If g is the acceleration due to gravity on the surface of the earth- the gain in potential energy of an object of mass m raised from the earth-s surface to a height equal to the radius R of the earth iscfrac-mgR-4-cfrac-mgR-2-mgR2mgR

Question

Toppr Admin · Accepted Answer

The object is at earth surface initially.
Initial potential energy, $P . E_{i} = \frac{- G M m}{R}$ where, $G M = g R^{2}$
$\Rightarrow$ $P . E_{i} = - m g R$

Final Potential energy, $P . E_{f} = \frac{- G M m}{2 R} = - \frac{m g R}{2}$

$∴$ Change in potential energy, $Δ P . E = - \frac{m g R}{2} - (- m g R) = \frac{m g R}{2}$

If g is the acceleration due to gravity on the surface of the earth, the gain in potential energy of an object of mass m raised from the earth's surface to a height equal to the radius R of the earth ismgR4mgR2mgR2mgR

The object is at earth surface initially.Initial potential energy, P.Ei=−GMmR where, GM=gR2⇒ P.Ei=−mgRFinal Potential energy, P.Ef=−GMm2R=−mgR2∴ Change in potential energy, ΔP.E=−mgR2−(−mgR)=mgR2

If $g$ is the acceleration due to gravity on the surface of the earth, the gain in potential energy of an object of mass $m$ raised from the earth's surface to a height equal to the radius $R$ of the earth is
$\frac{m g R}{4}$
$\frac{m g R}{2}$
$m g R$
$2 m g R$