Two identical particles each of mass m, move in circular path due to their own mutual gravitational force. Find the velocity of the particle if the radius of the circular path is a.
- \(\frac{\sqrt{4Gm}}{a}\)
- \(\frac{\sqrt{Gm}}{2a}\)
- \(\frac{\sqrt{2Gm}}{a}\)
- \(\frac{\sqrt{Gm}}{4a}\)
The Correct Option is D
Solution and Explanation
\(\frac{mv^2}{a}=\frac{Gm^2}{4a^2}\)
\(\Rightarrow v=\sqrt{\frac{Gm}{4a}}\)
\(\text{Hence the correct option is (D):}\) \(\frac{\sqrt{Gm}}{4a}\)
Top Questions on 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
- Two objects of masses 5 kg and 10 kg are placed 2 meters apart. What is the gravitational force between them?
(Use \(G = 6.67 \times 10^{-11}\, \mathrm{Nm^2/kg^2}\))- AP EAPCET - 2025
- Physics
- Gravitation
- The escape velocity from the surface of a planet is \(v_e\). What will be the escape velocity from a planet whose mass and radius are twice that of the original planet?
- BITSAT - 2025
- Physics
- Gravitation
- Consider a star of mass $ m_2 $ kg revolving in a circular orbit around another star of mass $ m_1 $ kg with $ m_1 \gg m_2 $. The heavier star slowly acquires mass from the lighter star at a constant rate of $ \gamma $ kg/s. In this transfer process, there is no other loss of mass. If the separation between the centers of the stars is $ r $, then its relative rate of change $ \frac{1}{r} \frac{dr}{dt} $ (in s$^{-1}$) is given by:
- JEE Advanced - 2025
- Physics
- Gravitation
- If the weight of an object is 200 Newtons, then the weight of the object at the midpoint between the Earth's surface and the Earth's center will be:
- UPCATET - 2025
- Physics
- Gravitation
Questions Asked in JEE Main exam
- The sum of all local minimum values of the function \( f(x) \) as defined below is:
\[ f(x) = \begin{cases} 1 - 2x & \text{if } x < -1, \\[10pt] \frac{1}{3}(7 + 2|x|) & \text{if } -1 \leq x \leq 2, \\[10pt] \frac{11}{18}(x-4)(x-5) & \text{if } x > 2. \end{cases} \]- JEE Main - 2025
- Functions
- Let \( \langle a_n \rangle \) be a sequence such that \( a_0 = 0 \), \( a_1 = \frac{1}{2} \), and \( 2a_{n+2} = 5a_{n+1} - 3a_n \).n= 0,1,2,3.... Then \( \sum_{k=1}^{100} a_k \) is equal to:
- JEE Main - 2025
- Sequences and Series
- The expression given below shows the variation of velocity \( v \) with time \( t \): \[ v = At^2 + \frac{Bt}{C + t}. \] The dimension of \( ABC \) is:
- JEE Main - 2025
- Thermodynamics terms
- The elemental composition of a compound is 54.2%C, 9.2%H, and 36.6%O. If the molar mass of the compound is 132 g/mol, the molecular formula of the compound is:
- JEE Main - 2025
- Thermodynamics
Let A be a 3 × 3 matrix such that \(\text{det}(A) = 5\). If \(\text{det}(3 \, \text{adj}(2A)) = 2^{\alpha \cdot 3^{\beta} \cdot 5^{\gamma}}\), then \( (\alpha + \beta + \gamma) \) is equal to:
- JEE Main - 2025
- Matrix Operations
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].