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Balancing coulomb and gravitational force

  1. May 14, 2006 #1
    I am supposed to balance the coulomb repulsive and gravitational force in a way that both forces between the earth and the moon have the same amount.
    For that I can just use

    [tex]F_C = F_G[/tex]
    [tex]1/(4\pi\epsilon_0) * (Q_1*Q_2)/(r^2) = f (m_1 * m_2)/r^2[/tex]

    and then put Q_1*Q_2 on one side.

    Now I have to assign a charge to the earth and the moon, which is where I am at a loss. My thought was that the coloumb repulsive force of the moon has to be equal to the earth's gravitational force and vice versa, so that I can do something like [tex]F_C_E / F_C_M = F_G_M / F_G_E[/tex] and with that calculate Q_1 and Q_2. However, I can just calculate the gravitational force between earth and moon and not only the force that the earth exerts on the moon, which I would need for the approach used above.

    Would it be sufficient to relate the masses of earth and moon to each other and the ratio of both coulomb forces would be equal to that?
     
  2. jcsd
  3. May 14, 2006 #2

    Andrew Mason

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    There is only one coulomb force between the charged Earth and the charged moon. Assume that you have the same charge on the earth and the moon. If the Coulomb force balances gravity:

    [tex]kQ^2 = GMm[/tex]

    [tex]Q = \sqrt{GMm/k}[/tex] where [itex]k = 9 x 10^9 Nm^2/C^2[/itex]

    AM
     
  4. May 14, 2006 #3
    I worded it wrong, that's of course what I meant. However, in another part of the exercise I have to balance the coulomb force between the earth and the moon, and for that I need some kind of ratio. Would this ratio be equivalent to the ratio earth's mass / moon's mass or doesn't it matter? I mean something like [tex]Q = Q_E / Q_M = Mass_M / Mass_E[/tex].
    They specifically asked for this and I've never seen them ask for something that doesn't matter, which is why I am asking.
     
    Last edited: May 14, 2006
  5. May 14, 2006 #4

    Andrew Mason

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    ?? You balance the force by making the coulomb force equal to the gravitational force. You can do that by adding the same charge Q to each of the moon and earth. Why would you add different charges?

    AM
     
  6. May 14, 2006 #5
    You're right. I'm stupid. ;)
    The question specifically asked for how Q has to be divided between earth and moon and I guess I got hung up on that.
    Thanks for the help! :)
     
  7. May 15, 2006 #6

    Andrew Mason

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    Well you certainly aren't stupid. I hope I did not make you feel that way. The question demonstrates the tremendous difference between the coulomb and gravitational forces. The charge required is on the order of 5x10^13 Coulomb.

    Since a Coulomb is 6.24x10^18 electrons and a mole has 6 x 10^23 atoms, one mole of ions each with a charge of +1 has a charge of 10^5 coulomb. This means that 5 x10^8 moles of ionized hydrogen, (about 5x10^5 Kg or 500 Tonnes of ionized hydrogen) on the earth and moon would do it.

    AM
     
  8. Feb 23, 2011 #7
    What would be the total potential energy of the system, seeing as there would be a +ve electrostatic potential energy and a -ve gravitational energy? Would the total be zero?
    James
     
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