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 on an object at an altitude equal to one-third of Earth's radius, Newton's law of universal gravitation is applied.

The initial gravitational force on the object at Earth's surface is:

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

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

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

The new gravitational force, \( F_2 \), is calculated as:

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

The ratio of the forces at the two positions is:

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

Using \( F_1 = 48 \, \text{N} \):

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

Therefore, the gravitational force on the object at this height is \( 27 \, \text{N} \).