Question

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

• A. 3R
• B. 4R
• C. 5R
• D. 15R

Answer 1

The correct option is A 3R

Given :The height at which the weight of a body becomes 1/16th1/16th, its weight on the surface of earth (radius R), is :-

Solution :

We have formula for gravitational acceleration as g=$\frac{GM}{R^2}$

Weight at surface of earth is (m$\frac{GM}{R^2}$)

Weight at distance x from surface that is (m$\frac{GM}{x^2}$ )

(m$\frac{GM}{x^2}$ )=$m\frac{GM}{R^2}\times \frac{1}{16}$

Solving this we get $x=4R$ (This distance is from center of earth)

We need height that is 4R-R=3R

The correct opt:A