Question

A body of mass m is lifted up from the surface of earth to a height 3 times radius of earth. then find the change in potential energy of body

Answer 1

Initial potential energy=GMm/r^2
final potential energy=GMm/(r+3r)^2
= GMm /(4r)^2
=GMm/16r^2
therefore change in potential energy=
final potential energy - initial potential energy =
GMm /r^2 (1/16-1)
=-15GMm/16r^2
Hope this helps

Answer 2

Given data states that the body of mass m thrown to a height 3 times radius of earth, therefore the potential energy defined in terms of earth as a planet is given by the terms $\frac { GMm }{ R^{ 2 } }$ where G is the Gravitation constant, M is the mass of earth and m is the mass of the body and R is the radius of earth. Therefore, initial potential energy be $\frac { GMm }{ R^{ 2 } }$ and final potential energy be $\frac { GMm }{ (R+r)^{ 2 } }$ where r is the new height where r = 3R. $\Rightarrow$ Change in potential energy = (final - initial) potential energy $=\frac { GMm }{ R^{ 2 } } -\frac { GMm }{ (R+3R)^{ 2 } } =\frac { GMm }{ R^{ 2 } } -\frac { GMm }{ 16R^{ 2 } } =\frac { 15GMm }{ R^{ 2 } } .$