The height of the point vertically above the earth’s surface, at which acceleration due to gravity becomes $1 %$ of its value at the surface is (Radius of the earth = R)

8 R

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9 R

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10 R

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20 R

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Solution

The correct option is B
9 R

Acceleration due to gravity at height h is given by $g^{'} = g {(\frac{R}{R + h})}^{2}$

$\Rightarrow \frac{g}{100} = g {(\frac{R}{R + h})}^{2} \Rightarrow \frac{R}{R + h} = \frac{1}{10} \Rightarrow h = 9 R$

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The height of the point vertically above the earth’s surface, at which acceleration due to gravity becomes 1% of its value at the surface is (Radius of the earth = R)

The correct option is B 9 R Acceleration due to gravity at height h is given by g′=g(RR+h)2 ⇒g100=g(RR+h)2⇒RR+h=110⇒h=9R

The height of the point vertically above the earth’s surface, at which acceleration due to gravity becomes $1 %$ of its value at the surface is (Radius of the earth = R)

The correct option is B
9 R

Acceleration due to gravity at height h is given by $g^{'} = g {(\frac{R}{R + h})}^{2}$

$\Rightarrow \frac{g}{100} = g {(\frac{R}{R + h})}^{2} \Rightarrow \frac{R}{R + h} = \frac{1}{10} \Rightarrow h = 9 R$