A test particle is moving in a circular orbit in the gravitational field produced by a mass density \(\rho(r) = \frac{K}{r^2}\). Identify the correct relation between the radius \(R\) of the particle's orbit and it's period \(T\) :
- $T/R^2$ is a constant
- $TR$ is a constant
- $T^2/R^3$ is a constant
- $T/R$ is a constant
The Correct Option is D
Solution and Explanation
$ m = 4\pi KR$
$ v \propto \sqrt{4\pi K} $
$ \frac{T}{R} = \frac{2\pi}{\sqrt{4\pi K}}$
Top Questions on Gravitation
Match the LIST-I with LIST-II
\[ \begin{array}{|l|l|} \hline \text{LIST-I} & \text{LIST-II} \\ \hline \text{A. Gravitational constant} & \text{I. } [LT^{-2}] \\ \hline \text{B. Gravitational potential energy} & \text{II. } [L^2T^{-2}] \\ \hline \text{C. Gravitational potential} & \text{III. } [ML^2T^{-2}] \\ \hline \text{D. Acceleration due to gravity} & \text{IV. } [M^{-1}L^3T^{-2}] \\ \hline \end{array} \]
Choose the correct answer from the options given below:
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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].