SATHEE: Gravitation Question 228

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) $[\sqrt{\frac{G M}{R} (1 + 2 \sqrt{2})}]$

B) $[\frac{1}{2} \sqrt{\frac{G M}{R} (1 + 2 \sqrt{2})}]$

C) $[\sqrt{\frac{G M}{R}}]$

D) $[\sqrt{2 \sqrt{2} \frac{G M}{R}}]$

Show Answer

Answer:

Correct Answer: B

Solution:

Net force on any one particle $[= \frac{G M^{2}}{{(2 R)}^{2}} + \frac{G M^{2}}{{(R \sqrt{2})}^{2}} \cos 45^{\circ} + \frac{G M^{2}}{{(R \sqrt{2})}^{2}} \cos 45^{\circ}]$ $[= \frac{G M^{2}}{R^{2}} [\frac{1}{4} + \frac{1}{\sqrt{}}]]$ This force will be equal to centripetal force so $[\frac{M u^{2}}{R} = \frac{G M^{2}}{R^{2}} [\frac{1 + 2 \sqrt{2}}{4}]]$ $[u = \sqrt{\frac{G M}{R} [1 + 2 \sqrt{2}]}]$ $[= \frac{1}{2} \sqrt{\frac{G M}{R} (2 \sqrt{2} + 1)}]$

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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:

Answer:

Solution:

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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