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Effect of water displacement on gravitation pull between ships

  1. Dec 24, 2009 #1
    I calculated, using Newton's formula F = GmM/r^2, that the gravitational attraction between two large, full oil tankers whose sides are touching would be 1,331 lb., given the mass of each ship as 6.5 x 10^8 kg, distance between their CG's of 69 m, G = 6.674 x 10^-11 N(m/kg)^2, and 1 N = 0.2248 lb.

    The question is this: Does the water displaced by the two ships add, subtract, or have no effect on the force I feel if I am trying to hold the ships apart? and why?
     
  2. jcsd
  3. Dec 24, 2009 #2
    Compared to what? It's a little like asking if the number 5 is happy.

    If the water were instead inside the ships and they were just barely afloat, you can recalculate a larger attraction. If you calculate the attraction with the ships out of the water on an ideal frictionless surface you'll get your same answer. If they are on a typical real world surface with friction you'd measure no attraction but maybe calculate the same theoretical figure as you did because they'd be stuck in place on a dock or other surface.

    It would be better to think of your question with the ships empty or less full...or refilled with a liquid of different density....then you can easily answer your own question.
     
  4. Dec 26, 2009 #3
    Sorry if my question seemed trivial. At first I thought that the "holes in the water" formerly occupied by the displaced water might be modeled as having a gravitational repulsion to each other, due to all the surrounding water's gravitational attraction to itself. But each hole actually has a ship of the same mass sitting in it. So there is no lateral pull on the water and no lateral pull of the water on the ships. Thanks for your help.
     
  5. Dec 26, 2009 #4

    A.T.

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    Even if the holes were not filled, they would rather attract than repulse each other.
     
  6. Dec 27, 2009 #5
    Nice thought! I enjoy wondering about holes in the water. Not sure, but I think the holes might repel. Check out this hand-waving:)

    Think of a volume of ocean with motionless water.

    At a given depth beneath the surface, we make appear two ideal spheres. Since these perfectly massless, rigid, and watertight spheres have ideally thin walls, no water has yet been disturbed.

    Since gravity is universal, the water inside one sphere is attracted to the water inside the other sphere with a force given by Newton’s F = GmM/r^2. But the spheres remain motionless because of offsetting forces of gravity from the surrounding environment.

    Next we tie each sphere to the ocean bottom, using ideally strong, thin, and massless ropes. Each rope is attached in its line between its sphere’s water’s center of mass (CM) and the earth’s CM.

    Then we make the water inside the spheres disappear, creating two holes in the water of mass zero, thereby removing the earlier mentioned force of attraction.

    Absent this force of attraction, I think the holes would be pulled apart by the “forces of gravity from the surrounding environment” mentioned earlier. When the system again comes to rest, I think each rope would be pulling against an upward buoyancy vector plus a smaller lateral vector of f = GmM/R^2, where R > r.
     
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