Does the moon have greater gravitational force than the Sun on the Earth?

  1. I'm going nuts trying to figure this out... in text books and online, everything i read says that the moon and earth have a much stronger gravitational force between them than the sun and the earth, and this is why the moon has greater effect on tides than the sun. They all say that this is because the moon, though much less massive than, is much closer to the earth than the sun.

    But... every time i do the calculations using the G(m1)(m2)/r^2, i get that the force between the Sun and the oceans is 8.37 x 10^18 N, and the force between the Moon and the oceans is 4.65 x 10^16 N

    so my calculations say that the Sun has a greater force on the oceans than the moon... if i changed out the value of the oceans' mass with the earths mass, i still get that the sun has a greater gravitational pull on the earth than the moon...

    so why do all my sources say that the earth-moon gravitational force is greater??????
  2. jcsd
  3. D H

    Staff: Mentor

    You are confusing the gravitational force with tidal forces. The gravitational force varies with the inverse square of the distance between two bodies. Tidal forces result from the gradient of the gravitational force and thusly vary with the inverse cube of the distance. So while the gravitational force exerted by the sun on the earth is much greater than that exerted by the moon, the situation is reverse for tidal forces.
  4. Doc Al

    Staff: Mentor

    That equation gives you the gravitational force between the bodies, but that's not the tidal force. The tidal force depends on the change in gravitational strength at different points: It's the variation in pull on one side of the earth compared to the pull on the other that creates the tidal force. The tidal force is inversely proportional to the distance cubed.

    Read this: Moon as Dominant Tidal Source
  5. thank you very much for the response...

    but now i must ask, what is the "gradient of gravitational force" ??

    why do you cube instead of square?
  6. ah, this makes good sense to me...
    so tell me, do we see the moon having a greater periodic effect on tides than the sun because as the points on the earth vary frequently with the moon, they experience a greater relative distance change with the moon than with the sun?

    i thought that while searching around, i had read sources saying that gravitational force was greater, but they must have all said "tide-generating" potential or something like that (as does the earth science text book sitting in front of me)
  7. Doc Al

    Staff: Mentor

  8. Because gravity in our universe is a three dimensional concept (4 but lets ignore time). Squaring would be x*x but gravity is divided into three dimensions, not two.
  9. russ_watters

    Staff: Mentor

    Not really, no. What you are saying implies that Newton's gravitational force equation itself should be a cube function, but it isn't. And the tidal force equation comes simply from the derivative of that equation, since it is just the difference between two different gravitational forces: F1-F2=dF
  10. First of all, hello, this is my first post in this forum.
    Second, excuse me everybody for opening such an old thread, but I have a question regarding Tidal forces.
    How exactly did the OP come to 8.37 x 10^18 and 4.65 x 10^16 N respectively?
    When I calculate the gravitational force of the Sun and of the Moon on the Earth I always get -3.5436^22 and -1.9830*10^20 rounded.
    I`m using [​IMG]
  11. mgb_phys

    mgb_phys 8,809
    Science Advisor
    Homework Helper

    They are probably using the mass of the oceans rather than the mass of the Earth.
    If you are just comparing forces it doesn't matter - it would be easier to compare the force on 1kg at the Earth

    edit: if you use 1.37E21kg as the mass of the oceans it comes out about those numbers
    Last edited: Dec 8, 2009
  12. Oh, I see. The second part of this particular problem wants me to show that the difference in gravitational force on both sides of the earth is dF = (-2dr/r)*F.
    I started at first using this , but it`s actually about acceleration and doesn`t get the job done. Any ideas would be greatly appreciated. :)

    I mean simply calculating the derivative of the force gives the required result, but that seems too simple to me. Can it be?
    Last edited: Dec 8, 2009
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