Question:
Find the height over the earth's surface at which the weight of a body becomes half of its value at the surface.
Solution:
Let height be hÂ
$m \frac{G M}{(R+h)^{2}}=\frac{1}{2} m \frac{G M}{R^{2}}$ $2^{R^{2}}=(R+h)^{2}$
$\sqrt{2} R=R+h$
$\mathrm{~h}=(\sqrt{2}-1) \mathrm{R}$
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