The acceleration due to gravity on the surfac | vedclass

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(C) The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the ma

Vedclass · Accepted Answer

$g/4$

Vedclass · Answer

(C) The acceleration due to gravity at the surface of the Earth is given by $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $R$ is the radius of the Earth.At a height $h$ above the surface,the acceleration due to gravity $g_h$ is given by the formula $g_h = \frac{GM}{(R+h)^2}$.Given that $h = R$,we substitute this into the formula:$g_h = \frac{GM}{(R+R)^2} = \frac{GM}{(2R)^2} = \frac{GM}{4R^2}$.Since $g = \frac{GM}{R^2}$,we can write $g_h = \frac{1}{4} \left( \frac{GM}{R^2} \right) = \frac{g}{4}$.Therefore,the acceleration due to gravity at a height $h = R$ is $g/4$.

The acceleration due to gravity on the surface of the Earth is $g$. What is the acceleration due to gravity at a height $h = R$ above the Earth's surface,where $R$ is the radius of the Earth?

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