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I have problem understanding transaction (operations and methods applied) for one equation to other equation. It is about gravitational potential. $${\vec F_{grav}=\frac{GMm_{obj}\vec R}{R^3}}$$ If we take $${\Phi(r)=-\frac{GM}{r}, where\space r=|\vec R|}$$ we can say that $${\vec F_{grav}=m\vec g(r), where\space \vec g(r)=-\nabla \Phi}$$ So then in the book it was said that ${\nabla^2\Phi=4\pi G\mu}$. So isit possible to tell me wht ${\mu}$ stand for and how the last formula is derived?

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  • $\begingroup$ en.wikipedia.org/wiki/Gauss's_law_for_gravity $\endgroup$
    – lemon
    Jun 20, 2015 at 17:39
  • $\begingroup$ Looks like they are using $\mu$ for the mass density in place of the more common $\rho$. Also, which "last equation" are you referring to, the $\mathbf{F}_{grav}$ one or the $\nabla^2\Phi$ one? $\endgroup$
    – Kyle Kanos
    Jun 20, 2015 at 18:07
  • $\begingroup$ ${\nabla^2}$ one $\endgroup$
    – Blake
    Jun 20, 2015 at 18:14
  • $\begingroup$ this article from wikipedia is right on this material but again it is hard for me to get the derivation, it would be great if somebody can explain it more explicit if it is possible of course. $\endgroup$
    – Blake
    Jun 20, 2015 at 18:16
  • $\begingroup$ Generally, proofs of relations are considered off topic. Doing a Google search of derive poisson's equation gravity actually brings up a few explicit derivations. $\endgroup$
    – Kyle Kanos
    Jun 20, 2015 at 18:27

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The purpose of all this is to calculate and derive potential equation of the gravitation field. Let us assume that we have symmetrical sphere object that "generates" gravitational field. We want to know if the field depends on the objects "homogeneity" (correct me if this is not quite the right term). So from Gauss theorem (it has different names, but they said it with that name in Stanford lactures) we have that the divergence of the object (gravitation field in it) is equal to the net flux exiting from spherical surface that surrounds the object. So we have: $${\int_{V} \nabla \vec g(\vec r)dV=\int_{S}\vec g(\vec r)d\vec S}$$ where ${d\vec S=\hat{n}.dS}$. It can be written in this way because if we consider two infinitesimally small vectors that are sides of ${dS}$, then ${dS=|d\vec x\times d\vec y|}$ and ${\hat{n}=\frac{d\vec x\times d\vec y}{|d\vec x\times d\vec y|}}$ that is why for simplicity we can write ${d\vec S}$ RHS: $${\int_{S}\vec g(\vec r)dS=\int_{S}-\frac{GM}{r^2}\hat{r}\times \hat{r}dS}$$ ${\hat{r}=\hat{n}}$ because both are normal and perpendicular to the surface of the sphere. $${\int_{S}-\frac{GM}{r^2}\hat{r}\times \hat{r}dS=-\frac{GM}{r^2}\int_{S}\hat{r}\times \hat{r}dS=-\frac{GM}{r^2}\int_{S}dS=-\frac{GM}{r^2}4\pi r^2=-4\pi GM}$$ So we got: $${\int_{V} \nabla \vec g(\vec r)dV=-4\pi GM}$$ Assuming ${M=\int_V\mu dV}$, we get $${\int_{V} \nabla \vec g(\vec r)dV=-4\pi G\int_V\mu dV=\int_V-4\pi G\mu dV}$$ We have similar integrals on both sides so: $${\nabla \vec g(\vec r)=-4\pi G\mu}$$ which is $${\nabla^2 \Phi=4\pi G\mu}$$

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