The radius of the earth is R and acceleration due to gravity its surface is -g- If a body of mass -m- is sent to a height of R-4 from the earth-s surface- the potential energy is-mgR-3mgR-44mgR-5mgR-16

Question

Toppr Admin · Accepted Answer

We know that-g-GMR2u-GMm-R-R4-x21D2-u-4GMm5Ru-4GMm5R-x21D2-u-4mgR5Then-Option C is correct answer-

The radius of the earth is $R$ and acceleration due to gravity at its surface is ' $g$ '. If a body of mass ' $m$ ' is sent to a height of $R / 4$ from the earth's surface, the potential energy is:
$m g R / 3$
$m g R / 4$
$4 m g R / 5$
$m g R / 16$

We know that,

$\begin{matrix} g = \frac{G M}{R^{2}} u = \frac{G M m}{(R + \frac{R}{4})} \Rightarrow u = \frac{4 G M m}{5 R} u = \frac{4 G M m}{5 R} \end{matrix}$

$\Rightarrow u = \frac{4 m g R}{5}$
Then,
Option $C$ is correct answer.

The radius of the earth is R and acceleration due to gravity at its surface is 'g'. If a body of mass 'm' is sent to a height of R/4 from the earth's surface, the potential energy is:mgR/3mgR/44mgR/5mgR/16

We know that,g=GMR2u=GMm(R+R4)⇒u=4GMm5Ru=4GMm5R⇒u=4mgR5Then,Option C is correct answer.

The radius of the earth is $R$ and acceleration due to gravity at its surface is ' $g$ '. If a body of mass ' $m$ ' is sent to a height of $R / 4$ from the earth's surface, the potential energy is:
$m g R / 3$
$m g R / 4$
$4 m g R / 5$
$m g R / 16$

We know that,

$\begin{matrix} g = \frac{G M}{R^{2}} u = \frac{G M m}{(R + \frac{R}{4})} \Rightarrow u = \frac{4 G M m}{5 R} u = \frac{4 G M m}{5 R} \end{matrix}$

$\Rightarrow u = \frac{4 m g R}{5}$
Then,
Option $C$ is correct answer.