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Caught Between Two Gravitational Fields

  1. Jul 4, 2013 #1
    If a body was between two gravitational fields, would it feel a tension force like the two gravitational forces were pulling it apart, or would the two fields cancel each other out and the body wouldn't feel anything?

    I'm assuming in this scenario that both fields are big enough so that there are no significant tidal forces to worry about. I'm thinking the two would cancel each other out, but I'm unfamiliar with general relativity so I'm not certain.
     
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  3. Jul 4, 2013 #2

    WannabeNewton

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    In Newtonian gravity the fields would superpose linearly but there is nothing that says a priori they must cancel out completely. It depends on the configuration of the two fields being superposed. In GR the fields don't superpose linearly as the EFEs are non-linear equations (unlike Newton's equations) so there will be complex non-linear interactions (gravitational waves may be released etc). Regardless, you have to be more specific about what field configurations you are considering because even in Newtonian gravity the fields don't have to exactly cancel as stated above (the most trivial counterexample is to consider two particles of different masses placed on the real line-examine what happens to a test mass placed to the left and right of the entire system as well as in between the two masses). In Newtonian gravity, if you take the vector sum of all the gravitational forces on a test mass at a given point, it will unequivocally get pulled in the direction of the resultant force at that given point.
     
    Last edited: Jul 4, 2013
  4. Jul 4, 2013 #3
    A good example of this is the gravitational tug between the Earth and the moon. You will feel a net gravitational tug from the Earth until you are about 40,000 miles from the moon, at which point you pass a Lagrangian point where the two gravitational forces are perfectly balanced. At this point, you would feel no net gravitational force. One you pass that Lagrangian, you are, in effect, falling towards the moon and under the principal tug of the moon's gravitation.

    http://www.thefullwiki.org/Lagrangian_point
     
  5. Jul 5, 2013 #4

    A.T.

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    Even at points where forces don't cancel, the bodies wouldn't "feel" anything, if the tidal forces are negligible. Gravity just causes coordinate acceleration. You need a gravity gradient to cause internal stresses, that can be "felt" locally.

    General Relativity also has points between two masses where the object would remain in an unstable equilibrium. But unlike Newton, General Relativity doesn't just predict the movement in free fall. An object at such a point would still experience gravitational time dilation compared to a distance clock. So not all gravitational effects cancel at this equilibrium point in General Relativity.
     
  6. Jul 5, 2013 #5

    HallsofIvy

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    The "tension force" is precisely what meant by "tidal forces" so saying "there are no significant tidal forces" you are a-priori saying there will be NO "tension force like the two gravitational forces were pulling it apart".
     
  7. Jul 5, 2013 #6

    pervect

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    If you have weak fields, you can use Newtonian mechanics (and/or linearize GR) and then say that the total tidal force "pulling apart" a body at the equilibrium point is the vector sum of the tidal forces due to each body.

    So midway between the earth and moon, the tidal forces along the earth-moon axis would be (2 G M_earth)/d_earth^3 + (2 G M_moon)/d_moon^3.

    (I hope the notation is self-explanatory).

    There would be compressive tidal forces of half this magnitude in the transverse direction.
     
  8. Jul 5, 2013 #7

    D H

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    That's a common misunderstanding. The first Lagrange point is *not* the point at which the gravitational forces cancel. The point at which gravitational forces cancel is between the first Lagrange point and the Moon. Other than it's value as a torture device for students (i.e., "find the point between the Earth and the Moon at which the net gravitational force on some object is zero"), that point at which gravitational forces cancel is of minimal physical interest.


    Units! You should know better. That expression has units of mass/time2. You need to multiply by the size of the object projected onto the line between the Earth and the Moon. Unless the object is very large this is going to be a very, very small force.
     
  9. Jul 5, 2013 #8

    Bill_K

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    And also times the mass of the other object. The Newtonian force is Gm1m2/r2. Because this is a force we're talking about, whereas the Riemann tensor gives you a differential acceleration.
     
  10. Jul 5, 2013 #9

    D H

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    Ouch.

    What I did is akin to correcting someones bad grammar but making a grammar mistake in the process.
     
  11. Jul 5, 2013 #10
    I like this answer.

    For the OP: 'Gravity just causes coordinate acceleration' means not one that is felt as a force...not physical. Not like accelerating in car.

    If the OP meant by 'no tidal forces' 'a uniform gravitational field from an infinite plane', then it would seem the only 'force' felt by the body would be elongation stresses between different parts of the body...perpendicular to the plane [like 1/2 the spaghettification discussed in black holes]...

    due to the 'gravity gradient' if the field weakens with distance from the plane.
    [Is this 'field weakening' even possible ??]

    Otherwise, it's just a body in free fall and feels no forces....
     
    Last edited: Jul 5, 2013
  12. Jul 5, 2013 #11

    D H

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    Same here. There's just not much of interest at that point. Not from an engineering perspective; there's nothing to take advantage of. Not from a Newtonian perspective; there's no interesting dynamics. Not from a relativistic perspective; there's nothing to feel, assuming tidal forces are small.

    The only interesting thing about this point is the huge number of times we at Physics Forums have to help introductory physics solve for the location of this point.
     
  13. Jul 5, 2013 #12
    Me too, that pretty much answers it, thanks. As long as the gravitational acceleration doesn't change significantly over the length of the body, it's pretty much just in free fall.
     
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