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. \[\left( \frac{T'}{2} \right)=\frac{T}{8\sqrt{2}}=\frac{T\sqrt{2}}{16}\]You need to login to perform this action.
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