Question:
Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to $ R^{-5/2} $ , then
Imagine a light planet revolving around a very massive star in a circular orbit of radius R with a period of revolution T. If the gravitational force of attraction between the planet and the star is proportional to $ R^{-5/2} $ , then
Updated On: June 02, 2025
- $T^2 $ is proportional to $ R^2 $
- $T^2 $ is proportional to $\, R^{7/2}$
- $T^2 $ is proportional to $ R^{3/2}$
- $T^2 $ is proportional to $ R^{3.75}$
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The Correct Option is B
Solution and Explanation
$\frac{mv^2}{R} \propto R^{-5/2}$
$ \therefore \, \, \, \, \, \, \, \, v \propto R^{-3/4}$
Now , $ \, \, \, \, \, \, \, \, \, T=\frac{2 \pi R }{v} \, \, \, or \, \, \, T^2 \propto \bigg(\frac{R}{v}\bigg)^2 $
$ or \, \, \, \, \, \, \, T^2 \propto \bigg( \frac{R}{R^{-3/4}} \bigg)^2 \, \, \, or \, \, T^2 \propto R^{7/2}$
$ \therefore \, \, \, \, \, \, \, \, v \propto R^{-3/4}$
Now , $ \, \, \, \, \, \, \, \, \, T=\frac{2 \pi R }{v} \, \, \, or \, \, \, T^2 \propto \bigg(\frac{R}{v}\bigg)^2 $
$ or \, \, \, \, \, \, \, T^2 \propto \bigg( \frac{R}{R^{-3/4}} \bigg)^2 \, \, \, or \, \, T^2 \propto R^{7/2}$
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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].