If g is the acceleration due to gravity on the earth's surface, the gain in potential energy of the body at a height equal to three times the radius R of the earth will be

Question

If g is the acceleration due to gravity on the earth's surface, the gain in potential energy of the body at a height equal to three times the radius R of the earth will be (A) mgR (B) (1/2)mgR (C) (1/3)mgR (D) (3/4)mgR

Tardigrade Admin · Accepted Answer

The gravitational potential energy at any point at a distance

x

from the centre of earth is

E = - \frac{GM m}{x}

On the surface of earth

x = R

,
so,

E_{1} = \frac{- GM m}{R} = - m g R

At a height

3 R

, from the surface of earth,

x = 4 R

So,

E_{2} = \frac{- GM m}{4 R} = - \frac{m g R}{4}

Increase in potential energy,

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

Q. If $g$ is the acceleration due to gravity on the earth's surface, the gain in potential energy of the body at a height equal to three times the radius $R$ of the earth will be

$m g R$

$\frac{1}{2} m g R$

$\frac{1}{3} m g R$

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

Q. If g is the acceleration due to gravity on the earth's surface, the gain in potential energy of the body at a height equal to three times the radius R of the earth will be

mgR

21​mgR

31​mgR

43​mgR

The gravitational potential energy at any point at a distance x from the centre of earth is E=−xGMm​ On the surface of earth x=R, so, E1​=R−GMm​=−mgR At a height 3R, from the surface of earth, x=4R So, E2​=4R−GMm​=−4mgR​ Increase in potential energy, =−4mgR​+mgR=43​mgR

Q. If $g$ is the acceleration due to gravity on the earth's surface, the gain in potential energy of the body at a height equal to three times the radius $R$ of the earth will be

$m g R$

$\frac{1}{2} m g R$

$\frac{1}{3} m g R$

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