Question

A man weighs $100\ Kgf$ on the surface of earth. At what height above the surface of earth his weight will be $50\ Kgf$? [ where R is radius of the earth]
a) $\dfrac{R}{4}$
b) $\dfrac{R}{2}$
c) $(\sqrt{3} - 1) R$
d) $(\sqrt{2} - 1) R$

Answer 1

Hint: The value of acceleration due to gravity lowers with an increase in an object's height, and the acceleration due to gravity becomes zero at an infinite length from the earth. As depth rises, the value of acceleration due to gravity (g) decreases. Acceleration due to gravity is higher at poles and smaller at the equator. So, if a person travels from the equator to the poles, his weight drops as the value of g drops.

Complete step by step solution:
Given: Weight of the man on the surface of earth, $ mg = 100\ Kgf$.
R is the radius of earth.
We have to find the height, h, above the surface of earth his weight will be, $ mg’ = 50\ Kgf$.
Acceleration due to gravity at height, $g’ = g \left( 1+ \dfrac{h}{R} \right)^{-2}$.
Multiply by m both sides.
$mg’ = mg \left( 1+ \dfrac{h}{R} \right)^{-2}$.
Put all the given values in the above formula.
$50 = 100 \left( 1+ \dfrac{h}{R} \right)^{-2}$.
$\implies 2 = \left( 1+ \dfrac{h}{R} \right)^{2}$.
$\implies \left( 1+ \dfrac{h}{R} \right) = \sqrt{2}$
$\implies \dfrac{h}{R} = \sqrt{2} – 1 $
$\therefore h = (\sqrt{2} – 1) R $
Option (D) is correct.

Note: As the earth is a spheroid, its radius closer to the equator is higher than its radius closer to the poles. Since for a mass, the acceleration due to gravity is inversely proportional to the square of the earth's radius, it changes with latitude due to the earth's shape.