Question

A satellite of mass m is orbiting the earth (radius R) at a height h from its surface. The total energy of the satellite in terms of g, the value of acceleration due to gravity at the earth’s surface, is 

1. \(\frac{{mg{R^2}}}{{2\left( {R + h} \right)}}\)
2. \(- \frac{{mg{R^2}}}{{2\left( {R + h} \right)}}\)
3. \(\frac{{2mg{R^2}}}{{R + h}}\)
4. \(- \frac{{2mg{R^2}}}{{\left( {R + h} \right)}}\)

Answer 1

Correct Answer - Option 2 : \(- \frac{{mg{R^2}}}{{2\left( {R + h} \right)}}\)

Concept:

For a satellite orbiting the earth’s surface at a height ‘h’ from the earth’s surface

The potential energy is given by

\(PE = - \frac{{GMm}}{{R + h}}\)

The kinetic energy is given by

\(KE = \frac{1}{2}m{v^2} = \frac{1}{2}m\left( {\frac{{GM}}{{R + h}}} \right) = \frac{{GMm}}{{2\left( {R + h} \right)}}\)

Calculation:

 Given A satellite of mass m is orbiting the earth (radius R) at a height h from its surface,

The total energy is given by

E = KE + PE

\(E = \frac{{GMm}}{{2\left( {R + h} \right)}} - \frac{{GMm}}{{R + h}} = - \frac{{GMm}}{{2\left( {R + h} \right)}}\)

\( \Rightarrow E = - \frac{1}{2}\frac{{GM}}{{{R^2}}}\frac{{m{R^2}}}{{\left( {R + h} \right)}}\) 

\(\Rightarrow E = - \frac{{1mg{R^2}}}{{2\left( {R + h} \right)}}\)