(C) The variation in acceleration due to gravity at a height $h$ above the Earth's surface is given by $g' = g(1 - 2h/R)$.Given $g' = 9 m s^{-2}$ an

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(C) The variation in acceleration due to gravity at a height $h$ above the Earth's surface is given by $g' = g(1 - 2h/R)$.Given $g' = 9 m s^{-2}$ and $h = R/20$,we have:$9 = g(1 - 2(R/20)/R) = g(1 - 1/10) = g(9/10)$.Therefore,$g = 10 m s^{-2}$.The variation in acceleration due to gravity at a depth $d$ below the Earth's surface is given by $g'' = g(1 - d/R)$.For an equal depth $d = h = R/20$,we have:$g'' = 10(1 - (R/20)/R) = 10(1 - 1/20) = 10(19/20) = 9.5 m s^{-2}$.

Class 11 Physics Gravitation Acceleration Due to Gravity and its Variation

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The acceleration due to gravity at a height $(1/20)^{th}$ of the radius of the Earth above the Earth's surface is $9 m s^{-2}$. Its value at an equal depth below the surface of the Earth is: (in $m s^{-2}$)

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A
$9$
B
$9.25$
C
$9.5$
D
$9.8$

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Gravitation

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Acceleration Due to Gravity and its Variation

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The acceleration due to gravity at a height $(1/20)^{th}$ of the radius of the Earth above the Earth's surface is $9 m s^{-2}$. Its value at an equal depth below the surface of the Earth is: (in $m s^{-2}$)

Similar Questions

At what distance above and below the surface of the earth will a body have the same weight? (Take the radius of the earth as $R$.)

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