Question

A satellite of mass $m$ is orbiting the Earth (of radius $R$) at a height $h$ from its surface. The total energy of the satellite in terms of ${{g}_{0}}$, the value of acceleration due to gravity at the earth’s surface is:
(A). $\dfrac{2m{{g}_{0}}{{R}^{2}}}{R+h}$
(B). $-\dfrac{2m{{g}_{0}}{{R}^{2}}}{R+h}$
(C). $\dfrac{m{{g}_{0}}{{R}^{2}}}{2(R+h)}$
(D). $-\dfrac{m{{g}_{0}}{{R}^{2}}}{2(R+h)}$

Answer 1

The total energy of a satellite is the sum of its potential energy and kinetic energy. The potential energy of a system due to Earth’s gravitational pull is always negative. The acceleration due to gravity is the constant acceleration acting on a body above the Earth’s surface.

Formulas used:
$E=-\dfrac{GMm}{2(R+h)}$
${{g}_{0}}=\dfrac{GM}{{{R}^{2}}}$

Complete step by step solution:
When a satellite is orbiting the Earth’s surface, it possesses two energies; kinetic energy due to its motion and potential energy due to the gravitational pull of the Earth. The total energy of a satellite orbiting around the Earth will be the sum of its potential energy and kinetic energy.
The total energy possessed by the satellite is-
$E=-\dfrac{GMm}{2(R+h)}$ ………………………. (1)
Here, $E$ is the total energy of the satellite
$G$ is the gravitational constant
$M$ is the mass of the Earth
$m$ is the mass of the satellite
$R$ is the radius of the Earth
$h$ is the height of satellite from the Earth’s surface
Acceleration due to gravity is the constant acceleration acting on a freely falling object near the surface of the Earth.
It is given by-
${{g}_{0}}=\dfrac{GM}{{{R}^{2}}}$ ………………... (2)
Here, ${{g}_{0}}$is the acceleration due to gravity
From eq (2), we get,
${{g}_{0}}{{R}^{2}}=GM$
When we substitute it in eq (1), we get,
$E=-\dfrac{{{g}_{0}}{{R}^{2}}m}{2(R+h)}$
Therefore, the total energy of a satellite orbiting the Earth at a height $h$ is $E=-\dfrac{{{g}_{0}}{{R}^{2}}m}{2(R+h)}$.

Hence, the correct option is (D).

Note:
The value of gravitational constant is $6.7\times {{10}^{-11}}N\,{{m}^{2}}\,k{{g}^{-2}}$. It is the constant of proportionality involved in the calculations of Newton’s law of gravitation. The satellite orbits in an elliptical orbit around the Earth. The distance of a satellite from the Earth is measured from the centre of the Earth.