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  4. The acceleration due to gravity becomes ( (g/2) ) g = acceleration due to gravity on the surface of the earth) at a height equal to

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Q. The acceleration due to gravity becomes $ \left( \frac{g}{2} \right) $ g = acceleration due to gravity on the surface of the earth) at a height equal to

MGIMS WardhaMGIMS Wardha 2008

Solution:

The acceleration due to gravity $ g=\frac{GM}{{{R}^{2}}} $ At a height h above the earth's surface, the acceleration due to gravity is $ g'=\frac{GM}{{{(R+h)}^{2}}} $ $ \therefore $ $ \frac{g}{g'}={{\left( \frac{R+h}{R} \right)}^{2}}={{\left( 1+\frac{h}{R} \right)}^{2}} $ $ \frac{g'}{g}={{\left( 1+\frac{h}{R} \right)}^{-2}} $ $ =\left( 1-\frac{2h}{R} \right) $ But $ g'=\frac{g}{2} $ (given) $ \therefore $ $ \frac{g/2}{g}=1-\frac{2h}{R} $ $ \frac{2h}{R}=\frac{1}{2} $ $ h=\frac{R}{4} $