Question

The additional kinetic energy to be provided to a satellite of mass $m$ revolving around a planet of mass $M$, to transfer it from a circular orbit of radius ${R_1}$ to another of radius ${R_2}$(${R_2} > {R_1}$) is
(A) $GMm\left( {\dfrac{1}{{{R_1}^2}} - \dfrac{1}{{{R_2}^2}}} \right)$
(B) $GMm\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
(C) $2GMm\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$
(D) $\dfrac{{GMm}}{2}\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$

Answer 1

The kinetic energy of a satellite is the negative of half its potential energy. Since energy is inputted into the system, the final energy is larger than the initial energy.
Formula used: In this solution we will be using the following formulae;
$KE = \dfrac{{GMm}}{{2R}}$ where $KE$ is the kinetic energy of a satellite or any object orbiting around a gravitational field, $G$ is the universal gravitational constant, $M$ is the mass of the gravitational body, $m$ is the mass of the satellite, and $R$ is the radius of the orbit.

Complete Step-by-Step Solution:
Generally an object orbiting a gravitational field has a kinetic energy that is given by
$KE = \dfrac{{GMm}}{{2R}}$ where $KE$ is the kinetic energy of a satellite or any object orbiting around a gravitational field, $G$ is the universal gravitational constant, $M$ is the mass of the gravitational body, $m$ is the mass of the satellite, and $R$ is the radius of the orbit.
Hence, for the case of a change in kinetic energy, which allows the satellite to move from one orbital radius ${R_1}$ to ${R_2}$ can be given as
$KE = - \left( {\dfrac{{GMm}}{{2{R_2}}} - \dfrac{{GMm}}{{2{R_1}}}} \right)$ which can be given as
$KE = \dfrac{{GMm}}{{2{R_1}}} - \dfrac{{GMm}}{{2{R_2}}}$
Hence by factorisation of all common variables, we have
$KE = \dfrac{{GMm}}{2}\left( {\dfrac{1}{{{R_1}}} - \dfrac{1}{{{R_2}}}} \right)$

Hence, the correct option is D

Note: For clarity, although the kinetic energy itself is regarded as a positive quantity, due to the inverse relation with the radius the negative is placed behind it for change in kinetic energy so that an increase in radius will result in an increase in kinetic energy as it really is.