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Newtonian Gravitation Paradox ?

  1. May 13, 2008 #1
    Consider two point mass particles m1 m2 (with m1 different from m2) initially stationary relative to each other a distance x apart . The only force acting on them is attractive Newtonian gravity, i.e. Gm1m2/x^2 acting equally in opposite directions along the line of separation between the particles.

    Using Newton's second law F= ma, the acceleration of m1 relative to m2 is Gm2/x^2; while the acceleration of m2 relative to m1 is Gm1/x^2. So they have different accelerations, and hence will always have different speeds, relative to each other. So the separation is reducing at 2 different rates for the 2 particles !

    But there can only be one speed of (reducing) separation between 2 particles ?

    Paradox or elementary mistake ?
  2. jcsd
  3. May 13, 2008 #2

    Doc Al

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    Staff: Mentor

    No, those are accelerations with respect to an inertial frame. To get the acceleration with respect to each other you would add them.
  4. May 20, 2008 #3
    Perhaps. Of course the addition may give a numerically correct answer - but we then need to consider what is "an inertial frame" ? As gravity cannot be turned off, do we then accept that all inertial frames have zero mass ?
  5. May 20, 2008 #4
    Yes!!! yay I had the exact same question!!! My physics tearcher and a another random physics teacher was saying how I was insane.... lol

    hmm so in my mind the accelerations should be added.... This can cause that not all object fall at the same speed!! (well or acceleration, let just say that there's a certain distance and the time it takes them to fall towards each other with no resistance is not the same!!) Hmm this is talked about in the forum as well I think.... and also in the Physics I of the MIT Open lectures... (last 10 min)

    Ps. I'm only in high school, don't take my word....
  6. May 20, 2008 #5
    Oh the experiment was done at a harvard tower.. but nothing came out...
  7. May 20, 2008 #6
    But those teachers do make similar stupid mistakes. A first year student was given this problem by his Prof. He posted that question on a homework help forum (not here on PF)

    The question was something like:

    The Earth collided with a Mars sized object (Theia) 4.5 billion years ago that gave rise to the Moon. The impact velocity relative to Earth was about 11 km/s. Assuming that the collision was inelastic, what was the change in the momentum of the Earth.

    A few more details were given. Now this was a question given to a first year student who are not physics students. They are medical students who have to take elementary physics classes.

    In the question they were asked to solve a simple conservation of momentum problem in case of an elastic collision and the Prof. simply made it more interesting. But by doing so he forgot that the 11 km/s came almost 100% from the gravitational acceleration of Theia. Total momentum is conserved and Theia can be said to have approached earth from infinity with zero momentum.

    So, told the student to understand this, write it up, and teach the teacher a lesson. :approve:
  8. May 20, 2008 #7
    Haha... I should teach my teacher about that!! lol (His a awsome teacher)

    I never really learn momentum in school though I know the basic. How did the 11km/s came about?
  9. May 20, 2008 #8

    11 km/s is the escape velocity. At that velocity you can just escape Earth's gravity. This means that in the limitof infinite distance to Earth your velocity would just be zero. If you look at it in everse yo then see that an object tha comes from infinity would have to crash into Earth with a speed of 11 km/s at least. The limit of 11 km/s is reached when the object has a zero velocity at infinity.

    In case of Theia we know that this is a good approximation. Theia formed atthe same distance as the Earth from the Sun. So Theia would not have a significant velocity relative to Earth "at infinity".
  10. May 20, 2008 #9
    Then don't you have to consider the escape velocity of earth?
  11. May 20, 2008 #10
    Elememntary mistake on your part Quan. Your basis for the equation (in bold above) is wrong.

    The 'x' used in your equation should be the distance to the CENTER of MASS of the system....(You are trying to use it as the distance BETWEEN the two bodies...which can only be done in the approximation).
    Two massive bodies rotate around a barycenter, ( or fall toward the center of mass of the system),...and 'x' is the distance of EACH from the Barycenter.....therefore you must use DIFFERENT x's in each equation ...which balances out the accelerations. .

    If you don't know the barycenter (which is usually the case) you simply use a force equation which includes what is commonly called the "reduced mass"....usually called 'mu'....and must be used for the 'effective' inertial mass when applying f = ma ....the problem is solved as a one-body problem....(which is done in binary systems).

    The correct formulation for 'reduced mass' to take BOTH masses into account is ...

    'mu' = Reduced mass = (M + m) / (M*m)

    Thus F = "mu"(a)

    See here for a description of "reduced (gravitational) mass"...
    (scroll to last section at bottom first)

    Or see here for a description of how to do it specifically.

    No paradox...just correct application of the appropriate equations. :biggrin:

    Last edited: May 20, 2008
  12. May 21, 2008 #11

    Doc Al

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    Staff: Mentor

    Realize that for the simple case offered by Quandarian, where the masses start from rest, the use of "reduced mass" is equivalent to simply adding the accelerations as I stated in my first post.
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