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Q.
A man weighs $80 \,kg$ on earth's surface. The height above ground where he will weigh $40 \,kg$, is : (Radius of earth is $6400 \,km$)
Jharkhand CECEJharkhand CECE 2005
Solution:
Value of acceleration due to gravity decreases on going above the earth's surface.
The weight of a body is given by $w=m g$ (on earth's surface)
At a height $h$ above the earth's surface
$w^{\prime}=\frac{m g}{\left(1+\frac{h}{R}\right)^{2}}$
$\therefore \frac{w}{w^{\prime}}=\left(1+\frac{h}{R}\right)^{2}$
Given, $w=80\, kg , w^{\prime}=40 \,kg$
$\therefore \frac{80}{40}=\left(1+\frac{h}{R}\right)^{2}$
$\Rightarrow 2=\left(1+\frac{h}{R}\right)^{2} $
$\Rightarrow \sqrt{2}=1.41=1+\frac{h}{R}$
$\Rightarrow h=0.41 \,R$
Given, $h=0.41 \,r$