A body of mass m is taken from earth surface to a height h equal to twice the radius of earth, the increase in potential energy will be ( g = acceleration due to gravity on earth's surface) (R -

Question

A body of mass m is taken from earth surface to a height h equal to twice the radius of earth, the increase in potential energy will be ( g = acceleration due to gravity on earth's surface) (R - radius of earth)

TestHub · Accepted Answer

Initial potential energy of the body on the surface of the earth is $U_{i} = - \frac{GM m}{R}$

Final potential energy of the body at height $h = 2 R$ is $U_{f} = - \frac{GM m}{R + h} = - \frac{GM m}{R + 2 R} = - \frac{GM m}{3 R}$

Increase in potential energy is $Δ U = U_{f} - U_{i} = - \frac{GM m}{3 R} - (- \frac{GM m}{R}) = \frac{2 GM m}{3 R}$

Using the relation $g = \frac{GM}{R ^{2}}$ , we have $GM = g R^{2}$

Substituting this in the expression for $Δ U$ , we get $Δ U = \frac{2 ( g R ^{2} ) m}{3 R} = \frac{2}{3} m g R$

Answer: $\frac{2}{3} mgR$

Physics - Gravitation Question with Solution | TestHub

Options:

Doubts & Discussion