Assuming the radius of the earth as R, the change in gravitational potential energy of a body of mass m, when it is taken from the earth's surface to a height 3 R above its surface, is

Question

Assuming the radius of the earth as R, the change in gravitational potential energy of a body of mass m, when it is taken from the earth's surface to a height 3 R above its surface, is (A) 3 mgR (B) (3/4) m g R (C) 1 m g R (D) (3/2) m g R

Tardigrade Admin · Accepted Answer

Gravitational potential energy (GPE) on the surface of earth,

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

GPE at

3 R, E_{2} = - \frac{GM m}{( R + 3 R )} = - \frac{GM m}{4 R}

∴

Change in GPE

= E_{2} - E_{1} = - \frac{GM m}{4 R} + \frac{GM m}{R} = \frac{3 GM m}{4 R}

= \frac{3 g R ^{2} m}{4 R} (∵ g = \frac{GM}{R ^{2}}) = \frac{3}{4} m g R

Q. Assuming the radius of the earth as $R$ , the change in gravitational potential energy of a body of mass $m$ , when it is taken from the earth's surface to a height $3 R$ above its surface, is

$3 m g R$

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

$1 m g R$

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

Q. Assuming the radius of the earth as R, the change in gravitational potential energy of a body of mass m, when it is taken from the earth's surface to a height 3R above its surface, is

3mgR

43​mgR

1mgR

23​mgR

Gravitational potential energy (GPE) on the surface of earth, E1​=−RGMm​ GPE at 3R,E2​=−(R+3R)GMm​=−4RGMm​ ∴ Change in GPE =E2​−E1​=−4RGMm​+RGMm​=4R3GMm​ =4R3gR2m​(∵g=R2GM​)=43​mgR

Q. Assuming the radius of the earth as $R$ , the change in gravitational potential energy of a body of mass $m$ , when it is taken from the earth's surface to a height $3 R$ above its surface, is

$3 m g R$

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

$1 m g R$

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