A body of mass m is placed on earth surface which is taken from earth surface to a height of h = 3R, then change in gravitational potential energy is

Question

A body of mass m is placed on earth surface which is taken from earth surface to a height of h = 3R, then change in gravitational potential energy is (A) (4/3) mgR  (B) (2/3) mgR  (C) (3/4) mgR  (D) (mgR/ 2)

Tardigrade Admin · Accepted Answer

Gravitational potential energy on earth's surface

= - \frac{GM m}{R}

, where

M

and

R

are the mass and radius of the earth respectively,

m

is the mass of the body and

G

is the universal gravitational constant.
Gravitational potential energy at a height

h = 3 R

= - \frac{GM m}{R + h} = - \frac{GM m}{R + 3 R} = - \frac{GM m}{4 R}

∴

Change in potential energy

= - \frac{GM m}{4 R} - (- \frac{GM m}{R})

= - \frac{GM m}{4 R} + \frac{GM m}{R} = \frac{3}{4} \frac{GM m}{R}

Again, we have,

\frac{GM m}{R ^{2}} = m g

(where

g

is acceleration due to gravity on earth's surface).

∴ \frac{GM m}{R} = m g R

.

∴

Change in potential energy

= \frac{3}{4} m g R

.

Q. A body of mass $m$ is placed on earth surface which is taken from earth surface to a height of $h = 3 R$ , then change in gravitational potential energy is

$\frac{4}{3} m g R$

$\frac{2}{3} m g R$

$\frac{3}{4} m g R$

$\frac{m g R}{2}$

Q. A body of mass m is placed on earth surface which is taken from earth surface to a height of h=3R, then change in gravitational potential energy is

34​mgR

32​mgR

43​mgR

2mgR​

Q. A body of mass $m$ is placed on earth surface which is taken from earth surface to a height of $h = 3 R$ , then change in gravitational potential energy is

$\frac{4}{3} m g R$

$\frac{2}{3} m g R$

$\frac{3}{4} m g R$

$\frac{m g R}{2}$