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Gravitational force between disk and particle

  1. Nov 22, 2012 #1
    1. The problem statement, all variables and given/known data
    Mass M is distributed uniformly over a disk of radius a. Find the gravitational force between this disk-shaped mass and a particle with mass m located a distance x above the center of the disk.


    2. Relevant equations
    The problem gives the hint to use the equation found in an earlier problem for the force of gravity between a ring and a particle. This equation is [itex]\frac{GmMx}{(x^2+a^2)^{3/2}}[/itex] where a is the radius of the ring and x is the distance of the particle with mass m from the ring of mass M.


    3. The attempt at a solution
    I switched out r (for the radius) for a, integrated with respect to r, and used 0 to a as my limits of integration. [itex]\int \frac{GmMx}{(x^2+r^2)^{3/2}}dr[/itex] I'm not sure how to get the limits of integration on there. Anyway, the answer I got was [itex]\frac{GmMa}{\sqrt{x^2+a^2}}[/itex].

    The back of the book has [itex]\frac{2GMm}{a^2}(1-\frac{x}{\sqrt{a^2+x^2}})[/itex]. I'm not sure what I'm doing wrong here. The radius of the ring is what is changing, which is why I integrated with respect to the radius r. Your help would be much appreciated.
     
  2. jcsd
  3. Nov 22, 2012 #2

    gneill

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    Staff: Mentor

    Before, M was the mass of a single ring. Now M represents the mass of the whole disk, so each of the many rings of thickness dr over the integration will have some portion of the whole mass. Call it dM.
     
  4. Nov 23, 2012 #3
    Consider a differential element of mass dm. If you draw a figure, you will realize (if you think about it long enough) that this must be related to the area by:

    $$ dm = 2\pi ada\sigma $$
    Since you can imagine you have a disc which you added a small width increase da. You'll end up with a slightly nasty integral, but you can simplify it by doing some clever trickery related to the trigonometry of the problem (there are electrostatic analogs to this problem which you might want to look into as well).
     
  5. Nov 23, 2012 #4
    Awesome, thank you so much. Totally overlooked that dM.
     
  6. Nov 23, 2012 #5
    Ok, so [itex] dM=ρ2\pi dr[/itex], where [itex]ρ=\frac{M}{\pi a^2}[/itex]. Then [itex]dM=\frac{2Mdr}{a^2}[/itex].

    So the integral becomes
    [itex]\int \frac{2GMmxdr}{a^2(x^2+r^2)^{3/2}}=\frac{2GMmx}{a^2}\int\frac{dr}{(x^2+r^2)^{3/2}}[/itex]. The bounds of integration are from 0 to a. Substitute [itex] r=xtan\theta[/itex] and [itex]dr=x(sec\theta)^2d\theta[/itex].

    [itex]\frac{2GMmx}{a^2}\int (x^2+x^2(tan\theta)^2)^{-3/2}x(sec\theta)^2d\theta=\frac{2GMm}{a^2x}sin\theta=\frac{2GMmr}{a^2x\sqrt{x^2+r^2}}[/itex] evaluated from 0 to a. The answer I get is [itex]\frac{2GMma}{a^2x\sqrt{x^2+a^2}}[/itex]. Could you please point out what I'm missing? Thanks.
     
  7. Nov 23, 2012 #6

    gneill

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    Staff: Mentor

    If ##\rho## is the density, then a differential mass element consisting of a ring of radius r and width dr is:

    ##dM = \rho \, 2 \pi r \, dr##

    but ##\rho = \frac{M}{\pi a^2}## so that

    ## dM = 2 \frac{M}{a^2} r \, dr ##

    Don't forget that r in the numerator.
     
  8. Nov 23, 2012 #7
    Great, finally got it. Thank you gneill.
     
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