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Question

(A) The acceleration due to gravity at a height $h$ above the surface of the earth is given by the formula: $g' = g \left(1 + \frac{h}{R}\right)^{-2}$

Vedclass · Accepted Answer

$2R$

Vedclass · Answer

(A) The acceleration due to gravity at a height $h$ above the surface of the earth is given by the formula: $g' = g \left(1 + \frac{h}{R}\right)^{-2}$.Given that $g' = \frac{g}{9}$,we substitute this into the equation:$\frac{g}{9} = \frac{g}{(1 + \frac{h}{R})^2}$.Canceling $g$ from both sides,we get: $\frac{1}{9} = \frac{1}{(1 + \frac{h}{R})^2}$.Taking the square root of both sides: $\frac{1}{3} = \frac{1}{1 + \frac{h}{R}}$.This implies: $1 + \frac{h}{R} = 3$.Solving for $h$: $\frac{h}{R} = 2$,which gives $h = 2R$.

The height above the surface of the earth where acceleration due to gravity becomes $\frac{g}{9}$ is ( $R$ is the radius of the earth,$g$ is the acceleration due to gravity at the surface).

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