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I have a number of queries about the shell theorem, specifically the external case, which states that the gravitational pull of a spherically symmetric body always acts at the centre of the sphere.

Isn't the case for outside the shell easily proven using the fact that forces always 'act' on the centre of mass of a body (shown by applying Newton's second law). And using Newton's third law to state that an equal and opposite force is applied on the test object, (in the direction of the centre of mass)? Is this line of reasoning flawed; does it even apply (that gravity pulls towards the C.O.M of an object)?

A rigorous proof is given on wikipedia: http://en.wikipedia.org/wiki/Shell_theorem#Outside_a_shell

Also, I have reason to doubt my 'proof' because they claim that the shell theorem shows the converse of a more general form of Newton's laws which appears to have a terms proportional to $r$. However, my statement should apply to any force. I'm not sure why this is.

https://en.wikipedia.org/wiki/Shell_theorem#Converses_and_generalizations

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  • $\begingroup$ Comment to the post (v3): It seems that OP's own doubt answers OP's question in the negative: Pick a force which doesn't have the shell-theorem property (say, a central power law with a wrong power), and use OP's "proof" to reach a contradiction. $\endgroup$
    – Qmechanic
    Feb 16, 2017 at 20:23
  • $\begingroup$ yeah, though i would like to know why my method was incorrect $\endgroup$
    – user86425
    Feb 16, 2017 at 21:18
  • $\begingroup$ well, i suppose a counter example would be a pair of parallel vectors which don't necessarily point towards each other $\endgroup$
    – user86425
    Feb 17, 2017 at 11:27
  • $\begingroup$ Let's assume that the strong version of Newton's 3rd law holds. Even with this assumption, the "proof" is still not valid. $\endgroup$
    – Qmechanic
    Feb 17, 2017 at 11:31
  • $\begingroup$ yeah, I was pointing out the flawed reasoning in my argument. $\endgroup$
    – user86425
    Feb 17, 2017 at 11:33

1 Answer 1

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TL:DR: OP's "proof" is not valid.

  1. Consider a system of $N$ point particles with masses $m_i$ and positions ${\bf r}_i$, where $i\in\{1, \ldots, N\}$. Let the only external force $${\bf F}_i~=~m_i{\bf g}({\bf r}_i)$$ on the $i$th particle be from an external field ${\bf g}({\bf r})$, which is not considered part of the system, and which may not necessarily be external Newtonian gravity from an external point source.

  2. Let $$M~:=~\sum_i m_i$$ be the total mass, and $${\bf r}_{CM}~:=~\frac{1}{M}\sum_i m_i{\bf r}_i$$ be the center of mass.

  3. It is then true from Newton's 2nd law that $$M\ddot{\bf r}_{CM}~=~ \sum_i{\bf F}_i~=~\sum_im_i{\bf g}({\bf r}_i),$$ where internal forces cancel by Newton's 3rd law, but that does not necessarily imply that $$\ddot{\bf r}_{CM}~=~{\bf g}({\bf r}_{CM}).\qquad(\longleftarrow\text{In general wrong!}) $$

  4. In contrast, Newton's shell theorem considers the combined Newtonian gravitational field of the $N$ point particle organized spherically symmetric.

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