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Homework Help: Gravitational force - point mass and circular platform

  1. May 31, 2016 #1
    1. The problem statement, all variables and given/known data
    Calculate the gravitational force between a point mass and a circular platform.

    2. Relevant equations

    3. The attempt at a solution

    The actual solution is different. They integrated by the angle between a/r (alpha) but i do not understand the cos(alpha) at the start, check this out:


    Can anyone explain this please? Thanks.
  2. jcsd
  3. May 31, 2016 #2

    Simon Bridge

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    Force is a vector... the cosine is one of the components.
  4. Jun 3, 2016 #3
    Yes, my mistake, thanks! What happens if the radius of the disc is infinite? The part I don't understand: σ(area density)=M/4πR^2, but isn't it just M/πR^2 (without "4" at the bottom)? I mean the whole mass of the disc would be M=πR^2σ, since πR^2 is the area of a cricle, with the 4 added its a sphere, but there is no sphere here just an infinite circle so where does the 4 come from?
  5. Jun 3, 2016 #4


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    I do not know the language (what is it, by the way?) but are you sure that the expression ##\sigma = M/4 \pi R^2 ## is referring to the same question? All I see used for that question is

    $$ dM = 2 \pi x \rho dx h $$ which, if you integrate over x from 0 to R gives indeed ## M = \pi R^2 h \rho## as you expected.
  6. Jun 4, 2016 #5
    It is not the same question but it is related to the first one (just with x going to infinity, not to R). Anyways, I have managed to use my google-fo and found a newer edition (1996); the one I have is 1988. This part is wrong. It states that F=8πGm (force between the particle and an infinite plane. The solution states specifically that you have to limit R towards infinity BUT with σ=M/4πR^2 being a constant), but the newer one is corrected to F=2πGmσ, which is what I got as well, so this is solved, thank you. Language is Slovenian. Heres a quick snap of the newer: http://i.imgur.com/m0j9s9q.png ("neskončno velika plošča" literally means "infinitely big plate").
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