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Argument with a co-worker about gravitation

  1. Dec 8, 2005 #1
    I'm having an argument with a co-worker about gravitation and need some "enlightened" advice.

    The argument is whether or not two objects dropped at the same time would impact the surface at EXACTLY the same time or not.

    I already presented the proof that when combining the law of gravitation:

    F = G*mo*me/r^2


    F= mo*ao

    That the mass of the object (mo) falls out and you are left with the acceleration of the object (ao) only being dependant upon the mass of the other body (earth) and the radius.

    She is arguing that if you actually calculate this, then both objects will accelerate towards the earth at the exact same rate, but that the earth will be accelerating towards the objects at different rates, as when you do that calc, the mass of the earth drops out and you are left with acceleration being dependant on the mass of the object, which isn't the same in both cases.

    I've tried to explain this as a frame of reference issue, but she isn't biting. Can anyone give me a better, solid reasoning, or at least explain in clearer english, why this is not the case.

    or maybe it is and I'm on crack.....
  2. jcsd
  3. Dec 8, 2005 #2

    Doc Al

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    Sounds right to me. Realize that the earth and object exert equal forces on each other, and thus both accelerate towards their center of mass. So a bigger object will produce a greater acceleration of the earth. (Realize that this is just an amusing thought experiment: The acceleration of the earth would be fantastically small for any non-astronomical object.)

    It's not just a frame of reference issue, since the time it takes for object and earth to collide is frame independent (ignoring relativity).
  4. Dec 8, 2005 #3
    First you have to specify if you are considring only earth and masses in your system ( no other planets and starts). Then it depneds on the mass of objetcs you are thinking...If masses are small like 1 KG or 10 KG or 100 KG ...then the time will be nearly the same.....since the value of G is very small ...But if you take mass of falling objects as mass of Jupiter planet then surely time will be diiferent.

  5. Dec 8, 2005 #4
    well the argument got to the point of standing on the moon and dropping a feather and the earth and seeing which one would impact first. So we did encounter pretty large disparities of masses. and yes we are ignoring all outside forces. So it's just the gravitational forces between the objects involved.

    So what you are saying is that gravitational acceleration is actually dependant upon the mass of the object falling.

    Sorry, was this discussed before?

    I won't beat my point into the ground. If I'm wrong, I'll admit I'm wrong and move on.
    Last edited: Dec 8, 2005
  6. Dec 8, 2005 #5


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    Why do I have a bad Deja Vu feeling about this?


  7. Dec 8, 2005 #6

    Doc Al

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    The acceleration (with respect to an inertial frame) of the object would be pretty much independent of mass, but the time it takes to fall would not be, since the earth is accelerating upward (at different rates) to meet it.

    Why don't you calculate the accelerations of the earth and object for two cases: A 0.1 gram feather and a 10 kg shot put.

    Oh, yeah.... (I tried to find the thread, but couldn't.)
  8. Dec 8, 2005 #7
    Is it because it's a vexing question that you've encountered before.:wink:
  9. Dec 8, 2005 #8


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    If you want to do this argument, it is important to note that the Earth is not a point mass. It is a distributed system. See for instance


    So we have to ask - are these two objects being dropped in the same place?

    If they are, then any "motion of the Earth", which will not be that of a rigid body, will be such that the surface of the Earth rises the same amount. Thus the two bodies will strike the surface of the Earth at the same time, assuming you mean the physical surface.

    If the two objects are being dropped in different places, then they probably won't strike at the same time, but it will probably be due to local differences in the force of gravity (which is not totally uniform along the surface of the Earth), rather than the very small effect of the deformation of the Earth's surface.

    Perhaps you mean to ask, if you drop two bodies from the same height and measure the time to impact at the same location, if it will be exactly the same?
  10. Dec 8, 2005 #9


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    try using tex markup. this ain't USENET or ASCII only.

    so which rate is it? it's one earth, is it accelerating upward at

    [tex] a_e = G \frac{m_0}{r^2} [/tex]


    [tex] a_e = G \frac{m_1}{r^2} [/tex] ?

    since it is actually only one earth.

    perhaps both objects are attracting the earth, and perhaps their force vectors (pointing in the same direction) add. then it's

    [tex] a_e = G \frac{m_0+m_1}{r^2} [/tex]

    so the earth accelerates upward at the rate above and both objects accelerate downward at

    [tex] a_0 = a_1 = G \frac{m_e}{r^2} [/tex].

    so what is her reason to believe they hit the surface at different times?
  11. Dec 8, 2005 #10

    Doc Al

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    You're right!

    She was comparing the time for each object to fall if they were dropped separately, not together. If you dropped them side by side at the same time, then the two objects would hit the ground together. (Given the usual simplifying assumptions.)

