Question:
An artificial satellite revolves around earth in circular orbit of radius r with time period of orbit T. The satellite is made to stop in the orbit which makes it fall into earth. Time of fall of the satellite onto earth is given by
An artificial satellite revolves around earth in circular orbit of radius r with time period of orbit T. The satellite is made to stop in the orbit which makes it fall into earth. Time of fall of the satellite onto earth is given by
Updated On: Jul 6, 2022
- $\sqrt3\frac{T}{6}$
- $\frac{\sqrt2}{8}T$
- $\frac{T}{\sqrt3}$
- $\sqrt\frac{2}{3}\frac{T}{\pi}$
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The Correct Option is B
Solution and Explanation
On stopping, the satellite will fall along the radius r of the orbit which can be regarded as a limiting case of an ellipse with semi major axis equal to $\frac{r}{2}$ Using Keple's third law $T^2 \infty r^3$ time of fall = $\frac{T}{2} = \frac{T}{2 \sqrt{8}} = \frac{\sqrt{2}T}{8}$
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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].