A body of mass m rises to a height 5 R from the surface of earth. If g is the acceleration due to gravity at the surface of earth, the increase in potential energy is (R= radius of earth )

Question

A body of mass m rises to a height 5 R from the surface of earth. If g is the acceleration due to gravity at the surface of earth, the increase in potential energy is (R= radius of earth ) (A) (4/5) m g R (B) (5/6) m g R (C) (6/7) m g R (D) m g R

Tardigrade Admin · Accepted Answer

Gravitational potential energy at any point at a distance

r

from the centre of the earth is

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

where

M

and

m

are the masses of earth and body respectively. At the surface of the earth,

r = R ∴ U_{1} = - \frac{GM m}{R}

At a height

h

from surface,

r = R + h = R + 5 R = 6 R

∴ U_{2} = - \frac{GM m}{6 R}

Increase in potential energy is

Δ U = U_{2} - U_{1} = - \frac{GM m}{6 R} - (- \frac{GM m}{R})

= \frac{5 GM m}{6 R}

= \frac{5}{6} m g R

(∵ g = \frac{GM}{R ^{2}})

Q. A body of mass $m$ rises to a height $5 R$ from the surface of earth. If $g$ is the acceleration due to gravity at the surface of earth, the increase in potential energy is $(R =$ radius of earth $)$

$\frac{4}{5} m g R$

$\frac{5}{6} m g R$

$\frac{6}{7} m g R$

$m g R$

Q. A body of mass m rises to a height 5R from the surface of earth. If g is the acceleration due to gravity at the surface of earth, the increase in potential energy is (R= radius of earth )

54​mgR

65​mgR

76​mgR

mgR

Q. A body of mass $m$ rises to a height $5 R$ from the surface of earth. If $g$ is the acceleration due to gravity at the surface of earth, the increase in potential energy is $(R =$ radius of earth $)$

$\frac{4}{5} m g R$

$\frac{5}{6} m g R$

$\frac{6}{7} m g R$

$m g R$