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

Hint:Gravitational potential energy of a body due to earth’s gravitation is proportional to the mass of both earth and the object and inversely proportional to the distance between that body and earth’s center. Also, on earth’s surface its gravitational force is the same as the weight of that object.

Formulae used:
Potential energy at any height h over earth’s surface is given by:
${U_h} = - \dfrac{{GmM}}{{(R + h)}}$.......................(1)
Where,
${U_h}$is the potential energy of the body at height h over earth’s surface,
G is gravitational constant,
m is mass of the object,
M is the mass of earth,
R is the radius of earth,
h is the object’s height over earth’s surface.

On earth’s surface relation between gravitational force and weight of a body is:
$\dfrac{{GmM}}{{{R^2}}} = mg$ ……………………….(2)
Where,
g is the acceleration due to gravity.

Complete step by step answer:
Given:
Mass of the body is m.
The body is raised at a height nR from earth’s surface i.e. h=nR.

To find: Change in potential energy due to the raise in height of the body.

Step 1
For earth’s surface h=0 and at a height nR, h=nR. So, substitute these values in eq.(1) to get the change in potential energy as:
$
  \Delta U = {U_{nR}} - {U_0} = - \dfrac{{GmM}}{{(R + nR)}} - \left( { - \dfrac{{GmM}}{{(R + 0)}}} \right) \\
   \rightarrow \Delta U = \dfrac{{GmM}}{R}\left( {1 - \dfrac{1}{{(n + 1)}}} \right) \\
  \therefore \Delta U = \left( {\dfrac{{n + 1 - 1}}{{n + 1}}} \right)\dfrac{{GmM}}{R} = \left( {\dfrac{n}{{n + 1}}} \right)\dfrac{{GmM}}{R} \\
 $..............................(3)

Step 2
Now, from eq.(2) find the relationship between M, G, R and g as:
$
  \dfrac{{GmM}}{{{R^2}}} = mg \\
  \therefore GM = g{R^2} \\
 $...............................(4)

Step 3
Use this relationship obtained in eq.(4) into eq.(3) to get the final expression for $\Delta U$as:
\[
  \Delta U = \left( {\dfrac{n}{{n + 1}}} \right)\dfrac{{GMm}}{R} = \left( {\dfrac{n}{{n + 1}}} \right)\dfrac{{g{R^2}m}}{R} \\
  \therefore \Delta U = \left( {\dfrac{n}{{n + 1}}} \right)mgR \\
 \]

Correct answer:
The change in gravitational potential energy is (a) \[\left( {\dfrac{n}{{n + 1}}} \right)mgR\].

Note:Here, potential energy of the body is always considered to be negative. This negative sign suggests that the state of the body is stable. As we take the body upwards from the surface, its potential energy actually increases. At infinite distance it becomes 0 and at the center of the earth its negative infinity.