    Given the actual statement of the problem ("would two objects dropped at the same time, hit the earth at the same time") my previous answers in this thread are wrong. D'oh! :redface:

    (And I'm sure we went through this same reasoning in the old thread and I still misread the problem! :rolleyes: )
  12. Dec 11, 2005 #11


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    Not true!

    You may have stated a simultanaiety in time, but not necessarily in space. You are assuming the Earth's acceleration toward the heavier object is also in the direction of the lighter object. If they are dropped far enough apart, the time for the heavier object to hit will still be shorter.

    In fact, if the two objects are dropped on opposite sides of the planet (we cannot assume you and your co-worker work in the same offices), the time discrepancy will be doubled rather than cancelling out.
    Last edited: Dec 11, 2005
  13. Dec 11, 2005 #12

    actually even if the two objects were dropped side by side I think the heavier object would still converge with the earth faster (assuming the earth and the two objects are perfect spheres). There would be a triangle effect. Two lines from the center of the earth directly to the centers of the two objects and also a line between the two objects pulling towards each other. The earth would still be more influenced by the line of movement that is directed at the heavier object.

    in other words, if the heavier object were on the right and the lighter on the left, the earth would move slightely towards the right as the objects are dropped.
  14. Dec 11, 2005 #13

    Doc Al

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    Note that I said "side by side", not just simultaneously. You could, of course, drop one object from a greater height. :rolleyes:

    But, yes, for an even higher order correction to an already ludicrously small effect, you'd have to worry about how their difference in position (after all they can't be exactly at the same place) affects the direction of the earth's acceleration.
  15. Dec 11, 2005 #14


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    You're going to have to tell her that you were wrong. There are two reasons for this.

    Since you're measuring the time at which they impact, the answer is clear. That is, you must calculate the acceleration of both the objects and the earth.

    A heavy object does attract the earth more than a light object, and so a heavy object will, in fact, fall to earth faster than a light object. For reasonable objects, like raindrops and aircraft carriers, the effect is quite small. But theoretically, the effect is there and can be calculated in exactly the way she suggested:

    [tex]F = G*m_o*m_e/r^2[/tex]

    [tex]F = F_o = m_o a_o -> a_o = G m_e/r^2[/tex]

    [tex]F = -F_e = m_e a_e -> a_e = -G m_o/r^2[/tex]

    So the total acceleration is:

    [tex]a_t = a_o - a_e = G (m_e + m_o)/r^2[/tex].

    In other words, if your heavier object is an aircraft carrier weighing [tex]6 \times 10^{7}kg[/tex], and your lighter object is a raindrop weighing, well, pretty much nothing, then the aircraft carrier will fall with an acceleration around 1 part in [tex]6 \times 10^{24} kg/ 6 \times 10^{7} kg = 10^{17}[/tex] larger.

    Since s = 0.5 a t^2, the time required for the aircraft carrier to impact the earth will be smaller by about two parts in [tex]10^{17}[/tex].

    Last edited: Dec 11, 2005
  16. Dec 11, 2005 #15
    Two different masses have different effect on the space-time distortion so I would say yes, the acceleration will be different
    Last edited: Dec 11, 2005
  17. Dec 12, 2005 #16

    Doc Al

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    Carl, you might want to check rbj's post above. While it's true that a heavier object will fall faster than a lighter one if dropped separately, that's not the case here. The two objects are dropped together, side by side, and thus hit the ground at the same time (ignoring the higher order correction due to the direction of the earth's acceleration slightly favoring the heavier mass).
  18. Dec 12, 2005 #17
    why would that be ignored?
  19. Dec 12, 2005 #18
    Because the effect is really really small.
  20. Dec 13, 2005 #19


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    You're right, I overlooked that. But the answer is still that the two masses don't hit at the same time.

    If the masses are dropped at the same time, it becomes a three body problem (in general relativity) and the math is a lot harder. But I'm pretty sure that when you're done crunching it all out, you will find that the collisions are not, in general, simultaneous.

  21. Dec 13, 2005 #20


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    Because the Earth is not an ideal uniform sphere, it's gravitational field varies slightly from place to place. (Geologists actually have maps of it).

    Thus if you drop two different masses from different places at the same time, they likely won't arive at the same time, but the dominant effect will be because of the difference in gravitational fields at the locations where they were dropped, not because of the unmeasurably tiny effect of the acceleration of the Earth by the test masses, as I remarked earlier in post #8.


    Also, as I mentioned, it would be highly inaccurate to view the Earth as being a point mass if one were to attempt to work out the effects on the Earth of a falling object.
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