The acceleration due to gravity 'g' for objects on or near the surface

The acceleration due to gravity ‘g’ for objects on or near the surface of earth is related to the universal gravitational constant ‘G’ as (‘M’ is the mass of the earth and ‘R’ is its radius):

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G = g $rac{ ext{M}}{ ext{R}^2}$

g = G $rac{ ext{M}}{ ext{R}^2}$

M = $rac{ ext{gG}}{ ext{R}^2}$

R = $rac{ ext{gG}}{ ext{M}^2}$

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Answer is Wrong!

This question was previously asked in

UPSC NDA-2 – 2015

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The correct option is B, g = G $\frac{\text{M}}{\text{R}^2}$.

The acceleration due to gravity ‘g’ on the surface of the Earth is caused by the gravitational force exerted by the Earth on an object. According to Newton’s Law of Universal Gravitation, the force (F) between the Earth (mass M) and an object (mass m) on its surface (distance R from the center, where R is the Earth’s radius) is given by F = G $\frac{\text{Mm}}{\text{R}^2}$, where G is the universal gravitational constant. This gravitational force is also the object’s weight, W, which is given by W = mg (by Newton’s second law, where ‘g’ is the acceleration due to this force). Equating the two expressions for the force (F = W), we get mg = G $\frac{\text{Mm}}{\text{R}^2}$. Dividing both sides by the mass of the object ‘m’ gives the relationship for the acceleration due to gravity: g = G $\frac{\text{M}}{\text{R}^2}$.

This formula shows that the acceleration due to gravity ‘g’ is independent of the mass of the object itself and depends only on the mass and radius of the planet (or celestial body) and the universal gravitational constant G. The value of ‘g’ varies slightly across the Earth’s surface due to factors like altitude, latitude, and local geological variations.

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