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Question re: Newton's 3rd law

  1. Apr 21, 2014 #1
    I'm looking for knowledgeable help to resolve an argument between myself and someone else.

    A bowling ball is positioned, say one hundred thousand miles from the Earth, motionless, relative to the Earth. It then begins to fall, finally crashing into the planet after some period of time.

    Another planet, of similar mass and dimensions of the Earth, is positioned at exactly the same distance as the ball, also motionless. The two planets begin to pull together and after some period of time, collide, spoiling a lot of people's day.

    Would it take the bowling ball longer to collide with the Earth than the other planet, the same amount of time, or less time?
     
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  3. Apr 21, 2014 #2

    Matterwave

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    It would take the bowling ball longer. The other planet will pull the Earth towards it, while the bowling ball's effect on the Earth would be negligible. It will take the bowling ball a factor of sqrt(2) longer than the planet.

    See: http://en.wikipedia.org/wiki/Free-fall_time
     
  4. Apr 21, 2014 #3

    Drakkith

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    The bowling ball accelerates towards the Earth at about 9.8 m/s2. The Earth accelerates towards the bowling ball at... well, for our discussion "virtually zero" is accurate enough.

    A planet similar to Earth would accelerate towards the Earth at about 9.8 m/s2 and the Earth would begin to accelerate towards it by about 9.8 m/s2.
     
  5. Apr 22, 2014 #4

    Nugatory

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    OP specified an original separation of about 25 earth radius, so the gravitational accelerations will be appreciably less than 1g for most of the travel... but this is more of an additional observation than a correction here. Drakkith's key point, that the earth and the ball/planet move towards their mutual center of gravity, holds.
     
  6. Apr 22, 2014 #5
    I appreciate the quick responses, thanks.

    I have been debating this issue with a guy who is better educated in physics than I am. He was telling me about Newton's third law of gravitation and that a 100 pound object would fall to Earth at exactly the same speed as a 1 pound object.

    I said yes, but in theory, the heavier object should fall a tiny bit faster since it exerted more gravitational pull upon the Earth than the smaller one. Of course that difference would be incredibly small and immeasurable in Newton's time.

    I then said, what if we replace those two objects with an entire planet and a bowling ball, and asked him the same question that I posed to the forum.

    His response was that the bowling ball would compensate for the greater gravitational pull of the planets, by accelerating faster, and that it would strike the Earth in the same amount of time as the planet.

    Any thoughts on that, pro or con?
     
  7. Apr 22, 2014 #6

    Andrew Mason

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    This is just a matter of applying the Law of Universal Gravitation. The forces of each of the two planets (earth and planet x) on the other would be:

    [itex]\vec{F} = -\frac{GM_eM_x}{R^2}\hat{R}[/itex] where [itex]\hat{R}[/itex] is the unit radial vector from the centre of the earth/planet x and R is the distance between the centres of the two bodies.

    The accelerations are, initially, the same. However, because the earth would move toward the planet x and vice versa, the force/acceleration would increase more rapidly than in the bowling ball case. Also, because planet x is bigger than the bowling ball, the separation between their surfaces is less so there is less initial distance to cover in the "fall".

    AM
     
  8. Apr 22, 2014 #7

    Pythagorean

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    Newton's third law is appropriate here:

    [tex]F_{21} = -F_{12}[/tex]

    and Newton's second:

    [tex]F = ma[/tex]

    which means:

    [tex]m_2 a_2 = -m_1 a_1[/tex]

    then, solving that for [itex]a_2[/itex] and substituting below, the relative acceleration between the two bodies would be:

    [tex]a_2 -a_1 = (1+\frac{m_1}{m_2})a_1 [/tex]

    If [itex]m_1 \ll m_2[/itex], as in the bowling ball case, then the relative acceleration is just the acceleration of the bowling ball, [itex]a_1 = G\frac{m_2}{r^2}[/itex] where [itex]m_2[/itex] is the mass of Earth. [itex]a_2[/itex] is essentially 0.

    Now... with two planets, [itex]m_2[/itex] is still the same, so [itex]a_1[/itex] is initially still the same, but in our coupling term above, you now can't neglect the additional [itex]\frac{m_1}{m_2}[/itex] term and you can see how their mutual acceleration can be twice as great, for instance, if [itex]m_1=m_2 [/itex].
     
    Last edited: Apr 22, 2014
  9. Apr 22, 2014 #8
    Your friend is correct for most cases of object's falling.There won't be a measurable amount of difference between objects of 1 pound and 100 pounds.
    What is different is when large bodies fall. When two Earth sized planets attract each other they both fall at the same rate towards each other, therefore there closing speed is greater, than say a bowling ball falling towards one of them.

