If the gravitational acceleration surface of earth is g- then increase in potential energy in lifting an object of mass m to a height equal to the radius R of earth will be-2mgRdfrac -mgR-2-mgRdfrac -mgR-4-

Question

If the gravitational acceleration surface of earth is  g- then increase in potential energy in lifting an object of mass m to a height equal to the radius R of earth will be-2mgRdfrac -mgR-2-mgRdfrac -mgR-4-

Toppr Admin · Accepted Answer

Gravitational petential energy at r- P-E-x2212-GMmr-xA0-As GM-gR2-x27F9-P-E-x2212-mgR2rFor r-R- -xA0- P-E-x2212-mgRFor r-2R- -xA0- P-E-x2032-x2212-mgR2Gain in P-E- P-E-x2032-x2212-P-E-mgR2

If the gravitational acceleration at surface of earth is $g$ , then increase in potential energy in lifting an object of mass $m$ to a height equal to the radius $R$ of earth will be.
$2 m g R$
$\frac{m g R}{2}$
$m g R$
$\frac{m g R}{4}$

Gravitational petential energy at r, $P . E = - \frac{G M m}{r}$
As $G M = g R^{2} ⟹ P . E = - \frac{m g R^{2}}{r}$
For $r = R$ , $P . E = - m g R$
For $r = 2 R$ , $P . E^{'} = - \frac{m g R}{2}$
Gain in P.E= $P . E^{'} - P . E = \frac{m g R}{2}$

If the gravitational acceleration at surface of earth is g, then increase in potential energy in lifting an object of mass m to a height equal to the radius R of earth will be.2mgRmgR2mgRmgR4

Gravitational petential energy at r, P.E=−GMmr As GM=gR2⟹P.E=−mgR2rFor r=R, P.E=−mgRFor r=2R, P.E′=−mgR2Gain in P.E= P.E′−P.E=mgR2

If the gravitational acceleration at surface of earth is $g$ , then increase in potential energy in lifting an object of mass $m$ to a height equal to the radius $R$ of earth will be.
$2 m g R$
$\frac{m g R}{2}$
$m g R$
$\frac{m g R}{4}$

Gravitational petential energy at r, $P . E = - \frac{G M m}{r}$
As $G M = g R^{2} ⟹ P . E = - \frac{m g R^{2}}{r}$
For $r = R$ , $P . E = - m g R$
For $r = 2 R$ , $P . E^{'} = - \frac{m g R}{2}$
Gain in P.E= $P . E^{'} - P . E = \frac{m g R}{2}$