above equation,

-7.186 x 10^6 = 1/2v^2 - (3.987 x 10^14) / ( 2.55 x 10^7)

v = 3.9542 x 10^3

v = 5.592 x 10^3 - 3.9542 x 10^3

v = 1.6378 x 10^3