Question

A body of mass is placed on the earth s surface. It is taken from the earth s surface to a height $h=3R$, R is the radius of earth. The change in gravitational potential energy of the body is

Answer 1

Step1: Given data

1. The radius of the earth is R.
2. The height of the body from the earth's surface is $h=3R$.

Step2: Gravitational potential energy

1. The gravitational potential energy at any point in a gravitational field of a system of masses is the amount of work done in bringing a unit mass from infinity to that point.
2. Gravitational potential energy due to a point mass m at a distance r is, $V=-{\int }_{\infty }^{r}GM\frac{m}{x^2}dx=-GM\frac{m}{r}$, where G is the universal gravitational constant and M is the mass of the earth.

Step3: Finding the change in Gravitational potential energy

We know that the gravitational potential energy of a body of mass m placed on the earth's surface is

$V=-G\frac{Mm}{R}$ ……………(1)

where M and R are the mass of the earth and the radius of the earth.

Now, gravitational potential energy at a height $h=3R$ from the earth's surface is

$V^{\text{'}}=-G\frac{mM}{R+3R}$ …………(2)

So, the change in gravitational potential energy is

$∆V=V-V\text{'}=-G\frac{Mm}{R}$$-$$\left(-G\frac{mM}{R+3R}\right)$. (From equations 1 and 2 )

or $∆V=-G\frac{Mm}{R}+G\frac{Mm}{4R}=-G\frac{3Mm}{4R}$

or $∆V=-G\frac{3mM}{4R}$

Therefore, the change in gravitational potential energy is $\mathbf{∆}\mathit{V}\mathbf{=}\mathbf{-}\mathit{G}\frac{\mathbf{3}\mathbf{m}\mathbf{M}}{\mathbf{4}\mathbf{R}}$.