having trouble with the following:

two identical stars of mass M orbit their C of M,show

T^2=2pi^2R^3/(GM)

when doing this we need to use F=GMM/((R/2)^2). Why is this?

i have found the distance from the star to C of M is R/2,so yes the radius of the orbit is R/2

But if they orbit the C of M then surely this point has mass M+M=2M so we would have F=2GM^2/{R/2}^2?

That is the planets orbit a point of mass 2M at a distance R/2 from the mass 2M. so im thinking along the lines of this

Physics Forums - View Single Post - Center of Mass & Binary Stars

secondly we have a massless boat of length L on frictionless water.

A man P1 (mass m1) stands at the left holding a snake (mass m3). He chucks it to a man P2 (mass m2) on the right of the boat.

the question asks to show the CoM is the same at end,when P2 has snake as at the start when P1 had the snake.

to do this i have found a velocity of the boat moving to the left and found it has shifted Lm3/(m1+m2+m3)

from here we can shift axis to get the result. but what is going on here?

If i looked at boat at start as left hand side has mass (m1+m2) C of M would be towards that end.

At the end if i came alomg,the right side has more mass so the C of M is towards that end.

but if i watch the event i would see the boat move to the left so if i stood in front of the C of M before the throw i would still be in front of the C of M after the throw due to the motion of the boat

what is going on here? does the boat know how far to move so its Cof M stays in the same place on my axis of co-ordinates? when we chuck the snake does this cause the boat to move to the left because the C of M changes? (or is is newtons laws in play (opposite and = reaction) i just assume the snake not on boat will give a new C of M, so does boat move to keep C of M in same place so i will always be standing in front of the C of M?

sorry if my questions dont ,make sense, i will try to simplify them when i get a reply.