A body of mass m is raised from the surface of the earth to a height nR ….

Q: A body of mass m  is raised from the surface of the earth to a height nR(R = radius of earth ). Magnitude of the change in the gravitational potential energy of the body is (g =acceleration due to gravity on the surface of earth)

(a) $ \displaystyle (\frac{n}{n+1} ) mgR $

(b)$ \displaystyle (\frac{n-1}{n} ) mgR $

(c) $ \displaystyle \frac{mgR}{n} $

(d) $ \displaystyle \frac{mgR}{n-1} $

Ans: (a)

Sol:
$ \displaystyle U_1 = -\frac{G M m}{R} $

$ \displaystyle U_2 = -\frac{G M m}{R+nR} $

$ \displaystyle U_2 – U_1 = \frac{G M m}{R}(1-\frac{1}{n+1}) $

$ \displaystyle = \frac{G M m}{R} \frac{n}{n+1} $

$ \displaystyle = \frac{g R^2 m}{R} \frac{n}{n+1} $

$\displaystyle = m g R \frac{n}{n+1} $