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:
1. \( \sqrt{2 \sqrt{2} \frac{{GM}}{{R}}} \)
2. \( \sqrt{\frac{{GM}}{{R}}(1+2 \sqrt{2})} \)
3. \( \frac{1}{2} \sqrt{\frac{{GM}}{{R}}(1+2 \sqrt{2})} \)
4. \( \sqrt{\frac{{GM}}{{R}}}\)

Four identical particles of mass \(M\) are located at the corners of a square of side ‘\(a\)’. What should be their speed if each of them revolves under the influence of another gravitational field in a circular orbit circumscribing the square?

The ratio of the weights of a body on the Earth’s surface to that on the surface of a planet is \(9:4\). The mass of the planet is \(\left(\dfrac{1}{9}\right)^\text{th} \) that of the Earth. If \(R\) is the radius of the Earth, what is the radius of the planet? (Take the planets to have the same mass density)

Two stars of masses \(m\) and \(2m\) at a distance \(d\) rotate about their common centre of mass in free space. The period of revolution is:
1. \(\frac{1}{2\pi}\sqrt{\frac{d^3}{3Gm}}\)
2. \(2\pi\sqrt{\frac{d^3}{3Gm}}\)
3. \(\frac{1}{2\pi}\sqrt{\frac{3Gm}{d^3}}\)
4. \(2\pi\sqrt{\frac{3Gm}{d^3}}\)

Four identical particles of equal masses \(1\) kg made to move along the circumference of a circle of radius \(1\) m under the action of their own mutual gravitational attraction. The speed of each particle will be:
1. \( \sqrt{\frac{G}{2}(1+2 \sqrt{2})} \)
2. \( \sqrt{G(1+2 \sqrt{2})} \)
3. \( \sqrt{\frac{G}{2}(2 \sqrt{2}-1)} \)
4. \( \frac{\sqrt{{(1+2 \sqrt{2}) G}}}{2}\)

A solid sphere of radius \(R\) gravitationally attracts a particle placed at \(3R\) form its centre with a force \(F_1\). Now a spherical cavity of radius \(\frac{R}{2}\) is made in the sphere (as shown in figure) and the force becomes \(F_2\). The value of \(F_1:F_2\) is:

  
1. \(25:36\)
2. \(36:25\)
3. \(50:41\)
4. \(41:50\)