Question

A body of mass m rises to height$h=\frac{R}{5}$from the earth's surface, where R is the earth's radius. If g is the acceleration due to gravity at the earth's surface, the increase in potential energy is

Answer 1

Step 1: Given data

1. The radius of earth is R.
2. Mass of the body is m.
3. The height of the body from the earth's surface is $h=\frac{R}{5}$ .

Step 2: 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.

Step 3: 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 $h=\frac{R}{5}$ from the earth's surface is,

$$\begin{gathered} V\text{'}=-G\frac{Mm}{R+{\displaystyle \frac{R}{5}}}=-G\frac{5Mm}{6R} \\ orV\text{'}=-G\frac{5Mm}{6R}..................\left(2\right) \end{gathered}$$

So, the change in gravitational potential energy is

$$\begin{gathered} ∆V=V\text{'}-V=-G\frac{5mM}{6R}-\left(-G\frac{mM}{R}\right). \\ or∆V=-G\frac{5mM}{6R}+G\frac{mM}{R}=G\frac{mM}{6R}. \\ or∆V=G\frac{mM}{6R}. \end{gathered}$$

(From equations 1 and 2 )

Therefore, the increase in gravitational potential energy is $G\frac{Mm}{6R}$.