Singularity in Newton's gravitational law

Question

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?

Answer 1

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 $r$.

Answer 2

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.

Answer 3

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.