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

• A. $\sqrt{\frac{GM}{R}}$
• B. $\sqrt{ 2 \sqrt 2 \frac{GM}{R}}$
• C. $\sqrt{\frac{GM}{R} (1+2 \sqrt2)}$
• D. $ \frac{1}{2} \sqrt{\frac{GM}{R} (1+2 \sqrt2)}$

Answer 1

The Correct Option is D

\(\frac{F}{\sqrt{2}}+\frac{F}{\sqrt{2}}+F'=\frac{M v^{2}}{R}\)

\(\frac{2 \times G M^{2}}{\sqrt{2}(R \sqrt{2})^{2}}+\frac{G M^{2}}{4 R^{2}}=\frac{M v^{2}}{R}\)

\(\frac{G M^{2}}{R}\left[\frac{1}{4}+\frac{1}{\sqrt{2}}\right]=M v^{2}\)

\(v=\sqrt{\frac{G m}{R}\left(\frac{\sqrt{2}+4}{4 \sqrt{2}}\right)}=\frac{1}{2} \sqrt{\frac{G m}{R}(1+2 \sqrt{2})}\)

The Correct Option is (D): \(\frac{1}{2} \sqrt{\frac{GM}{R} (1+2 \sqrt2)}\)