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(B) The acceleration due to gravity at the surface of the earth is $g = \frac{GM}{R^2}$.At a height $h$,the acceleration due to gravity is $g_h = \fra

Vedclass · Accepted Answer

$\frac{2}{3} R$

Vedclass · Answer

(B) The acceleration due to gravity at the surface of the earth is $g = \frac{GM}{R^2}$.At a height $h$,the acceleration due to gravity is $g_h = \frac{GM}{(R+h)^2}$.Given that the value of $g$ is reduced by $64 \%$,the remaining value is $g_h = (100 \% - 64 \%) 	ext{ of } g = 36 \% 	ext{ of } g = 0.36g$.Therefore,$\frac{g_h}{g} = 0.36 = \frac{36}{100}$.Substituting the expressions for $g$ and $g_h$:$\frac{R^2}{(R+h)^2} = \frac{36}{100}$.Taking the square root on both sides:$\frac{R}{R+h} = \frac{6}{10} = \frac{3}{5}$.Cross-multiplying gives:$5R = 3(R+h) = 3R + 3h$.$2R = 3h$.$h = \frac{2}{3} R$.

The height $h$ from the surface of the earth at which the value of $g$ will be reduced by $64 \%$ from the value at the surface of the earth is ($R=$ radius of the earth).

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