Question

The height at which the weight of a body becomes $1 / 16^{\text {th }}$, its weight on the surface of earth (radius R), is

• A. 5R
• B. 15R
• C. 3R
• D. 4R

Answer 1

The Correct Option is C

To find the height at which the weight of a body becomes \( \frac{1}{16} \)th of its weight on the Earth's surface, we need to utilize the concept of gravitational force.

The weight of a body at a distance \( r \) from the center of the Earth is given by:

W = \frac{GMm}{r^2}

where:

• W is the weight of the body.
• G is the universal gravitational constant.
• M is the mass of the Earth.
• m is the mass of the body.
• r is the distance from the center of the Earth.

On the Earth's surface, the weight \( W_0 \) is:

W_0 = \frac{GMm}{R^2}

where \( R \) is the radius of the Earth.

Now, we want the weight to be \( \frac{1}{16} \) of \( W_0 \) at a height \( h \):

\frac{GMm}{(R + h)^2} = \frac{1}{16} \times \frac{GMm}{R^2}

Canceling out the common terms \( \frac{GMm}{R^2} \), we get:

\frac{1}{(1 + \frac{h}{R})^2} = \frac{1}{16}

This simplifies to:

(1 + \frac{h}{R})^2 = 16

Taking the square root on both sides:

1 + \frac{h}{R} = 4

Therefore,

\frac{h}{R} = 4 - 1 = 3

Hence, \( h = 3R \).

The correct answer is 3R.