Question

A satellite moves around the Earth in an elliptical orbit such that at perigee (closest approach) it is two Earth radii above the Earth’s surface. At apogee (farthest position), it travels with one-fourth the speed it has at perigee. In terms of the Earth’s radius R, what is the maximum distance of the satellite from the Earth’s surface ?

Answer 1

Let r_p and r_a be the distances of the satellite from the centre of the Earth at perigee and apogee, respectively. Let v_p and v_a be its linear (tangential) velocities at perigee and apogee.

Data : r_p = 2R + R = 3R, v_a = \(\frac14 V_p\)

Let L_p and L_a be the angular momenta of the satellite about the Earth’s centre. Because the gravitational force (\(\vec F\)) on the satellite due to the Earth is always radially towards the centre of the Earth, its direction is opposite to that of the position vector  (\(\vec r\)) of the satellite relative to the centre of the Earth, so that the torque \(\vec \tau\) = \(\vec r\)x \(\vec F\) = 0 . Hence, the angular momentum of the satellite about the Earth’s centre is constant in time.

\(\therefore\) L_p = L_a

If m is mass of the matellite,

At apogee, the distance of the satellite from the Earth’s surface is 12R – R = 11R.