Question

A body weighs $72N$ on the surface of the earth. What is the gravitational force on it due to the Earth at a height equal to half the radius of earth?

• A. $48N$
• B. $32N$
• C. $30N$
• D. $24N$

Answer 1

The correct option is B

$32N$

Step 1: Given data

1. The weight of the body is $72N$.
2. The height of the body is $\frac{R}{2}$.

Step 2: Gravitational force

1. The gravitational force of the earth is defined by the force exerted on a body due to the earth.
2. Gravitational force, $F=G\frac{mM}{R^2}$, where, M and m are the mass of the earth and the body respectively, G is the universal gravitational constant, and R is the radius of the earth.
3. Weight W is a gravitational force, which is equivalent to F.

Step 3: Finding the gravitational force

Gravitational force at a distance R (on the surface of the earth) is

$F=G\frac{mM}{R^2}$ which is equal to the body weight.

So,

$F=G\frac{mM}{R^2}..................\left(1\right)$

Now, gravitational force at a distance $\frac{R}{2}$ from the earth's surface is

$$\begin{gathered} F\text{'}=G\frac{mM}{{\left(R+{\displaystyle \frac{R}{2}}\right)}^2}=G\frac{mM}{{\left({\displaystyle \frac{3R}{2}}\right)}^2} \\ orF\text{'}=G\frac{4Mm}{9R^2}..............\left(2\right) \end{gathered}$$

Dividing equation 2 by equation 1

$$\begin{gathered} \frac{F\text{'}}{F}=\frac{G{\displaystyle \frac{4Mm}{9R^2}}}{G{\displaystyle \frac{Mm}{R^2}}}=\frac{4}{9} \\ orF\text{'}=\frac{4}{9}\times F=\frac{4}{9}\times 72 \\ orF\text{'}=32N \end{gathered}$$

The gravitational force exerted on the body at a height is $32N$. So, option (B) is correct.