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.
Top Questions on Gravitation
- Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R):
(A) A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet.
(R) The mass of the pendulum remains unchanged at Earth and the other planet.
In light of the above statements, choose the correct answer from the options given below:- JEE Main - 2025
- Physics
- Gravitation
A small point of mass \(m\) is placed at a distance \(2R\) from the center \(O\) of a big uniform solid sphere of mass \(M\) and radius \(R\). The gravitational force on \(m\) due to \(M\) is \(F_1\). A spherical part of radius \(R/3\) is removed from the big sphere as shown in the figure, and the gravitational force on \(m\) due to the remaining part of \(M\) is found to be \(F_2\). The value of the ratio \( F_1 : F_2 \) is:
- JEE Main - 2025
- Physics
- Gravitation
- Given below are two statements, one is labelled as Assertion (A) and the other is labelled as Reason (R): \begin{itemize} \item[(A)] A simple pendulum is taken to a planet of mass and radius, 4 times and 2 times, respectively, than the Earth. The time period of the pendulum remains same on earth and the planet. \item[(R)] The mass of the pendulum remains unchanged at Earth and the other planet. \end{itemize} In light of the above statements, choose the correct answer from the options given below:
- JEE Main - 2025
- Physics
- Gravitation
- "The height from Earth's surface at which acceleration due to gravity becomes \( g/4 \), where \( g \) is acceleration due to gravity on the surface of Earth and \( R \) is the radius of Earth?"
- MHT CET - 2024
- Physics
- Gravitation
The height from Earth's surface at which acceleration due to gravity becomes \(\frac{g}{4}\) is \(\_\_\)? (Where \(g\) is the acceleration due to gravity on the surface of the Earth and \(R\) is the radius of the Earth.)
- MHT CET - 2024
- Physics
- Gravitation
Questions Asked in AP EAMCET exam
Arrange the following in increasing order of their pK\(_b\) values.
- AP EAMCET - 2024
- Molecular Polarity
What is Z in the following set of reactions?
- AP EAMCET - 2024
- Molecular Polarity
Acetophenone can be prepared from which of the following reactants?
- AP EAMCET - 2024
- Molecular Polarity
What are \(X\) and \(Y\) in the following reactions?
- AP EAMCET - 2024
- Molecular Polarity
What are \(X\) and \(Y\) respectively in the following reaction?
- AP EAMCET - 2024
- Molecular Polarity
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].