If g is the acceleration due to gravity at the Earth's surface, the gain of the potential energy of an object of mass m raised from the surface of the Earth to the height equal to the radius R of the Earth is

Question

If g is the acceleration due to gravity at the Earth's surface, the gain of the potential energy of an object of mass m raised from the surface of the Earth to the height equal to the radius R of the Earth is (A) (m g R/4) (B) (m g R/2) (C) mgR (D) 2 mgR

Tardigrade Admin · Accepted Answer

Potential energy of an object at the surface of the Earth is

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

..... (i)
Potential energy of the object at a height,

h = R

from the surface of the Earth.

U_{2} = - \frac{GM m}{R + h} = - \frac{GM m}{R + R}

Gain in potential energy

Δ U = U_{2} - U_{1} = (- \frac{GM m}{2 R}) - (- \frac{GM m}{R}) = \frac{1}{2} \frac{GM m}{R}

Also,

GM = g R^{2}

Δ U = \frac{1}{2} \frac{g R ^{2} m}{R} = \frac{m g R}{2}

Q. If $g$ is the acceleration due to gravity at the Earth's surface, the gain of the potential energy of an object of mass $m$ raised from the surface of the Earth to the height equal to the radius $R$ of the Earth is

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

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

$m g R$

$2 m g R$

Q. If g is the acceleration due to gravity at the Earth's surface, the gain of the potential energy of an object of mass m raised from the surface of the Earth to the height equal to the radius R of the Earth is

4mgR​

2mgR​

mgR

2mgR

Q. If $g$ is the acceleration due to gravity at the Earth's surface, the gain of the potential energy of an object of mass $m$ raised from the surface of the Earth to the height equal to the radius $R$ of the Earth is

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

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

$m g R$

$2 m g R$