A ''double star'' is a composite system of two stars rotating about their center of mass under their mutual gravitational attraction. Let us consider such a ''double star'' which has two stars of masses 'm' and '2m' at a separation '1'. If T is the time period of rotation about their enter of mass then.
- $T=2\pi\sqrt\frac{l^3}{mG}$
- $T=2\pi\sqrt\frac{l^3}{2mG}$
- $T=2\pi\sqrt\frac{l^3}{3mG}$
- $T=2\pi\sqrt\frac{l^3}{4mG}$
The Correct Option is C
Solution and Explanation
Top Questions on Gravitation
- The value of \( g \) at height \( h \) above Earth's surface is \( \frac{g}{\sqrt{3}} \). Find \( h \) in terms of the radius of the Earth.
- MHT CET - 2025
- Physics
- Gravitation
- At a height \( h \) above the Earth's surface, the acceleration due to gravity becomes \( \frac{g}{\sqrt{3}} \). What is the value of \( h \) in terms of the Earth's radius \( R \)?
- MHT CET - 2025
- Physics
- 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
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].