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) $\left( {\dfrac{n}{{n + 1}}} \right)mgR$
B) $\left( {\dfrac{n}{{n - 1}}} \right)mgR$
C) $nmgR$
D) $\dfrac{{mgR}}{n}$

Answer 1

The question above discuss about Gravitational Potential energy which is mathematically given as:
${U_e} = - \dfrac{{GMm}}{R}$ (G is the gravitational constant, M is the mass of object one and m is the mass of another object, R is the distance between them).
When an object is raised at a height h above the surface of the earth Gravitational Potential becomes:
${U_h} = - \dfrac{{GMm}}{{R + h}}$(Where h is the height to which an object is raised).
Using the above two relations we will conclude the problem.

Complete step by step solution:
First we will define gravitational potential and Gravitational potential energy:
Gravitational potential at a point in the gravitational field is the amount of work done in bringing a unit mass from infinity to that point.
Gravitational potential energy: When mass m is moved from one position to another position away from it in the gravitational field of another mass M, some work is done by the external force against the force of gravitational attraction between M and m. The work gets associated with mass m as its gravitational potential energy.
Now, we will prove the change in gravitational potential occurred due to change in height of the object (raised above the surface of the earth).
We have Gravitational potential as:
${U_e} = - \dfrac{{GMm}}{R}$.................1
When object is raised to some height above the surface of the earth, then Gravitational potential becomes;
${U_h} = - \dfrac{{GMm}}{{R + h}}$.................2
In the question, h = nR, therefore we can write equation 2 as:
${U_h} = - \dfrac{{GMm}}{{R + nR}}$............3
We can find the change when we subtract equations 1and 3.
$\Rightarrow \Delta U = - \dfrac{{GMm}}{{R + nR}} - \left( { - \dfrac{{GMm}}{R}} \right) \\\ \Rightarrow \Delta U = - \dfrac{{GMm}}{R}\left( {\dfrac{1}{{n + 1}} - 1} \right) \\\$( taking $- \dfrac{{GMm}}{R}$common out )
Now multiplying the minus sign inside the bracket;
$\Rightarrow \Delta U = \dfrac{{GMm}}{R}\left( { - \dfrac{1}{{n + 1}} + 1} \right)$
$\Rightarrow \Delta U = \dfrac{{GMm}}{R}\left( {1 - \dfrac{1}{{n + 1}}} \right) \\\ \Rightarrow \Delta U = \dfrac{{GMm}}{R}\left( {\dfrac{{n + 1 - 1}}{{n + 1}}} \right) \\\$ (Taking LCM inside the bracket)
$\Rightarrow \Delta U = \dfrac{{GMm}}{R}\left( {\dfrac{n}{{n + 1}}} \right)$

Thus, option 1 is correct.

Note: Law of Gravitation states that: Every particle of matter in the universe attracts every other particle with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between them. The force with two particles attracting each other is called the force of gravitation.