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The ratio of the gravitational force between electron and proton

  1. Sep 10, 2013 #1
    1. The problem statement, all variables and given/known data
    Some of the earliest atomic models held that the orbital velocity of an electron in an atom could be correlated with the radius of the atom. If the radius of the hydrogen atom is 10^−10 m and the electrostatic force is responsible for the circular motion of the electron, what is the ratio of the gravitational force between electron and proton to the electrostatic force? How does this ratio change if the radius of the atom is doubled? Explain {Answer: Fg/Fe = 4.39 x 10-40}.

    2. Relevant equations

    fe = (1/4πε0)*(q^2/r^2)
    fg = G m^2/r^2

    3. The attempt at a solution
    The answer says Fg/Fe, so I divided fg/fe to get (Gm^2)/((9*10^9)q^2). I tried every different way possible but cannot manage to get the correct answer. Can anyone help me solve this problem, I've been stuck for a long time. I preferred you show me how to do it and how you got to the final answer.
  2. jcsd
  3. Sep 10, 2013 #2
    What are you using as the masses? One of them would be the mass of a proton; the other would be the mass of an electron.
  4. Sep 10, 2013 #3
    Maybe my formula is wrong . Do you think this formula would be correct? G*me*mp/(9*10^9 n m^2/c^2)*q1*q2?
  5. Sep 10, 2013 #4
    Oh never mind. I got the answer. How does the ratio change if radius if atom is doubled?
  6. Sep 10, 2013 #5
    What do you think - does it look like the formula depends on the radius?
  7. Sep 10, 2013 #6
    Since the radius isn't included in the formula. I'm assuming it doesn't matter?
  8. Sep 10, 2013 #7
  9. Sep 10, 2013 #8
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