Question

Four particles, each of mass M and equidistant from each other, move along a circle of radius R under the action of their gravitational attraction, the speed of each particle is \(\{\sin (45)^{0}=\cos (45)^{\circ }=1/(2)^{0.5}\}\)

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

Hint:

The particles sit at the corners of a square, and the net gravitational force on each supplies the centripetal force.

Answer 1

The Correct Option is D

Step 1: Understanding the Concept
The four particles form a square inscribed in a circle of radius R.
Side \(=\sqrt{2}R\), diagonal \(=2R\).

Step 2: Net force on one particle
Each neighbour pulls with \(\dfrac{GM^2}{2R^2}\) at \(45^{\circ}\) to the radius.
Two neighbours give \(2\times\dfrac{GM^2}{2R^2}\cos45^{\circ}=\dfrac{GM^2}{\sqrt{2}R^2}\)
The opposite particle gives \(\dfrac{GM^2}{4R^2}\).

Step 3: Total force
\(F=\dfrac{GM^2}{4R^2}\left(2\sqrt{2}+1\right)\)

Step 4: Centripetal condition
\(\dfrac{Mv^2}{R}=F\)
\(v^2=\dfrac{GM}{4R}\left(1+2\sqrt{2}\right)\)
\(v=\dfrac{1}{2}\sqrt{\dfrac{GM}{R}\left(1+2\sqrt{2}\right)}\)

Final Answer:
The speed is half of the square root of GM by R times one plus two root two. \[ \boxed{\text{(D) }\dfrac{1}{2}\sqrt{\dfrac{GM}{R}\left(1+2\sqrt{2}\right)}} \]