Question

A satellite is revolving in a circular orbit at a height h from the earth’s surface, such that h << R, where R is the radius of the earth. Assuming that the effect of earth’s atmosphere can be neglected, what is the minimum increase in speed required so that the satellite could escape from the gravitational field of earth?
(A) $\sqrt {gR} \left( {\sqrt 2 - 1} \right)$
(B) $\sqrt {2gR}$
(C) $\sqrt {gR}$
(D) $\sqrt {\dfrac{{gR}}{2}}$

Answer 1

In this problem,we are going to apply the concept of escape velocity and orbital velocity. The minimum speed required by a body to escape the gravitational field of the earth is $\sqrt {2gR}$ and the speed required to orbit the earth is $\sqrt {gR}$.

Complete step by step answer:
The satellite is revolving in a circular orbit around the earth at a height of h, such that h << R. Since h is negligible compared to the radius of the earth, it will not affect the acceleration of the satellite due to gravity. So, the acceleration due to gravity at a height of h will remain g.
Since the satellite is orbiting earth in a circular orbit, it already has some velocity. That velocity is the orbital velocity. This velocity is given by the formula,
${V_o} = \sqrt {\dfrac{{GM}}{R}} = \sqrt {gR}$, since $g = \dfrac{{GM}}{{{R^2}}}$
To escape the gravitational field of the earth, the minimum speed required is known as the escape velocity, which is given by the formula,
${V_e} = \sqrt {\dfrac{{2GM}}{R}} = \sqrt {2gR}$
The minimum increase in speed required by the satellite to escape earth’s gravitational field is to increase its speed to the minimum escape velocity. This minimum increase in speed can be found out by finding out the difference between the orbital velocity and the escape velocity, which is,
$v = {V_e} - {V_o} = \sqrt {2gR} - \sqrt {gR} = \sqrt {gR} \left( {\sqrt 2 - 1} \right)$

Hence, the correct answer is option A.

Additional Information:
The escape velocity for an object to leave the earth’s gravitational field is 11.19 kms$^{ - 1}$ or 11.2 kms$^{ - 1}$. The orbital velocity around earth depends on the altitude of the satellite. For geostationary satellites at a height of 35786 km, the orbital velocity is 3.07 kms$^{ - 1}$.

Note: If the height h is comparable to the radius of earth, R, then the expression,
$g = \dfrac{{GM}}{{{R^2}}}$ cannot be used in the values of orbital and escape velocity. This substitution is only applicable in places where the change in the force of gravitation with change in height is not appreciable.