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Variation of g (Acceleration Due to Gravity)

If the value ...

Question

If the value of g at the surface of the earth is $9.8 m s^{- 2}$ , then the value of g at a place $480$ km above the surface of the earth will be (radius of earth $= 6400 k m$ )

4.2 m s^{- 2}

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7.2 m s^{- 2}

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8.5 m s^{- 2}

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9.8 m s^{- 2}

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Solution

The correct option is C $8.5 m s^{- 2}$
$\begin{matrix} g^{'} = \frac{G M}{{(R + h)}^{2}} = \frac{G M}{R^{2}} (1 - \frac{2 h}{R}) = g (1 - \frac{2 h}{R}) = 9.8 [1 - \frac{2 \times 480}{6400}] = 8.5 m / s^{2} \end{matrix}$
$∴$ Option $C$ is correct.

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