Q. If a man weighs 90 kg on the surface of earth, the height above the surface of the earth of radius $R,$ where the weight is 30 kg is
A) $0.73\ R$
B) $R/\sqrt{3}$
C) $R/3$
D) $\sqrt{3}R$
Explanation
(a) Given, $\frac{mg'}{mg} = \frac{30}{90}$ or $\frac{g'}{g} = \frac{1}{3}$
Now, $g^{'} = g\frac{R^{2}}{(R + h)^{2}}$ or $\frac{g'}{g} = \frac{R^{2}}{(R + h)^{2}} = \frac{1}{3}$
or $\frac{R}{R + h} = \frac{1}{\sqrt{3}}$ or $(R + h) = \sqrt{3}R$
or $h = \left( \sqrt{3} - 1 \right)R = 0.73R$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.