    The bowling ball attracts the earth, and the earth attracts the bowling ball but the attraction of the bowling ball on the earth, is no way as strong as another earth sized planet.

    The two earth sized objects only have to travell half the distance before the collide so they take less time.
     
    Last edited: Apr 22, 2014
  10. Apr 22, 2014 #9

    rcgldr

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    This depends on if you define "fall" as the change in speed of the object relative to it's initial state, in which case the size of the object doesn't matter, or if you define "fall" as the change in closure rate between the object and earth, in which case the size matters. The time to impact will be a function of closure rate.
     
  11. Apr 22, 2014 #10
    Thanks again for the detailed replies. But the crux of the argument with my friend is my original claim that a 100 lb object would fall faster than a 1 lb object, for the same reason that a planet would.

    Is there a dividing line somewhere? Would the moon fall faster than the bowling ball?

    How about a one mile wide asteroid?

    It seems to me, in my infinite ignorance, that the same rule should apply in all cases.
     
  12. Apr 22, 2014 #11

    Pythagorean

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    I think if you look at my analysis you can see that yes, the bigger the object (m1) for a fixed m2 (earth), the greater the acceleration.
     
  13. Apr 22, 2014 #12
    Would the two planets accelerate towards each other at 9.8 m/s2 as they have twice the mass of one earth combined.
     
  14. Apr 22, 2014 #13
    ModusPwnd, I understand (sort of) what Newton is saying and that a feather would fall to the ground at the same, measurable speed as a hammer.

    What I am suggesting is that since the hammer is pulling slightly harder on the Earth than the feather, that it would fall slightly faster - the difference of course, being too small to measure.

    If that is correct, then the difference becomes much greater when we consider very large objects, like moons, asteroids, planets, etc.

    It seems that these various objects have one thing in common. They are each pulling on the Earth and the Earth is pulling on them. Therefore, shouldn't the same rule apply in each case?

    If so, then if a planet will collide with the Earth faster than a bowling ball, a 100 lb object should collide faster than a 1 lb object.
     
  15. Apr 22, 2014 #14

    Pythagorean

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    That's theoretically correct. The feather pulls on Earth. The hammer pulls on Earth more.
     
  16. Apr 22, 2014 #15

    Andrew Mason

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    I think you meant to say:

    Newton's third law is appropriate here:

    [tex]F_{21} = -F{12}[/tex]

    and Newton's Second:

    [tex]F = ma[/tex] ...
    ??? You are confusing us! The force is greater but the acceleration is the same relative to an inertial point.
     
  17. Apr 22, 2014 #16

    Pythagorean

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    Oops, yeah, wrong numbers on Newton's

    But we're talking about the relative acceleration. I explicitly took the difference between a1 and a2 as such. Robert Harris is considering the motion of the Earth, too. So it is indeed larger for a larger m1, given that m2 is fixed (Earth).

    Remember the OP's question:

     
  18. Apr 22, 2014 #17

    Andrew Mason

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    If you raise a kilogram mass by one metre, the centre of the earth has to move away from its initial position by 1/.594e25 = 1.7e-25 m.

    To give you an idea of how much this is, a hydrogen atom is about 10e-10m in diameter and a proton diameter is about 10e-15 m (femtometre). So a kg. mass falling 1 m. is going to move the earth about one ten-billionth of the diameter of a proton.

    Note: If you happen to drop a hammer and a feather at the same time, they take identical times to fall.

    Also: if you drop a hammer at the same time that someone else is doing the same thing on the diametrically opposite side of the earth, the earth, even in theory, does not move at all.

    Since there are things being raised and lowered all the time around the earth, whether the feather or hammer takes a longer time to drop, when dropped separately, really depends on what else is happening around the earth when you do the experiment.

    AM
     
  19. Apr 22, 2014 #18

    Pythagorean

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    As rcgldr pointed out, it depends on how you define fall. I get the impression OP and I are defining fall differently than you (relative distance vs. inertial reference frame).
     
  20. Apr 22, 2014 #19

    Andrew Mason

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    If they are dropped at the same time from the same height why would one reach the surface of the earth before the other? The earth cannot accelerate differently for each if they are dropped at the same time.

    AM
     
  21. Apr 22, 2014 #20

    Pythagorean

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    Oh, I see what you mean. In both definitions, they would contact earth at the same time under those initial conditions, despite giving different values for the total acceleration. I agree.
     
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