A body of mass 2m is placed on earth's surface. Calculate the change in gravitational potential energy, if this body is taken from earth's surface to a height of h, where h = 4R.

$\frac{2 mgh}{R}$

$\frac{2}{3} mgR$

$\frac{8}{5} mgR$

$\frac{mgR}{2}$

Solution

$\frac{8}{5} mgR$

Potential energy,

U = $- \frac{G Mm}{r}$

= $\frac{- 2 G Mm}{r}$ [ $∵$ m = 2m]

At earth's surface, r = R

∴ U_e = $- \frac{2 GMm}{R}$

Now, if a body is taken to height, h = 4R, then the potential energy is given by

U_h = $- \frac{2 GMm}{R + h}$ [ r = h + R ]

= $- \frac{2 GMm}{5 R}$

Thus, change in gravitational potential energy,

ΔU = U_h $-$ U_e

= $- \frac{2 GMm}{5 R} + \frac{2 GMm}{R}$

= $\frac{8}{5} \frac{GMm}{R}$

ΔU = $\frac{8}{5} \frac{{gR}^{2} m}{R}$ [ $∵$ Gm = gR² ]

ΔU = $\frac{8}{5} mgR$

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