Problem 13 - A Complete Resource Book in Physics for JEE Main

Question

Four particles of equal mass \(M\) move along a circle of radius \(R\) under the action of their mutual gravitational attraction. The speed of each particle is (A) \(\frac{G M}{R}\) (B) \(\sqrt{\left(\frac{G M}{R}\right)}\) (C) \(\sqrt{\left[\frac{G M}{R}\left(\frac{2 \sqrt{2}+1}{4}\right)\right]}\) (D) \(\sqrt{\left[\frac{G M}{R}(\sqrt{2}+1)\right]}\)

Step-by-Step Solution

Answer

The speed of each particle is \(v = \sqrt{\left[\frac{G M}{R}\left(\frac{2 \sqrt{2}+1}{4}\right)\right]}\). The correct choice is (C).

Step 1: Write down the expression for gravitational force and centripetal force

For two particles with masses M and M' separated by a distance r, the gravitational force F between them is given by the formula: \(F = G\frac{M M'}{r^2}\), where G is the gravitational constant. For a particle moving in a circle with radius R and speed v, the centripetal force is given by the formula: \(F_{centripetal} = M\frac{v^2}{R}\).

Step 2: Calculate the total gravitational force acting on one particle

To find the total gravitational force acting on one particle, consider the interaction with three other particles. For simplicity, let's designate the center of the circle as O, and the four particles as A, B, C, and D, with particle A being the one we are interested in (because they are all identical, the force acting on A will be the same for all of them). Gravitational force is a vector quantity, so we need to calculate the resultant of these forces acting on particle A due to particles B, C, and D. To facilitate this computation, we can find the distance between the particles as follows: - AB and AD distances are easy as they are on the same circular path, with a distance of equal to the diameter: \(d_{AB} = d_{AD} = 2R\). - The distance between particle A and C (AC) is more complicated because they are located diagonally in the problem. To find this distance, notice that ABCD form a square (since the particles are positioned symmetrically), and use the Pythagorean theorem: \(d_{AC}^2 = (2R)^2 + (2R)^2\), resulting in \(d_{AC} = 2\sqrt{2}R\). Now, we calculate the gravitational forces due to each particle acting on particle A: - \(F_{AB} = G\frac{M^2}{(2R)^2} = \frac{GM^2}{4R^2}\) - \(F_{AD} = \frac{GM^2}{4R^2}\) (since it has the same distance as AB) - \(F_{AC} = G\frac{M^2}{(2\sqrt{2}R)^2} = \frac{GM^2}{8R^2}\)

Step 3: Find the net gravitational force acting on particle A

Since we have found the gravitational forces acting on particle A due to particles B, C, and D, now we need to find the net gravitational force (vector sum) acting on particle A. Both \(F_{AB}\) and \(F_{AD}\) have the same magnitude and are acting perpendicular to each other. Hence their vector sum can be calculated using the Pythagorean theorem: \(F_{AB+AD} = \sqrt{F_{AB}^2 + F_{AD}^2} = \sqrt{2F_{AB}^2} = \sqrt{2 \cdot \frac{GM^2}{4R^2}} = \frac{GM}{R\sqrt{2}}\) Now, we can find the net gravitational force acting on particle A due to all three particles B, C, and D. Notice that in this case, \(F_{AB+AD}\) and \(F_{AC}\) are also acting perpendicular to each other: \(F_{net} = \sqrt{F_{AB+AD}^2 + F_{AC}^2} = \sqrt{\left(\frac{GM}{R\sqrt{2}}\right)^2 + \left(\frac{GM}{R\sqrt{8}}\right)^2} = \sqrt{\frac{GM^2}{R^2}(\frac{1}{2} + \frac{1}{4})} = \frac{GM}{R}\sqrt{\frac{3}{4}}\)

Step 4: Equate centripetal force with net gravitational force and solve for the velocity

Now that we have the net gravitational force acting on particle A, we can equate it to the centripetal force and solve for the speed v: \(M\frac{v^2}{R} = \frac{GM}{R}\sqrt{\frac{3}{4}}\) By canceling out the common terms (M, R) in the equation, we get: \(v^2 = G\sqrt{\frac{3}{4}}\) So the speed of each particle is: \(v = \sqrt{G \sqrt{\frac{3}{4}}}\) Comparing this result with the given choices, we find that it is closest to option (C): \(v = \sqrt{\left[\frac{G M}{R}\left(\frac{2 \sqrt{2}+1}{4}\right)\right]}\) Therefore, the correct choice is (C).