Question

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

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

Answer 1

The correct option is C

$(\sqrt{2}-1)R$
Weight of the man on the surface of the earth, W = 80 N

Weight of the man at height h from the surface of the earth, $W_h=40N$

Weight of the man of mass m at the surface of the earth, $W=mg=m\frac{GM}{R^2}$

Here, M = Mass of the earth, R = Radius of the earth and G = Gravitational constant

Weight of the man at height h from the surface of the earth, $W_h=m{g}_h=m\frac{Gm}{(R+h{)}^2}$

Taking ratio of $W_{h}andW:\frac{W_h}{W}={\left(\frac{R}{R+h}\right)}^2\Rightarrow {\left(\frac{R}{R+h}\right)}^2=\frac{40}{80}=\frac{1}{2}\Rightarrow \frac{R}{R+h}=\frac{1}{\sqrt{2}}\Rightarrow \sqrt{2}R=R+h\Rightarrow h=\sqrt{2}R-R\Rightarrow h=(\sqrt{2}-1)R$

Thus, at height $(\sqrt{2}-1)R,$ the man weighs 40 N.

Hence, the correct answer is option (c).