Acceleration due to gravity and its variation

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Question

A body weighs 64 N on the surface of the earth. What is the gravitational force (in N) on it due to the earth at a height equal to one-third of the radius of the earth ?

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Solution

Here, $mg = 64 N; h = R / 3$
The value of acceleration due to gravity at a height h (when h is not negligible as compared to R),
$\begin{array}{l} g^{'} = g \frac{R^{2}}{(R + h)^{2}} \\ Or {mg}^{'} = mg \frac{R^{2}}{(R + h)^{2}} = 64 \times \frac{R^{2}}{(R + R / 3)^{2}} = 64 \times \frac{9}{16} = 36 N \end{array}$

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Acceleration due to gravity and its variation

Remember concepts with our Masterclasses.

A body weighs 64 N on the surface of the earth. What is the gravitational force (in N) on it due to the earth at a height equal to one-third of the radius of the earth ?

Here, mg=64N; h=R/3The value of acceleration due to gravity at a height h (when h is not negligible as compared to R),g′=gR2(R+h)2Or mg′=mgR2(R+h)2=64×R2(R+R/3)2=64×916=36 N

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