Acceleration due to gravity g for a body of m | vedclass

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(B) The gravitational force $F$ acting on a body of mass $m$ at the surface of the Earth is given by Newton's law of gravitation: $F = \frac{G M m}{R^

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$m^0$

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(B) The gravitational force $F$ acting on a body of mass $m$ at the surface of the Earth is given by Newton's law of gravitation: $F = \frac{G M m}{R^2}$.According to Newton's second law,the force is also $F = m g$,where $g$ is the acceleration due to gravity.Equating the two expressions: $m g = \frac{G M m}{R^2}$.Canceling $m$ from both sides,we get $g = \frac{G M}{R^2}$.Since $G$ and $M$ are constants for the Earth,$g$ is independent of the mass of the body $m$ (which can be written as $m^0$).Thus,$g$ is proportional to $m^0$.

Acceleration due to gravity $g$ for a body of mass $m$ on the Earth's surface is proportional to (Radius of Earth $= R$,mass of Earth $= M$)

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