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

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The radius of the Earth is smaller at the poles than at the equator.

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(B) The acceleration due to gravity is given by the formula $g = \frac{GM}{R^2}$,where $G$ is the gravitational constant,$M$ is the mass of the Earth,and $R$ is the distance from the center of the Earth.Since the Earth is not a perfect sphere but is flattened at the poles and bulges at the equator,the radius $R$ is smaller at the poles compared to the equator.Because $g$ is inversely proportional to the square of the radius $(g \propto \frac{1}{R^2})$,a smaller radius at the poles results in a higher value of $g$.Conversely,the larger radius at the equator results in a lower value of $g$.

Give a reason for the statement that the value of $g$ is greater at the poles than at the equator.

Similar Questions

The weight of an object:

Mona weighs $750 \, N$ on Earth:
$(i)$ On the planet Mars,the force of gravity is $38 \%$ of that of Earth. How much will Mona weigh on Mars?
$(ii)$ What will be Mona's mass on Earth (take $g = 10 \, m s^{-2}$)?

When dropped from the same height,a body reaches the ground quicker at the poles than at the equator. Why?

$(a)$ What do you mean by acceleration due to gravity?
$(b)$ What is the value of $g$ on the surface of the earth?
$(c)$ On what factors does the value of $g$ depend?

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