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Why does Moon cause larger tides than Sun?

  1. Dec 11, 2007 #1
    1. The problem statement, all variables and given/known data
    The gravitational attraction of the Sun and Moon on the Earth produces tides. The Sun's tidal effect is about half as great as the Moon's. The direct pull of the Sun on the Earth, however, is about 175 times that of the Moon. Why is it that the Moon causes larger tides?

    2. Relevant equations

    3. The attempt at a solution

    I read on several web pages that the tides are due to the difference in gravitational field, not the direct gravitational force, but I have no idea what that means. I think the "direct pull of the Sun on the Earth" is the gravitational force between the sun and the center of the earth. Somehow the tides need to be caused by something else, maybe the way the Earth is orbiting or spinning? Clearly I'm missing something here....
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Dec 11, 2007 #2
    The tidal force produced by a massive object (Moon, hereafter) on a small particle located on or in an extensive body (Earth, hereafter) is the vector difference between the gravitational force exerted by the Moon on the particle, and the gravitational force that would be exerted on the particle if it were located at the center of mass of the Earth. Thus, the tidal force depends not on the strength of the gravitational field of the Moon, but on its gradient. The gravitational force exerted on the Earth by the Sun is on average 179 times stronger than that exerted on the Earth by the Moon, but because the Sun is on average 389 times farther from the Earth, the gradient of its field is weaker. The tidal force produced by the Sun is therefore only 46% as large as that produced by the Moon.

    mmm i got that from wikipedia.com so im not taking credit for it or anything
     
  4. Dec 11, 2007 #3
    Gravitational force falls off as (distance)^2 (From Newton's Law). Thus, even though the sun is much more massive than the moon, it's gravitational strength falls off as 1/(distance)^2 so it can still have a greater gravitational effect on the Earth than moon. Tidal forces however, fall off as 1/distance cubed--this means that distance plays a much larger role, and allows the moon to have a greater tidal force on the earth just because it is so much closer.
     
  5. Dec 11, 2007 #4
    W(attraction between two bodies)=G*(m1+m2)/d^2
    Where G = 6.7 * 10^-11

    Use this formula, as the others explained earlier.
     
  6. Dec 11, 2007 #5
    Thank you all for your replies. I still don't understand why tidal force falls off at 1/distance cubed. Is there a formula for that? Can you please explain?
     
  7. Dec 12, 2007 #6
    http://en.wikipedia.org/wiki/Tidal_force

    r is the radius of the earth for our example and R is the distance from the sun to the center of the earth.

    What they're doing is calculating the difference in gravitational force on opposite sides of the body (the earth in our example). So, how much more strongly is the sun pulling on the side of the earth that is facing the sun as opposed to the side that is farthest from the sun. When they pull out the R^2 term in the denominator, that leaves a delta r divided by R, and in our example the distance to the sun is much greater than the radius of the earth so it is ok for them to do a taylor expansion which leaves an R cubed in the denominator when you take the important terms. This is why tidal forces fall off as R^3
     
  8. Jan 25, 2012 #7
    How much of Earth's Tides is caused by the rotation Moon and Earth around an axis and the Earth Orbiting the Sun?

    I've seen tides explained both the way everybody talks about it above as well as describing it as an effect of the inertia from the rotation, or what people mistakenly call centrifugal force.

    Help Please?

    Thanks in advanced
     
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