question_answer

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.

A) \[\frac{T\sqrt{2}}{8}\]  

B) \[\frac{T\sqrt{2}}{16}\]

C) \[\frac{T\sqrt{2}}{4}\]  

D) \[\frac{T\sqrt{2}}{32}\]

Correct Answer: B

Solution :

[b] \[{{T}_{O}}=\frac{2\pi r}{\omega }\]         \[{{T}_{I}}=\frac{2\pi r}{2\omega }=\frac{{{T}_{O}}}{2}\] Consider an imaginary comet moving along an ellipse. The extreme points of this ellipse are located on orbit of 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. \[\frac{T{{'}^{2}}}{{{(r/4)}^{3}}}=\frac{{{T}^{2}}_{I}}{{{(r/2)}^{3}}}\] \[T{{'}^{2}}=\frac{8}{64}{{T}^{2}}_{I}=\frac{{{T}^{2}}_{0}}{32}\]         \[(\because {{T}_{1}}={{T}_{0}}/2)\] \[T'=\frac{T}{4\sqrt{2}}\] (T'/2) represents time in which inner planet will fall into star. \[\left( \frac{T'}{2} \right)=\frac{T}{8\sqrt{2}}=\frac{T\sqrt{2}}{16}\]