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. $\frac{mg_0 R^2}{2(R - h)}$
• B. $-\frac{mg_0 R^2}{2(R + h)}$
• C. $\frac{2mg_0 R^2}{2(R + h)}$
• D. $- \frac{2 mg_0 R^2}{(R + h)}$

Answer 1

The Correct Option is B

To determine the total energy of a satellite orbiting the Earth, we need to consider both its potential energy and kinetic energy. The satellite of mass $m$ is orbiting at a height $h$ from the Earth's surface, where the Earth's radius is $R$.

The gravitational potential energy (U) of the satellite at a height $h$ is given by:

U = -\frac{G M m}{R + h}

Here, $G$ is the gravitational constant, and $M$ is the mass of the Earth. Since g_0 represents the gravitational acceleration at the Earth's surface, we use the relation:

g_0 = \frac{G M}{R^2}

We can express GM from this equation:

G M = g_0 R^2

Substituting G M in the potential energy formula:

U = -\frac{g_0 R^2 m}{R + h}

The kinetic energy (K) of the satellite in orbit is equal to half the magnitude of the potential energy (by the energy conservation in orbital mechanics):

K = \frac{1}{2} \left| U \right| = \frac{1}{2} \frac{g_0 R^2 m}{R + h}

The total energy (E) of the satellite is the sum of its kinetic and potential energies:

E = K + U

Substituting the values of K and U:

E = \frac{1}{2} \frac{g_0 R^2 m}{R + h} - \frac{g_0 R^2 m}{R + h} = -\frac{1}{2} \frac{g_0 R^2 m}{R + h}

Thus, the total energy of the satellite is:

-\frac{mg_0 R^2}{2(R + h)}

This matches the correct option: -\frac{mg_0 R^2}{2(R + h)}.