A body of mass m is placed on the earth's surface. It is taken from the earth's surface to a height h=3R when R is the radius of the earth. The change in gravitational potential energy of the body is

Question

A body of mass m is placed on the earth's surface. It is taken from the earth's surface to a height h=3R when R is the radius of the earth. The change in gravitational potential energy of the body is (A) (3/2)mgR (B) (3/4)mgR (C) (1/2)mgR (D) (1/4)mgR

Tardigrade Admin · Accepted Answer

PE at the surface of earth PE=(- G M m/R) PE at a height 3Re PE=(- G M m/R + 3 R)=(- G Mm/4 R) Δ PE=PEf-PEi=(- (G M m/4 R) )-((- G Mm/R) )=(3/4) (G Mm/R) ⇒ Δ PE=(3/4) mgR

Q. A body of mass $m$ is placed on the earth's surface. It is taken from the earth's surface to a height $h = 3 R$ when $R$ is the radius of the earth. The change in gravitational potential energy of the body is

$\frac{3}{2} m g R$

$\frac{3}{4} m g R$

$\frac{1}{2} m g R$

$\frac{1}{4} m g R$

PE at the surface of earth
$PE = \frac{- GM m}{R}$
PE at a height $3 R_{e}$
$PE = \frac{- GM m}{R + 3 R} = \frac{- G M _{m}}{4 R}$
$Δ PE = P E_{f} - P E_{i} = (- \frac{GM m}{4 R}) - (\frac{- G M _{m}}{R}) = \frac{3}{4} \frac{G M _{m}}{R}$
$\Rightarrow Δ PE = \frac{3}{4} m g R$

Q. A body of mass m is placed on the earth's surface. It is taken from the earth's surface to a height h=3R when R is the radius of the earth. The change in gravitational potential energy of the body is

23​mgR

43​mgR

21​mgR

41​mgR