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Gravitation (just to check if calculation is right)

  1. Jul 24, 2006 #1
    I was wondering if someone could tell me if I did this right:

    Two neutron stars are separated by a distance of 10^10m. They each have a mass of 10^30kg and a radius of 10^5m. They are initially at rest relative to each other. How fast are they moving when they collide?

    R = 2r + d = 2(10^5)+10^10

    v = sqrroot (2Gm2/R^2), sub in all the numbers and v = 1.2x10^5m/s
     
  2. jcsd
  3. Jul 24, 2006 #2

    HallsofIvy

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    Where did you get "v= sqrroot(2Gm2/R2"? Shouldn't that be v= sqrroot(2Gm2/R)?
     
  4. Jul 24, 2006 #3
    Hmm I think I rearranged the equation wrong then. Thanks.
     
  5. Jul 24, 2006 #4
    I did Fg = Gm1m2/r^2 = 1/2mv^2
     
  6. Jul 24, 2006 #5

    Andrew Mason

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    The total work done is the integral of the force from initial r to final r:

    [tex]\int_{r_i}^{r_f}Fdr = \int_{r_i}^{r_f}\frac{Gm^2}{r^2}dr = Gm^2\left(\frac{1}{r_i} - \frac{1}{r_f}\right)[/tex]

    This becomes the kinetic energy of both stars, each of which has half this energy:

    [tex]KE_{star} = \frac{1}{2}mv^2 = \frac{1}{2}Gm^2\left(\frac{1}{r_i} - \frac{1}{r_f}\right)[/tex]

    [tex]v = \sqrt{Gm\left(\frac{1}{r_i} - \frac{1}{r_f}\right)}[/tex]

    AM
     
  7. Jul 24, 2006 #6
    We haven't done integrals in this unit so I don't know how to solve that method =(
     
  8. Jul 24, 2006 #7

    Andrew Mason

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    All you need to know is that the potential energy is [itex]U = -GmM/r[/itex]. The change in potential energy between two positions is just [itex]\Delta U = -GmM(1/r_f - 1/r_i)[/itex]. This is the change in potential energy of the system. In this case, the system consists of two stars which move toward each other with equal speed.

    AM
     
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