Final answer:
The speed of each particle moving in a circle under mutual gravitational attraction is found to be √(3Gm/r), by equating the gravitational force with the centripetal force and solving for the velocity.
option c is the correct
Step-by-step explanation:
The specified scenario involves four particles of equal masses 'm' moving in a circle of radius 'r' due to their mutual gravitational attraction. To find the speed of each particle, we must set the gravitational force acting between the particles equal to the centripetal force needed to keep a particle in circular motion.
This is analogous to the motion of a satellite orbiting Earth, where the gravitational force provides the necessary centripetal acceleration. Using the formula for gravitational force (Gm1m2/r^2) and centripetal force (mv^2/r), and equating these, we find that the particle mass 'm' cancels out, resulting in the formula GM = v^2. Here, 'G' is the gravitational constant, and 'M' is the combined mass of the other three particles, which for symmetry can be assumed to act as if it were concentrated at the center of the circle.
Solving for the speed 'v', we get v = √(GM/r). Since the combined mass of the three other particles is '3m', we substitute this into our equation: v = √(3Gm/r), which simplifies to v = √(3) √(Gm/r). Therefore, the correct speed of each particle is given by option (c) √(3Gm/r).