Question

What will be gained in potential energy of a body of mass $'m'$ at a height equal to three times the radius $'R'$ of the earth?
(A) $mgR$
(B) $2mgR$
(C) $\dfrac{{mgR}}{3}$
(D) $\dfrac{{3mgR}}{4}$

Answer 1

In order to answer this question, first we will rewrite the given facts in the question, and then we will apply the formula for the gain or change in potential energy in terms of mass, height and the radius of the earth, i.e.. $\therefore \vartriangle U = \dfrac{{mgh}}{{1 + \dfrac{h}{R}}}$ .

Complete step by step solution:
Mass of a body $= m$
Radius of the Earth $= R$
According to the question, height is equal to three times of the radius $'R'$ :
$h = 3R$
Now, the change in potential energy in terms of mass, height and the radius of the earth will be:
$\therefore \vartriangle U = \dfrac{{mgh}}{{1 + \dfrac{h}{R}}}$
As we mentioned above- $h = 3R$
We will put $3R$ instead of $h$ in the above equation:-
$\Rightarrow \vartriangle U = \dfrac{{mg(3R)}}{{1 + \dfrac{{3R}}{R}}} \\\ \therefore \vartriangle U = \dfrac{{3mgR}}{4} \\\$
Therefore, $\dfrac{{3mgR}}{4}$ will be gained in potential energy of a body of mass $'m'$ at a height equal to three times the radius $'R'$ of the earth.
Hence, the correct option is (D) $\dfrac{{3mgR}}{4}$ .

Note: When two like charges (two protons or two electrons) are brought together, the system's potential energy rises. When two opposite charges, such as a proton and an electron, are brought together, the system's electric potential energy diminishes.