The value of acceleration due to gravity g is maximum at?
The value of acceleration due to gravity g is maximum at?
Solution and Explanation
The value of acceleration due to gravity (g) is slightly greater at the poles compared to the equator. This is because the Earth is not a perfect sphere but rather an oblate spheroid, meaning it is slightly flattened at the poles and bulges at the equator due to its rotation.
The variation in gravity is caused by two main factors: the centrifugal force due to the Earth's rotation and the variation in distance from the Earth's center.
At the equator, the centrifugal force due to the Earth's rotation is at its maximum because the distance from the Earth's axis of rotation is greatest. This centrifugal force counteracts the force of gravity, slightly reducing the effective acceleration due to gravity at the equator. As a result, the value of g at the equator is slightly lower than at the poles.
Conversely, at the poles, the distance from the Earth's axis of rotation is minimal, resulting in a smaller centrifugal force. Consequently, the effective acceleration due to gravity is slightly higher at the poles compared to the equator.
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Concepts Used:
Gravitation
In mechanics, the universal force of attraction acting between all matter is known as Gravity, also called gravitation, . It is the weakest known force in nature.
Newton’s Law of Gravitation
According to Newton’s law of gravitation, “Every particle in the universe attracts every other particle with a force whose magnitude is,
- F ∝ (M1M2) . . . . (1)
- (F ∝ 1/r2) . . . . (2)
On combining equations (1) and (2) we get,
F ∝ M1M2/r2
F = G × [M1M2]/r2 . . . . (7)
Or, f(r) = GM1M2/r2
The dimension formula of G is [M-1L3T-2].
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