Question

A satellite of mass m is in elliptical orbit around the Earth. The speed of the satellite is at its nearest position is $= \sqrt {\dfrac{{6GM}}{{5r}}}$ where r is the perigee (nearest point) distance from the center of the Earth. It is desired to transfer the satellite to the circular orbit of radius equal to its apogee (farthest point) distance from the center of the Earth. The change in orbital speed required for this purpose is
A. $0.35\sqrt {\dfrac{{G{M_e}}}{r}}$
B. $0.075\sqrt {\dfrac{{G{M_e}}}{r}}$
C. $\sqrt {\dfrac{{2G{M_e}}}{r}}$
D. Zero

Answer 1

Hint: When an object moves in a circle with radius r, angular momentum is produced which is the product of mass, velocity and radius of the circle.
Also the point of orbit of the satellite nearest to the Earth is called perigee and the point of satellite farthest from Earth is called apogee.

Complete step-by-step answer:
Step I:
Given the satellite is to be transferred to its apogee. It can be done if the semi major axis, that is by increasing the velocity. For this energy is to be provided to the orbit.
Let radius at perigee be ‘r’ and radius at apogee be ‘R’.
Given Speed at perigee of elliptical orbit$= \sqrt {\dfrac{{6GM}}{{5r}}}$

Step II:
Orbital velocity is the minimum velocity required to keep the satellite moving. It is the balance between the gravitational pull of the satellite and inertia of satellite motion. Angular momentum is produced due to rotation of satellite. The momentum can not be destroyed or created, so by law of conservation of angular momentum
Velocity at perigee,${v_p} = \sqrt {\dfrac{{6GM}}{{5r}}}$
Total energy of the elliptical orbit at apogee is given by=$- \dfrac{{GMm}}{{2a}}$where 2a is semi major axis
Total Energy of Ellipse = Kinetic Energy + Potential Energy
$- \dfrac{{GMm}}{{2a}} = \dfrac{1}{2}m{v_p}^2 - \dfrac{{GMm}}{r}$
Substituting value of velocity, and solving
$- \dfrac{{GM}}{{2a}} = \dfrac{1}{2}\dfrac{{6GM}}{{5r}} - \dfrac{{GM}}{r}$
$- GM[\dfrac{3}{{5r}} - \dfrac{1}{r}] = - GM[\dfrac{1}{{2a}}]$
$\dfrac{3}{{5r}} - \dfrac{1}{r} = \dfrac{1}{{2a}}$
$a = \dfrac{{5r}}{4}$

Step III:
Radius of the axis is given by
$2a = {r_a} + r$
${r_a} = 2a - r$
${r_a} = 2\dfrac{{5r}}{4} - r$
${r_a} = \dfrac{{3r}}{2}$

Step IV:
By law of conservation of angular momentum at perigee
$m{v_p}r = m{v_a}r$
${v_p}r = {v_a}r$
Substitute value of r and solve
${v_a} = \dfrac{2}{3}{v_p}$

Step V:
Velocity of the new circular orbit at apogee is ${v_o} = \sqrt {\dfrac{{GM}}{r}}$
Substitute value of r velocity is ${v_o} = \sqrt {\dfrac{{2GM}}{3}}$
Velocity of orbit at perigee is ${v_a} = \dfrac{2}{3}\sqrt {\dfrac{{6GM}}{{5r}}}$

Step VI:
$\Delta v = {v_a} - {v_o}$
$\Delta v = \sqrt {\dfrac{{6GM}}{{5r}}} - \sqrt {\dfrac{{GM}}{r}}$
$\Delta v = 0.095\sqrt {\dfrac{{GM}}{r}}$
Substitute value of r and solve
$\Delta v = 0.095\sqrt {\dfrac{{2GM}}{{3r}}}$
$\Delta v = 0.075\sqrt {\dfrac{{GM}}{r}}$

Step VII:
Option B is the correct answer.

Note: When an object or body moves around other in an elliptical path then angular momentum is produced. But if there is no external force or torque acting on the body, then angular momentum will not change. The total angular momentum is however free of external torque and is constant.