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Tidal Forces

  1. Jun 2, 2005 #1
    Why are tidal forces inversely proportional to the cube of the distance, and not to the square, as normal gravity force?
    Where can I find more ifo about this?
    Thanks in advance.
  2. jcsd
  3. Jun 3, 2005 #2
    Who said that? The tidal force within a small region of a local Cartesian system in free-fall has forces which are linear in the force. Those forces along the radial axis point away from the origin while those perpendicular to the radial point towards the origin. It is the origin of the coordinates which varies as the inverse cube. See Eq. (4) in http://www.geocities.com/physics_world/mech/tidal_force_tensor.htm

    The gravitational potential is proportional to 1/r so the second derivative is proportional to 1/r3.

  4. Jun 7, 2005 #3


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    See an explanation http://stommel.tamu.edu/~baum/reid/book1/book/node36.html
    Since they vary with the cube of the distance, tidal forces are very weak. That is why the small Moon exerts on the Earth a much stronger tidal force than the gigantic Sun. If they were linear, as pmb_phy says, the Sun would prevail.
    Last edited by a moderator: Apr 21, 2017
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