The height vertically above the earth's surface at which the acceleration duo to gravity becomes 1% of its value at the surface is (It is the radius of the earth)

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

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The height vertically above the earth's surface at which the acceleration duo to gravity becomes 1% of its value at the surface is (It is the radius of the earth)

The correct option is B 9 Rg′=g(1+hR)2⇒g100=g(1+hR)2(1+hR)2=100h=9R

The correct option is B 9 R
$g^{'} = \frac{g}{{(1 + \frac{h}{R})}^{2}} \Rightarrow \frac{g}{100} = \frac{g}{{(1 + \frac{h}{R})}^{2}}$
${(1 + \frac{h}{R})}^{2} = 100$
$h = 9 R$