Question

PLEASE ANSWER MY QUESTION
Two particles of equal mass (m) move in a circle of radius (r) under the action of their mutual gravitational attraction . Find the speed of each particle

Answer 1

Answer:

Mutual force of gravitational attraction = GM^2/(2R)^2. Thus, GM^2/4R^2 = Mv^2/R. We get speed of each particle, v =(GM/4R)^(1/2).

Answer 2

Answer:

Masesof two particles m1 =m2=m

Radius of two particles r1=r2=r

so, gravitational force of attraction between particles F=G × m x m ÷ by R x R -[1]

ANS: Centripetal force F= m × v × v ÷ R -[2]

Explanation:from condition [1] and [2]

Gmm / 2R x 2R= mv x v / R

So speed of each particle is v = 1/2√Gm /R