The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R from the surface of the earth is (g = acceleration due to gravity at the surface of the earth)

$\frac{g}{9}$

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$\frac{g}{3}$

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$\frac{g}{4}$

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$g$

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Solution

The correct option is A
$\frac{g}{9}$

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

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The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R from the surface of the earth is (g = acceleration due to gravity at the surface of the earth)

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The acceleration of a body due to the attraction of the earth (radius R) at a distance 2R from the surface of the earth is (g = acceleration due to gravity at the surface of the earth)

The correct option is A g9 g′g=(RR+h)2=(RR+2R)2=19 ∴g′=g9.

The correct option is A
$\frac{g}{9}$

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