Question

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

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

Answer 1

The Correct Option is B

To solve the problem of finding the gravitational force on a body at a height equal to half the radius of the Earth, we can apply the formula for gravitational force. The weight of the body on the Earth's surface is given by:

• \(W = mg\)

where:

• \(W\) is the weight (72 N)
• \(m\) is the mass of the body
• \(g\) is the acceleration due to gravity on the Earth's surface

Given that:

• The weight of the body, \(W = 72 \, \text{N}\)
• The gravitational force at a height is given by:
• \(F = \frac{G M m}{(R+h)^2}\)

where:

• \(G\) is the gravitational constant
• \(M\) is the mass of the Earth
• \(R\) is the radius of the Earth
• \(h\) is the height above the Earth's surface

Given that \(h = \frac{R}{2}\) (half the radius of the Earth), we have:

\((R + h)^2 = (R + \frac{R}{2})^2 = (\frac{3R}{2})^2 = \frac{9R^2}{4}\)

The gravitational force at this height is:

\(F_h = \frac{G M m}{(R + h)^2} = \frac{G M m}{\frac{9R^2}{4}}\)

The gravitational force on the surface of the Earth is:

\(F_0 = \frac{G M m}{R^2}\)

The ratio of forces is:

\(\frac{F_h}{F_0} = \frac{\frac{G M m}{\frac{9R^2}{4}}}{\frac{G M m}{R^2}}\)

After simplifying, we get:

\(\frac{F_h}{F_0} = \frac{4}{9}\)

Therefore:

• \(F_h = \frac{4}{9} \times W = \frac{4}{9} \times 72 \, \text{N} = 32 \, \text{N}\)

Thus, the gravitational force on the body at the height of half the Earth's radius is 32 N. Hence, the correct answer is 32 N.