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Electrical Attraction

  1. Jan 16, 2009 #1
    Here is my problem:

    Suppose that electrical attraction, rather than gravity, were responsible for holding the Moon in orbit around the Earth.
    If equal and opposite charges were placed on the Earth and the Moon, what should be the value of Q to maintain the present orbit? Use these data: mass of Earth=5.98*10^24 kg, mass of Moon = 7.35*10^22 kg, radius of orbit=3.84*10^8 m. Treat the Earth and Moon as point particles.


    I know that I use Gm1m2/r^2 = kQ2/r^2. But I cannot get the right answer!

    Help please :)
     
  2. jcsd
  3. Jan 16, 2009 #2

    rl.bhat

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    I know that I use Gm1m2/r^2 = kQ2/r^2. But I cannot get the right answer!
    You have to use m2v^2/r = kQ2/r^2 where m2 is the mass of the moon and v is the velocity of the moon around the earth.
     
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