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Variation of G Due to Height and Depth

The value of ...

Question

The value of acceleration due to gravity at a depth of $1600 k m$ is equal to [Radius of earth $= 6400 k m$ ]

9.8 {m s}^{- 2}

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4.9 {m s}^{- 2}

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7.35 {m s}^{- 2}

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19.6 {m s}^{- 2}

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Solution

The correct option is C $7.35 {m s}^{- 2}$
Given : $d = 1600 k m$ $R = 6400 k m$
Acceleration due to gravity at a depth $d$ , $g_{d} = g_{s} (1 - \frac{d}{R})$
where $g_{s} = 9.8 m s^{- 2}$ is acceleration due to gravity at earth's surface.
$∴$ $g_{d} = 9.8 (1 - \frac{1600}{6400}) = 7.35 m s^{- 2}$

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