A mass ‘m’ on the surface of the Earth is shifted to a target equal to the radius of the Earth. If ‘R’ is the radius and ‘M’ is the mass of the Earth, then work done in this process is

Question

A mass ‘m’ on the surface of the Earth is shifted to a target equal to the radius of the Earth. If ‘R’ is the radius and ‘M’ is the mass of the Earth, then work done in this process is (A) (mgR/2) (B) mgR (C) 2 mgR (D) (mgR/4)

Tardigrade Admin · Accepted Answer

UE = (-GMm/R) UT = (- GMm/2R) W = UT - UE = (-GMm/2R) + (GMm/R) = (GMm/2R) = (gR2m/2R) = (mgR/2)

Q. A mass ‘m’ on the surface of the Earth is shifted to a target equal to the radius of the Earth. If ‘R’ is the radius and ‘M’ is the mass of the Earth, then work done in this process is

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

$m g R$

$2 m g R$

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

$U_{E} = \frac{- GM m}{R}$
$U_{T} = \frac{- GM m}{2 R}$
$W = U_{T} - U_{E} = \frac{- GM m}{2 R} + \frac{GM m}{R}$
$= \frac{GM m}{2 R} = \frac{g R ^{2} m}{2 R} = \frac{m g R}{2}$

Q. A mass ‘m’ on the surface of the Earth is shifted to a target equal to the radius of the Earth. If ‘R’ is the radius and ‘M’ is the mass of the Earth, then work done in this process is

2mgR​

mgR

2mgR

4mgR​

UE​=R−GMm​ UT​=2R−GMm​ W=UT​−UE​=2R−GMm​+RGMm​ =2RGMm​=2RgR2m​=2mgR​

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

$m g R$

$2 m g R$

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

$U_{E} = \frac{- GM m}{R}$
$U_{T} = \frac{- GM m}{2 R}$
$W = U_{T} - U_{E} = \frac{- GM m}{2 R} + \frac{GM m}{R}$
$= \frac{GM m}{2 R} = \frac{g R ^{2} m}{2 R} = \frac{m g R}{2}$