The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the earth is (here R is the radius of the earth)

Question

The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the earth is (here R is the radius of the earth) (A) ((n/n+1))mg R (B) ((n/n-1))mg R (C) nmgR (D) (mgR/n)

Tardigrade Admin · Accepted Answer

Gravitational potential energy of mass m at any point at a distance r from the centre of earth is

U = - \frac{GM m}{r}

At the surface of earth r = R,

∴ U_{s} = - \frac{GM m}{R} = - m g R (∵ g = \frac{GM}{R ^{2}})

At the height h = nR from the surface of earth
r = R + h = R + nR = R(1 + n)

∴ U_{h} = - \frac{GM m}{R ( 1 + n )} = - \frac{m g R}{( 1 + n )}

Change in gravitational potential energy is

Δ U = U_{h} - U_{s} = - \frac{m g R}{( 1 + n )} - (- m g R)

= - \frac{m g R}{1 + n} + m g R

= m g R (1 - \frac{1}{1 + n}) = m g R (\frac{n}{1 + n})

Q. The change in the gravitational potential energy when a body of mass $m$ is raised to a height $n R$ above the surface of the earth is (here $R$ is the radius of the earth)

$(\frac{n}{n + 1}) m g R$

$(\frac{n}{n - 1}) m g R$

$nm g R$

$\frac{m g R}{n}$

Q. The change in the gravitational potential energy when a body of mass m is raised to a height nR above the surface of the earth is (here R is the radius of the earth)

(n+1n​)mgR

(n−1n​)mgR

nmgR

nmgR​

Q. The change in the gravitational potential energy when a body of mass $m$ is raised to a height $n R$ above the surface of the earth is (here $R$ is the radius of the earth)

$(\frac{n}{n + 1}) m g R$

$(\frac{n}{n - 1}) m g R$

$nm g R$

$\frac{m g R}{n}$