The acceleration due to gravity on the surfac | vedclass

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

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$4g$

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(D) The acceleration due to gravity on the surface of Earth is given by $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of Earth,and $R$ is the radius of Earth.Since $M$ is constant,we have $g \propto \frac{1}{R^2}$.Let the initial radius be $R_1 = R$ and the final radius be $R_2 = \frac{R}{2}$.Then,the ratio of the new acceleration $g_2$ to the initial acceleration $g_1$ is given by:$\frac{g_2}{g_1} = \frac{R_1^2}{R_2^2} = \frac{R^2}{(R/2)^2} = \frac{R^2}{R^2/4} = 4$.Therefore,$g_2 = 4g_1 = 4g$.

The acceleration due to gravity on the surface of Earth is $g$. If the diameter of Earth reduces to half of its original value and mass remains constant,then the acceleration due to gravity on the surface of Earth would be:

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