Question

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.

• A. $\frac{mgR}{2}$
• B. $2mgR$
• C. $mgR$
• D. $\frac{mgR}{4}$

Answer 1

Answer: (A)

$\frac{mgR}{2}$

Gravitational petential energy at r, $P.E=-\frac{GMm}{r}$

As $GM=g{R}^2⟹P.E=-\frac{mg{R}^2}{r}$

For $r=R$, $P.E=-mgR$

For $r=2R$, $P.E^{\prime }=-\frac{mgR}{2}$

Gain in P.E= $P.E^{\prime }-P.E=\frac{mgR}{2}$