Question

A man weighs $80kg$ on the surface of earth of radius $R$. At what height above the surface of earth his weight will be $40kg$?

• A. $\frac{R}{2}$
• B. $\sqrt{2}R$
• C. $(\sqrt{2}-1)R$
• D. $(\sqrt{2}+1)R$

Answer 1

Answer: (C)

$(\sqrt{2}-1)R$

Let $r$ be the distance from center of earth where it weighs 40 kg.

$W_{surface}=\frac{GMm}{R^2}$

$W_r=\frac{GMm}{r^2}$

Since $W_r=\frac{1}{2}{W}_p$,

$r=\sqrt{2}R$.

Hence the height of this point$=r-R=\sqrt{2}R-R=(\sqrt{2}-1)R$