The proof of the Gauss's law for gravity provided by Wikipedia takes use of the divergence theorem.

  • Is it possible to arrive at the integral form of the Gauss's law in a way which doesn't require the use of divergence?

I'd like to derive it from the Newton's law. The main idea is: Assume that two points masses $m_1,m_2$ act on each other with force $F=\frac{Gm_1m_2}{r^2}$. Do something. Arrive at the integral form of the Gauss's law.

So it doesn't have much in common with the "duplicates"

I had one semester of analysis, so don't know anything about the Delta Dirac function and family

  • $\begingroup$ I am not sure I follow. Gauss' Law is a general mathematical property of fields that has nothing to do with physics. Which of Newton's laws do you want to use to derive Gauss? The three laws of motion don't say anything about fields and his law of universal gravitation is an ad-hoc force law, which, strictly speaking, also doesn't require us to think of fields. $\endgroup$ – CuriousOne Jan 6 '16 at 10:18
  • $\begingroup$ I mean the law of universal gravitation. About the divergence: I simply want to avoid the mathematical concept of divergences, since I'm not familiar with them $\endgroup$ – marmistrz Jan 6 '16 at 10:36
  • $\begingroup$ Newton's law of universal gravitation is an ad-hoc force law. It doesn't say anything about a field. You can, if you like, define the gravitational potential as a field and then check that it is a special case of Gauss' Law, but you can't derive a general law from a special case. We are talking about this, right? en.wikipedia.org/wiki/Divergence_theorem $\endgroup$ – CuriousOne Jan 6 '16 at 10:40
  • $\begingroup$ Yes, but we define the field magnitude as $E = F/m = \frac{GM}{r^2}$, where the $m$ is the mass acted on. And the Gauss law combines the mass $M$ with the field magnitude $E$ (so force too). Yes, I'd like to avoid the divergence theorem linked $\endgroup$ – marmistrz Jan 6 '16 at 10:43
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    $\begingroup$ Duplicate of physics.stackexchange.com/q/73028/5739 which itself is flagged as a duplicate. $\endgroup$ – garyp Jan 6 '16 at 14:20