two particles of equal mass m go around a circle of radius R under the action of their mutual gravitational attraction . the speed v of each particle is

1)1/2*(Gm/R)^1/2

2)1/2R*(1/Gm)^1/2

The gravitational acceleration that each one feels towards the other is then: 

a = G m / d², 

where d is the distance to the other one. This question of "what is d?" seems to be your question. The answer is that they both orbit on opposite sides of this circle of radius R, thus d = 2 R. 

Thus: 

a = G m / (4 R²) = R ω² 

v² = R² ω² = G m / (4 R) 

v = ½ √(G m / R).

Thx
 

Please refer to the picture below if any dbts ask.

 Well, they will both be orbiting a center of mass, which happens to be halfway between them (because they both have the same mass). Thus their circular motions are described by: 

r = [R cos ωt, R sin ωt], |r| = R 
v = dr/dt = [-R ω sin ωt, R ω cos ωt], |v| = ω R 
a = dv/dt = [-R ω² cos ωt, - R ω² sin ωt], |a| = R ω² 

The gravitational acceleration that each one feels towards the other is then: 

a = G m / d², 

where d is the distance to the other one. This question of "what is d?" seems to be your question. The answer is that they both orbit on opposite sides of this circle of radius R, thus d = 2 R. 

Thus: 

a = G m / (4 R²) = R ω² 

v² = R² ω² = G m / (4 R) 

v = ½ √(G m / R).
hope it helps
 

define the matter

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