Sum

Find the height over the Earth's surface at which the weight of a body becomes half of its value at the surface.

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#### Solution

Assume that at height h, the weight of the body becomes half.

Weight of the body at the surface = mg

Weight of the body at height h above the Earth's surface = mg', where g' is the acceleration due to gravity at height h

\[\text { Now }, g' = \frac{1}{2}g\]

\[\therefore \left( \frac{1}{2} \right) \frac{GM}{R^2} = \frac{GM}{\left( R + h \right)^2} \left[ \because g = \frac{GM}{R^2} \right]\]

\[ \Rightarrow 2 R^2 = \left( R + h \right)^2 \]

\[ \Rightarrow \sqrt{2}R = R + h\]

\[ \Rightarrow h = \left( \sqrt{2} - 1 \right)R\]

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