Question

What happens to the weight of the body if the weight becomes \( \frac{1}{16} \) at a certain height? Also, consider the radius of the Earth to be \( R \).

• A. \( 4R \)
• B. \( 15R \)
• C. \( 5R \)
• D. \( 3R \)

Hint:

When weight reduces with height, apply the inverse square law: \( W_h = W \left( \frac{R}{R + h} \right)^2 \) to find the height or distance.

Answer 1

The Correct Option is A

The weight of a body at a height \( h \) above the Earth's surface is given by: \[ W_h = W \left( \frac{R}{R + h} \right)^2 \] Given \( \frac{W_h}{W} = \frac{1}{16} \), so: \[ \left( \frac{R}{R + h} \right)^2 = \frac{1}{16} \Rightarrow \frac{R}{R + h} = \frac{1}{4} \Rightarrow 4R = R + h \Rightarrow h = 3R \] Hence, the height is \( 3R \) above the surface, making the total distance from the center of Earth: \[ R + h = R + 3R = 4R \]