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Thread: centre of mass questions

  1. #1
    Dec 2009

    centre of mass questions

    having trouble with the following:

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


    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.
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  2. #2
    A Plied Mathematician
    Jun 2010
    CT, USA
    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 incorrect thinking. In a two-body problem, what's usually done is to consider the reduced mass of the system, defined as

    $\displaystyle \displaystyle{\mu=\frac{m_{1}m_{2}}{m_{1}+m_{2}}=\ frac{m^{2}}{2m}=\frac{m}{2}.}$

    At that point, we "have therefore formally reduced the problem of the motion of two bodies to an equivalent one-body problem in which we must determine only the motion of a 'particle' of mass $\displaystyle \mu$ in the central field described by the potential function $\displaystyle U(r)$." - Marion and Thornton, Classical Dynamics of Particles and Systems, 4th Ed., p. 293.

    The potential $\displaystyle U(r)$ referred to is the potential energy corresponding to the gravitational force between the two bodies. It only depends on the distance between the two bodies.

    Here's my question for you: at this point in your course, what do you know? Are you expected to derive the period from first principles like the Euler-Lagrange equation and conservation of angular momentum? Or from Newton's Second Law? Or do you start from the equations

    $\displaystyle \displaystyle{F(r)=-\frac{Gm_{1}m_{2}}{r^{2}}=-\frac{k}{r^{2}},}$ and

    $\displaystyle \displaystyle{T^{2}=\frac{4\pi^{2}\mu}{k}\,a^{3}}$?

    Here, $\displaystyle a$ is the semimajor axis of an elliptic orbit (in your case, it's just the distance from the C of M to one of the stars), $\displaystyle k$ is defined by the previous equation, and $\displaystyle \mu$ is the reduced mass defined earlier.

    Concerning the boat problem: so, what's going on here is that the first guy throws the snake to the other guy. As he throws it, the snake achieves a momentum to the right, forcing the thrower (and the boat that is fixed relative to him) to the left. When the catcher catches the snake, the snake then stops going right. Its momentum is arrested, which gets transferred through the catcher to the boat. The boat, you will recall, was moving left. This additional change in momentum is exactly the momentum earlier incurred. Hence, the system stops again when the catcher catches the snake. The entire problem of why the boat stops again is entirely solved by conservation of momentum considerations.

    The boat-snake-man-man system acts as a whole from the point of view of an observer such that its center of mass does the same thing before and after the throw. In this case, the system is at rest before the throw. Therefore, the center of mass is going to be at rest after the throw is completed.

    Does this shed more light on the matter?
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