The height h above the earth's surface at whi | vedclass

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(B) The acceleration due to gravity at a height $h$ above the earth's surface is given by the formula: $g_h = g \left( \frac{R}{R+h} \right)^2$. Given

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$(\sqrt{3}-1) R$

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(B) The acceleration due to gravity at a height $h$ above the earth's surface is given by the formula: $g_h = g \left( \frac{R}{R+h} \right)^2$. Given that $g_h = \frac{g}{3}$,we substitute this into the equation: $\frac{g}{3} = g \left( \frac{R}{R+h} \right)^2$. Dividing both sides by $g$: $\frac{1}{3} = \left( \frac{R}{R+h} \right)^2$. Taking the square root of both sides: $\frac{1}{\sqrt{3}} = \frac{R}{R+h}$. Rearranging the terms to solve for $h$: $R + h = \sqrt{3} R$. $h = \sqrt{3} R - R$. $h = (\sqrt{3} - 1) R$.

The height $h$ above the earth's surface at which the value of acceleration due to gravity $g$ becomes $\frac{g}{3}$ is ($R=$ radius of the earth).

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