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(B) The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM_e}{R_e^2}$.At a height $h = R_e$ above the surface,the accel

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$g/4$

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(B) The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM_e}{R_e^2}$.At a height $h = R_e$ above the surface,the acceleration due to gravity $g^{\prime}$ is given by the formula $g^{\prime} = \frac{GM_e}{(R_e + h)^2}$.Substituting $h = R_e$ into the equation,we get $g^{\prime} = \frac{GM_e}{(R_e + R_e)^2} = \frac{GM_e}{(2R_e)^2}$.This simplifies to $g^{\prime} = \frac{GM_e}{4R_e^2}$.Since $g = \frac{GM_e}{R_e^2}$,we can write $g^{\prime} = \frac{g}{4}$.

If the value of acceleration due to gravity on the surface of the Earth is $g$,what will be the value of $g$ at a height equal to the radius of the Earth from the surface?

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