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

Gravitational Potential Energy and Gravitational Field

We are working with the gravitational potential energy and gravitational field for a mass \(m\) at a distance \(r\) from the center of a planet with mass \(M\). 

Step 1: Gravitational Potential Energy

The gravitational potential energy \( T.E \) of a mass \( m \) in a gravitational field is given by the formula:

\(T.E = - \frac{GMm}{2r}\)

Where: - \( G \) is the gravitational constant, - \( M \) is the mass of the planet, - \( m \) is the mass of the object, - \( r \) is the distance from the center of the planet to the object.

Step 2: Gravitational Field at the Surface

The gravitational field strength \( g_0 \) at the surface of the planet is given by the formula:

\(g_0 = g_{surface} = \frac{GM}{R^2}\)

Where \( R \) is the radius of the planet, and \( g_0 \) is the gravitational field strength at the surface.

Step 3: Relating \( GM \) and \( g_0 \)

Rearranging the equation for \( g_0 \), we get the relationship:

\(GM = g_0 R^2\)

Step 4: Gravitational Potential Energy for Height \( h \)

If the object is at a height \( h \) above the surface, the gravitational potential energy \( T.E \) becomes:

\(T.E = - \frac{g_0 R^2 m}{2 (R + h)}\)

Where: - \( R \) is the radius of the planet, - \( h \) is the height above the surface of the planet, - \( m \) is the mass of the object.

Conclusion:

The gravitational potential energy at a height \( h \) above the surface is given by:

\(T.E = - \frac{g_0 R^2 m}{2 (R + h)}\)