Acceleration due to gravity on the surface of | vedclass

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(A) The acceleration due to gravity on the surface of the Earth is given by the formula: $g = \frac{GM}{R_e^2}$ where $G$ is the gravitational constan

Vedclass · Accepted Answer

$9$

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(A) The acceleration due to gravity on the surface of the Earth is given by the formula: $g = \frac{GM}{R_e^2}$ where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $R_e$ is the radius of the Earth. Given that the diameter is reduced to $1/3$ of its original value,the radius $R_e$ also reduces to $1/3$ of its original value,i.e.,$R' = R_e / 3$. The mass $M$ remains unchanged. The new acceleration due to gravity $g'$ is given by: $g' = \frac{GM}{(R_e/3)^2} = \frac{GM}{R_e^2 / 9} = 9 \left( \frac{GM}{R_e^2} \right) = 9g$ Therefore,the new acceleration due to gravity is $9g$.

Acceleration due to gravity on the surface of Earth is $g$. If the diameter of Earth is reduced to one-third of its original value and mass remains unchanged,then the acceleration due to gravity on the surface of the Earth is . . . . . . $g$.

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