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The G force is not infinity at distance zero(in free fall)

  1. Apr 27, 2003 #1
    i'm talking about droping a ball into a free fall under assumption that it can pass thru the earth's center all the way to minus height point.

    case 1:if you say that the force at the dropping point is F=G(/m/R)(M/R)<>0 then the equilibrium point is the center of the earth.but when the ball gets at the equilibrium point then the force is zero.

    case 2:if you say that the droping point is the equilibrium one then F at that moment is zero.but when the ball gets in the center of the earth then F=G(/m/R)(M/R)<>0 where R is the distance that the ball has passed to get there.

    actually there is another way of explaining this:
    E = F x D = energy = force x (equlibrium distance) = const.
    F x dD = - dF x D = D x dF > 0

    this means that all three vectors E,F,D rotate around the equilibrium point and the energy vector in the direction of the force.normatively all the vectors remain the same.only if you push the ball (i.e. invest some force in the system) then the normas change.pure and simple dynamics.the projections of the vectors depend on the starting conditions.let n be the vector of your aspect
    (if when the projection of D is max the projection of F is zero then n is parallel with F
    (if when the projection of D is max the projection of F is also max then n is parallel with E))
    then you should project all the vectors E,F,D on the plane normal with n.
  2. jcsd
  3. Apr 27, 2003 #2


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    I don't know why you posted this here, It really has nothing to do wtih theoretical Physics, it's pretty straight forward.

    F = GMm/r² works perfectly well as long as M is a point mass or in a case like the Spherical Earth all of M is located in a volume smaller then r.

    In the case of dropping an object through the Earth, this is not the case, as you fall deeper, more and more of the mass of the Earth is further from the center of "r" than you are. As a result the the Portion of "M" attributing to the force of gravity felt by the object gets smaller. If we call this portion M1 then we get

    F = GM1m/r²

    With r being the distance from the center at any given point.

    Since M1 approaches zero as we approach the center of the Earth, so does F.

    In a simple case, where we consider the density of the Earth as constant thoughout, we can find M1 by multiplying 4[pi]r³/3 by the density (d) and get:

    F = 4G[pi]rd/3

    Which of course approaches zero as r approaches zero.

    Again, just straight forward physics.
  4. Apr 28, 2003 #3


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    The question for a point mass singularity could be more interesting. On one hand, force goes to infinity. On the other hand, energy is preserved, thus the infinites of acceleration and deceleration somehow cancel, and surely you can even calculate the time it takes to cross throught a point mass.

    I have read that there is, at least for the three body problem, a change of coordinates which does not contain the singularity.

    What about classical electromagnetism? There, force is dependent of
    the velocity, and there are not a conservative potential.
  5. Apr 28, 2003 #4
    member i was talking about punctual (a point filled with) mass.the gravity law don't work with spherical mass but it assumes that all the mass is in concretated in one point=the center.whe you approach the center then you decrease the distance and since the falling object doesn't exchange energy, the force should aim for infinity.

    this is the case i'm observing.in this case i say:
    "it's not like that.when the falling object reaches the center then the force will be the one you calculate for distance = the one the object has passed starting from the release point to the center.in the moment of release the force was zero".
  6. Apr 28, 2003 #5


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    The assumption of all the mass being concentrated in the center of mass is only valid for points outside of the mass.

    IOW, it can only be used for that part of the mass that is closer to the center of mass than the point you are measuing the force of gravity at.
  7. May 11, 2003 #6
    And the wierd thing is, if the planet is Hollow, things inside it are weightless, assuming that it is a spherical shell of homogeneous density.
  8. May 14, 2003 #7


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    Craptastical. At the moment of release, the force was most definitely not zero, since if it was zero, the object wouldn't fall. Simple.
  9. May 27, 2003 #8
    The even weirder thing is that right at the center of gravity of earth, things are weightless. The earth does not need to be hollow (and isn't hollow) for that.
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