Differential form of the law of gravitation potential

Question

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?

Answer 1

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 $${\nabla^2 \Phi=4\pi G\mu}$$