Question

If g is the acceleration due to gravity on earth’s surface, what will be the gain in the potential energy of an object of mass m raised from the surface of the earth to a height equal to the radius R of the earth?

1. ½ mgR
2. 2 mgR
3. ¼ mgR
4. mgR

Answer 1

Correct Answer - Option 1 : ½ mgR

The gravitational potential energy of the two particles system is given by

\(U\left( r \right)\; = \; - \frac{{G{m_1}{m_2}}}{r}\)

Where r is the separation between the two particles having mass equal to m₁ and m₂ respectively.

The change in potential energy as the distance between the particles changes from r₁ to r₂ is given by

\({\rm{\Delta }}U\left( r \right)\; = \;U\left( {{r_2}} \right) - U\left( {{r_1}} \right)\; = \;G{m_1}{m_2}\left( {\frac{1}{{{r_1}}} - \frac{1}{{{r_2}}}} \right)\)

Here gravitational potential energy of the particle at earth surface, \({U_1}\; = \; - \frac{{GmM}}{R}\)

When raised at height ‘h’ from the surface; \({U_2}\; = \; - \frac{{GmM}}{{R + h}}\)

When h = R (as given)

\(\begin{array}{l} {U_2}\; = \; - \frac{{GmM}}{{2R}}\\ \therefore {\rm{\Delta }}U\; = \;{U_2} - {U_1}\; = \; - \frac{{GmM}}{R}\left( {\frac{1}{2} - 1} \right)\; = \;\frac{{GmM}}{{2R}} \end{array}\)