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Gravitational force from a square plate

  1. Dec 8, 2013 #1
    I ordinarily would put this up in a homework/ coursework forum, but this isn't either one, its just something I was curious about given that I am in a Mechanics class.

    So, I have done calculations for finding the gravitational force from a sphere and from a ring, and a flat circular plate. All these are symmetrical, tho. I was thinking about what happens with a square plate.

    Now, here's the thing. Let's put our particle at a distance z from the surface of the plate. We wil put it in the center, to make the forces as symmetrical as possible.

    Kind of intuitively, I would say that the way to do that calculation is to pretend you are dealing with a wire, (kind of like what you do calculating the electrical fields with a square wire). But this is a plate, so it's different (I think).

    We have the mass of the plate. At any "radius" b from the center, if the plate is density ρ, with a side of "a" units, the mass is going to be [itex]2b^2ρ[/itex]. So I thought we could have a dM = 4bρ.

    That wold say to me that th eintegral would look like this: [tex] \int dF_z = \int_0^a \frac{mG4d\rho dM}{s^2}[/tex] where s^2 is the distance to that dM, and that will be [itex]\sqrt{b^2+z^2}[/itex] so the integral would end up as: [tex] \int dF_z = mG\rho \int_0^a \frac{4b db}{\sqrt{b^2+z^2}}[/tex]

    but anyway, I'd be curios to know if I am approaching this right. I was thinking of the old science fiction tropes with these huge space stations, and wondering what wold happen if one was constructed as a plane square -- also Stephen Baxter -- a scientist himself -- wrote a novel called Raft which takes place in a universe where gravitational forces are a billion times stronger, and a rat of metal plates actually has enough "pull" to live on. (there's more to it than that, but y'all get the idea).
  2. jcsd
  3. Dec 8, 2013 #2
    I'm gonna go out on a limb on this one and put out what I got after quickly doing it out on paper. I didn't actually check the integral, so I'm not sure it's 100% right.

    You have the wrong expression for ##dm##, it should be ##dm=\rho dA = \rho dx dy##.

    I think this problem is easier to work out by not treating it as a bunch of lines.

    I think the integral should be $$Gm\rho \int ^{b/2}_{-b/2} \int ^{b/2}_{-b/2} \frac{dxdy}{(x^2+y^2+z^2)} cos\theta= Gm\rho z \int ^{b/2}_{-b/2} \int ^{b/2}_{-b/2} \frac{dxdy}{(x^2+y^2+z^2)^{3/2}} $$

    I think this makes sense because $$dF = \frac{Gm\rho dA}{r^2}cos\theta$$

    Don't take my word for it though, you should evaluate it and check to see if it makes sense.
  4. Dec 8, 2013 #3


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    There are fairly simple expressions for the field and potential of a charged rod, in its midplane. See http://www.lightandmatter.com/area1sn.html , section 10.3.1. I would try breaking up the square into parallel line segments, taking the expression for the potential and integrating it. The integral would probably be ugly, but you could put it into software such as maxima (which is free and open source).
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