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Q.
A man weighs $60\, kg$ at earth surface. At what height above the earths surface weight becomes $30\, kg$? (Given, radius of earth is $6400\, km$.)
J & K CETJ & K CET 2002
Solution:
From Newtons law of gravitation
$F=\frac{G M_{e} m}{R_{e}^{2}}$ ...(i)
where $G$ is gravitational constant,
$M_{e}$ the mass of earth,
$R_{e}$ the radius and
$m$ the mass of man.
Also, $F = mg$ = weight ...(ii)
From Eqs. (i) and (ii), we get
$W=\frac{G M_{e} m}{R_{e}^{2}}$
$\therefore \frac{W_{1}}{W_{2}}=\frac{R_{2}^{2}}{R_{1}^{2}}$
$=\frac{\left(R_{e}+h\right)^{2}}{R_{e}^{2}}$
Given, $W_{1}=60\, kg,\, W_{2}=30\, kg$
$\therefore \frac{60}{30}=\left(\frac{R_{e}+h}{R_{e}}\right)^{2}$
Taking square root $\sqrt{2}=\frac{R_{e}+h}{R_{e}}$
$\Rightarrow \sqrt{2} R_{e}=R_{e}+h$
$\Rightarrow h =\sqrt{2} R_{e}-R_{e} h=1.41\, R_{e}-R_{e} h=0.41\, R_{e}$
Given, $R_{e}=6400\, km$
$\therefore h=0.41 \times 6400=2624\, km$