An object is taken to height 2R above the surface of earth, the increase in potential energy is [R is radius of earth]
2 mgR
$\frac{m g R}{2}$
$\frac{m g R}{3}$
$\frac{2 m g R}{3}$

Question

An object is taken to height 2R above the surface of earth- the increase in potential energy is -R is radius of earth-2 mgRdfrac-mgR-2-dfrac-mgR-3-dfrac-2mgR-3-

Toppr Admin · Accepted Answer

Change in gravitational potential energy at a height h above the earth-apos-s surface is -x25B3-U-mgh1-hr-mg-2R1-2RR -here-h-2R-2mgR3

An object is taken to height 2R above the surface of earth, the increase in potential energy is [R is radius of earth]2 mgRmgR2mgR32mgR3

Change in gravitational potential energy at a height h above the earth's surface is △U=mgh1+(hr)=mg.2R1+2RR [here,h=2R]=2mgR3

An object is taken to height 2R above the surface of earth, the increase in potential energy is [R is radius of earth]
2 mgR
$\frac{m g R}{2}$
$\frac{m g R}{3}$
$\frac{2 m g R}{3}$

Change in gravitational potential energy at a height h above the earth's surface is $△ U = \frac{m g h}{1 + (\frac{h}{r})}$
$= \frac{m g .2 R}{1 + \frac{2 R}{R}}$ [here,h=2R]
= $\frac{2 m g R}{3}$