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I am struggling to derive the gravitational field strength within a solid sphere. I am considering a point a position vector $\textbf{r}$, and a small mass element of the sphere within, at a position vector $\textbf{r}_m$. The mass of this element would be $dm=\rho dV$ or $dm=\rho r_m^2sin\theta dr_md\theta d\phi$ in spherical polar coordinates. Then I have the small contribution to the gravitational field strength at $\textbf{r}$ from the mass being

$d\textbf{g}=-\frac{Gdm}{|\textbf{r-r}_m|^2} \frac{\textbf{r-r}_m}{|\textbf{r-r}_m|}$

$\int d\textbf{g}=-\int_{V} \frac{G\rho (\textbf{r-r}_m)}{|\textbf{r-r}_m|^3}dV$

And now I am stuck. I don't know how to integrate such a vector expression or even if it can be done... I would greatly appreciate some help!

Qmechanic
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Meep
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  • Hint: use the shell theorem – John Dvorak Apr 30 '17 at 10:40
  • @JanDvorak Thank you for your reply! I was hoping to arrive at the result without invoking some other theorem, but perhaps I underestimated the number of steps required to prove this result? I was hoping that there was a way to progress with this integral, which doesn't seem to lend itself to simplification by the divergence theorem which is another approach I considered. – Meep Apr 30 '17 at 10:44
  • Are you aware of Gauss's law? – pmal Apr 30 '17 at 10:47
  • Possible duplicates:https://physics.stackexchange.com/q/18446/2451 and links therein. – Qmechanic Apr 30 '17 at 10:51
  • @PaulMalinowski I know about Gauss'/divergence theorem, but have never used Gauss' law of gravitation before- I am seeing it now for the first time when looking for a solution to my problem! – Meep Apr 30 '17 at 10:52
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    @21joanna12 I think Gauss's law as applied to gravity is essential for solving this problem efficiently. For instance, it took Issac Newton many pages of tedious calculus to show that a hollow sphere has no gravitational field within it, while with Gauss's law the proof is 1 maybe 2 lines. – pmal Apr 30 '17 at 10:58
  • @PaulMalinowski In fact, Newton's version of that proof is a paragraph. – rob Apr 30 '17 at 20:57

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Gauss' law states that, for a closed surface $S$, $$ \iint_S \vec{g} \,d\vec{A} = -4\pi G m_{encl}$$ where $m_{encl}$ is the total mass enclosed by the surface. Say we want to find $g$ at radius $r$; then let $S$ be the sphere of radius $r$. Due to the symmetry of the sphere, this can be rewritten as $$\iint_S g \, dA = 4\pi G m_{encl}$$ giving us $$g\cdot 4\pi r^2 = 4\pi G M\frac{r^3}{R^3}$$ $$\implies g = \frac{GM}{R^3}r = \frac{4\pi G\rho}{3}r$$ The integral in your original post isn't really doable as far as I know. A proof of Gauss' law in $n$ dimensions can be found here, as the second reply.

florence
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