Question

A body weighs 48 N on the surface of the earth. The gravitational force experienced by the body due to the earth at a height equal to one-third the radius of the earth from its surface is:

• A. \( 27 \text{ N} \)
• B. \( 32 \text{ N} \)
• C. \( 36 \text{ N} \)
• D. \( 16 \text{ N} \)

Hint:

Remember that gravitational force (and hence weight) is inversely proportional to the square of the distance from the center of the earth. If the distance increases by a factor of \(n\), the force decreases by a factor of \(n^2\). Here, the distance becomes \(4/3\) times the radius, so the force becomes \( (3/4)^2 = 9/16 \) times the weight on the surface.

Answer 1

The Correct Option is A

To determine the gravitational force experienced by the body at a height equal to one-third the radius of the Earth, we begin by using Newton's law of universal gravitation.

Initially, the force of gravity on the body at the Earth's surface is given by:

\[ F_1 = \frac{GMm}{R^2} = 48 \, \text{N} \]

where \( F_1 \) is the gravitational force at the Earth's surface, \( G \) is the gravitational constant, \( M \) is the mass of the Earth, \( m \) is the mass of the body, and \( R \) is the radius of the Earth.

When the body is at a height of \( \frac{R}{3} \) above the Earth's surface, the total distance from the center of the Earth is \( R + \frac{R}{3} = \frac{4R}{3} \).

The new force, \( F_2 \), can be calculated as follows:

\[ F_2 = \frac{GMm}{\left(\frac{4R}{3}\right)^2} = \frac{GMm}{\frac{16R^2}{9}} = \frac{9GMm}{16R^2} \]

Now, we set up the ratio between the forces at the two positions:

\[\frac{F_2}{F_1} = \frac{\frac{9GMm}{16R^2}}{\frac{GMm}{R^2}} = \frac{9}{16}\]

Substituting the value of \( F_1 = 48 \, \text{N} \):

\[ F_2 = \frac{9}{16} \times 48 \, \text{N} = 27 \, \text{N} \]

Thus, the gravitational force experienced by the body at that height is \( 27 \, \text{N} \).