Gravitation Question 228
Question: Four particles, each of mass $ [M] $ and equidistant from each other, move along a circle of radius $ [R] $ under the action of their mutual gravitational attraction. The speed of each particle is
Options:
A) $ [\sqrt{\frac{GM}{R}(1+2\sqrt{2})}] $
B) $ [\frac{1}{2}\sqrt{\frac{GM}{R}(1+2\sqrt{2})}] $
C) $ [\sqrt{\frac{GM}{R}}] $
D) $ [\sqrt{2\sqrt{2}\frac{GM}{R}}] $
Show Answer
Answer:
Correct Answer: B
Solution:
- Net force on any one particle $ [=\frac{GM^{2}}{{{(2R)}^{2}}}+\frac{GM^{2}}{{{(R\sqrt{2})}^{2}}}\cos 45{}^\circ +\frac{GM^{2}}{{{(R\sqrt{2})}^{2}}}\cos 45{}^\circ ] $ $ [=\frac{GM^{2}}{R^{2}}\left[ \frac{1}{4}+\frac{1}{\sqrt{{}}} \right]] $ This force will be equal to centripetal force so $ [\frac{Mu^{2}}{R}=\frac{GM^{2}}{R^{2}}\left[ \frac{1+2\sqrt{2}}{4} \right]] $ $ [u=\sqrt{\frac{GM}{R}[1+2\sqrt{2}]}] $ $ [=\frac{1}{2}\sqrt{\frac{GM}{R}(2\sqrt{2}+1)}] $
BETA
