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Escape Velocity of the Milky Way

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  1. Nov 18, 2011 #1
    1. The problem statement, all variables and given/known data

    In the Solar neighborhood, the Milky Way has a flat rotation curve, with V(r)= Vc where Vc is a constant, implying a mass desnity profile ρ(r) ~ r^-2

    Assume there is a cutoff radius R beyond where the mass density is zero. Prove that the velocity of escape from the galaxy from any radius r<R is:

    Ve^2= 2Vc^2(1+ln R/r)


    2. Relevant equations

    Integral needs to be done in two parts


    3. The attempt at a solution

    The 1/2 mv^2 provides the energy needed to do the work of moving the mass m against the force of gravity from a radius r to infinity.
    I believe the integral needs to be evaluated at both r and R, however, I do not know what equation to integrate because integrating 1/2 mv^2 doesn't seem like it will yield the above equation.
     
  2. jcsd
  3. Nov 18, 2011 #2

    gneill

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    Staff: Mentor

    A body orbiting a central mass M in a circular orbit at radial distance r has orbital velocity
    [tex] V_c = \sqrt{\frac{G M}{r}} [/tex]
    For the galaxy then, if the velocity profile is flat then Vc is a constant from r to R, and you can obtain an effective mass w.r.t. radius by solving the above for M and calling it M(r).

    With M(r) you are in a position to determine the change in PE for an object taken from r to R.

    Next, consider that at distance R the circular orbital velocity is still Vc, and that since all the effective mass is "below" R, it will behave as a point mass at the center and the escape velocity there must be [itex]\sqrt{2}V_c[/itex]. That gives you the required KE at radius R.

    For escape, the starting KE will be the required final KE plus the loss in PE. Find the escape velocity from the starting KE.
     
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