Gravitation Question 226
Question: Two planets revolve with same angular velocity about a star. The radius of orbit of outer planet is twice the radius of orbit of the inner planet. If Tis time period of the revolution of outer planet, find the time in which inner planet will fall into the star. If it was suddenly stopped.
Options:
A) $ [\frac{T\sqrt{2}}{8}] $
B) $ [\frac{T\sqrt{2}}{16}] $
C) $ [\frac{T\sqrt{2}}{4}] $
D) $ [\frac{T\sqrt{2}}{32}] $
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Answer:
Correct Answer: B
Solution:
- $ [T _{O}=\frac{2\pi r}{\omega }] $ $ [T _{I}=\frac{2\pi r}{2\omega }=\frac{T _{O}}{2}] $ Consider an imaginary comet moving along an ellipse.
The extreme points of this ellipse are located on the orbit of the inner planet and the star.
Semimajor axis of orbit of such comet will be half of the semi-major axis of the inner planet's orbit.
According to Kepler’s law, if T is the time period of the comet, then the square of the time period is proportional to the cube of the semi-major axis of its orbit.
$ \[\frac{T'^{2}}{{(r/4)}^{3}}=\frac{T^{2} _{I}}{{{(r/2)}^{3}}}\] $
$ [T{{’}^{2}}=\frac{1}{8}T^{2} _{I}=\frac{T^{2} _{0}}{32}] $ $ [(\because T _{1}=T _{0}/2)] $
$ [T’=\frac{T}{4\sqrt{2}}] $ (T’) represents time in which inner planet will fall into star. $ [\left( \frac{T’}{2} \right)=\frac{T}{8\sqrt{2}}=\frac{T\sqrt{2}}{16}] $
BETA
