1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Required charge of Q for planets to produce observed magnitude of gravitational force

  1. Jan 9, 2013 #1
    1. The problem statement, all variables and given/known data
    The similarity of form of Newton’s law of gravitation and Coulomb’s Law caused some to speculate that the force of gravity is related to the electrostatic force. Suppose that gravitation is entirely electrical in nature—that an excess charge Q on the Moon and an equal and opposite excess charge –Q on the Earth are responsible for the gravitational force that causes the observed orbital motion of the Moon about the Earth. What is the required size of Q to reproduce the observed magnitude of the gravitational force?

    2. Relevant equations
    Fe=kQ1Q2/(r^2)
    Fg=GM1M2/(r^2)

    I used these values:
    The gravitational constant (G) = 6.67 x 10^-11 Nm^2 / Kg^2 .
    The mass of the earth (m1) = 6.0 x 10^24 Kg .
    The mass of the moon (m2) = 7.35 x 10^22 Kg .
    The average distance from the center of the earth to the center of the moon is 384,400,000 m
    k=9 x 10^9 Nm^2/C^2

    3. The attempt at a solution
    I set Fg equal to Fe and using the above values found the value of Q to be 1.63 * 10^27 C, but it is incorrect. Any idea where I went wrong? Thanks
     
    Last edited: Jan 9, 2013
  2. jcsd
  3. Jan 9, 2013 #2

    gneill

    User Avatar

    Staff: Mentor

    Re: Required charge of Q for planets to produce observed magnitude of gravitational f

    Hi dfetnum. Welcome to Physics Forums.

    You'll have to show more of your work so that we can see what's gone wrong. How did you arrive at the number you found?

    Note that if you equate your two force expressions that you can cancel out the distance (r2) variable right away.
     
    Last edited: Jan 9, 2013
  4. Jan 9, 2013 #3
    Re: Required charge of Q for planets to produce observed magnitude of gravitational f

    Thanks!

    So first I set Fe=Fg:

    G*M(earth)*M(moon)/(r^2) = k*(QE)(QM)/(r^2)

    radius's cancel out

    G*M(earth)*M(moon) = k*(QE)(QM)

    The problem stated that the charges are equal so:

    G*M(earth)*M(moon) = k*2Q

    Then I plugged in:

    (6.67 x 10^-11 Nm^2 / Kg^2) * (6.0 x 10^24 Kg) * (7.35 x 10^22 Kg) = (9 x 10^9 Nm^2/C^2) *2Q

    I used a calculator to find Q
     
  5. Jan 9, 2013 #4

    gneill

    User Avatar

    Staff: Mentor

    Re: Required charge of Q for planets to produce observed magnitude of gravitational f

    Ah. Q*Q ≠ 2Q :wink:
     
  6. Jan 9, 2013 #5
    Re: Required charge of Q for planets to produce observed magnitude of gravitational f

    WOWWWWWWWWWW, how could I have done that. I spent way too long on this problem.

    thanks!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Required charge of Q for planets to produce observed magnitude of gravitational force
Loading...