If $r=0$ in the well know equation $F= G\dfrac{m_1\cdot m_2}{r^2}$, it will not follow that the force will be infinite?
May someone please clarify it to me?
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4If $r = 0$, you are putting two masses on the same spot. Does that make sense to you? – ACuriousMind Jul 15 '14 at 21:03
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Technically speaking, $1/0$ is an undefined operation and not infinity. – Kyle Kanos Jul 15 '14 at 21:05
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@ACuriousMind correct, but if $r$ tends to zero, then the two bodies will very very close, then $F$ may tend to infinity? – Voyager Jul 15 '14 at 21:09
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Yes, F tends towards infinity as r approaches 0. So? – Mark Janssen Jul 15 '14 at 23:57
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Assuming that $m_1$ and $m_2$ take up a finite amount of space (e.g., two spheres of mass with radius $r_0$), that equation isn't even valid for $r < r_0$, so there's no inconsistency.
The derivation follows from Gauss' law; it is analogous to the application of Gauss' law in electrostatics; the $m_1$ and $m_2$ are the mass enclosed at some distance $r$.
user3814483
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@ticster I believe it does address the OP's concern by pointing out a conceptual/physical error in the question itself. – BMS Jul 15 '14 at 22:21
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Not really, you're allowed to ask about point particles. It's not a conceptual error to do so. – ticster Jul 15 '14 at 22:24
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@ticster Yes, but how can two point particles coexist (same physical location) in the classical sense? I figured that was obvious and pointed out by the comment beneath the question so that left one alternative to which I provided an explanation. Besides, the OP didn't specifically ask about point particles and the answer's been accepted. – user3814483 Jul 15 '14 at 23:44
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@user3814483 I don't see what's hard point particles coexisting in the classical sense. Adding certain symmetries could indeed lead to the kind of question he's asking. On a cartesian grid $(x,y,z)$ imagine two gravitating masses $m_1$ at $(-1,0,0)$ and $m_2$ at $(1,0,0)$. What happens when they intersect ? – ticster Jul 15 '14 at 23:48
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1@ticster You're welcome to write your own answer, and explain the limits and breakdown of classical mechanics, how point particles are idealized and as such can't physically exist so it doesn't make sense to answer that question unless you bring in QM. I stated my assumptions in answering the question. – user3814483 Jul 16 '14 at 00:15
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True point masses and other singularities can wreak all kinds of havoc in Newtonian physics. A couple of examples:
- Particles can attain infinite velocity in finite time: Saari, D., and Zhihong J. (1995), "Off to infinity in finite time." Notices of the AMS 42:5.
- Particles can exhibit non-deterministic behavior. See this question, Norton's dome and its equation, and also see Norton, John D. (2008) "The dome: An unexpectedly simple failure of determinism." Philosophy of Science 75:5, 786-798.
Fortunately, true point masses and singularities such as those exhibited by Norton's Dome don't exist in reality.
David Hammen
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Assuming two mass particles, there would be a miniumum r>0 represented by the particles themselves therefore: F= G*(m1*m2)/[r(m1)+r(m2)]^2.
Tim D
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