How to understand vector calculus results with regard to elementary example in Gravitation Hello I have just began to learn about vector calculus.
I am confused on what is going on in my notes and I am wondering if someone can help walk through what is going on and how to think about it.
In my notes I have:
Consider a point mass M at the origin. A point $P$, $\rho$ away-
Then $$F=\frac{-GM}{\rho^2}u_{\rho}$$
$$F=\nabla \psi= \frac{\partial \psi}{ \partial \rho} u_{\rho}+\frac{1}{\rho^2 sin\phi}\frac{\partial \psi}{ \partial \theta} u_{\theta}+\frac{1}{\rho}\frac{\partial \psi}{ \partial \phi} u_{\phi}$$
$\frac{\partial \psi}{\partial \rho}=\frac{-GM}{\rho^2} \to \psi=\frac{GM}{\rho}+f(\phi,\theta)$ , the other partials are zero, and so $f(\phi,\theta)=$ a constant
So $\psi=\frac{GM}{\rho}+C$
In conclusion $F=\frac{-GM}{\rho^2}u_{\rho}$ is a conservative vector field with $\psi=\frac{GM}{\rho}$
What I dont get:
Well I just don't understand how it was derived. I understand the first part as gravity is inversely proportional to the distance squared. So, then did we just take the partial derivatives to see what the gradient of the force was? Why did the other two partial derivatives just become zero?
Thank you all.
 A: Actually, it’s going in the other direction—they’re not computing the gradient of the force, they’re trying to see if this force is the gradient of some unknown potential $\psi$.  
Since $F$ is spherically symmetric, they expand $\nabla\psi$ in spherical coordinates and then match up the coefficients with those in the expression for $F$. That doesn’t have any terms in $u_\theta$ or $u_\phi$, so those terms drop out of $\nabla\psi$ as well.  
One other piece of the puzzle that you might need: When you integrate with respect to one of the variables, instead of having a constant of integration as you do in elementary calculus, you’ll have an arbitrary function of the other variables only. That’s where $f(\phi,\theta)$ in your notes comes from.
A: Notice that the unit vectors $u_\theta$ and $u_\phi$ do not appear in the force law, i.e. the force is spherically symmetric. So those other two partials in the gradient must be zero, or you'd get a component along either the $\theta$ or $\phi$ directions in the force. Thus $\psi = H(\rho)$ for some function $H$ smooth everywhere except the origin. Then it's just a matter of integrating the equation $dH/d\rho = -GM\rho^{-2}$. 
Which gives the answer stated.
Is that clear?
A: You're trying to find the function $\psi$, which is called the gravitational potential. It is so constructed (indeed, defined) so that $\vec{F} = -\nabla \psi$. This approach (dealing with potentials as opposed to forces) turns out to be quite handy mathematically in a variety of problems. Don't worry about it too much at the moment. But that is what is happening. 
The subject of how to calculate gradients is basic in vector calculus. It is usually simple but depends, as you can see, on the coordinate system used. I'd look it up on Wikipedia.
A: The gradient operator in spherical coordinates $(r,\theta,\phi)$ with unit vectors $(\hat r ,\hat \theta,\hat \phi)$ is given by
$$\nabla = \hat r\frac{\partial }{ \partial r} +\hat \theta \frac{1}{r }\frac{\partial }{ \partial \theta} +\hat \phi \frac{1}{r \sin\theta}\frac{\partial }{ \partial r} \tag 1$$
Now, suppose that $\nabla \psi = \hat r f(r)$ where $f$ is independent on $\theta$ and $\phi$.  Then, we have from $(1)$
$$\begin{align}
\frac{\partial \psi }{ \partial r}&=f(r)\tag 2\\\\
\frac{1}{r }\frac{\partial \psi}{ \partial \theta}&=0 \tag 3\\\\
\frac{1}{r \sin\theta}\frac{\partial \psi}{ \partial r}&=0 \tag 4
\end{align}$$
The general solution to $(2)$ is given by 
$$\psi =\int f(r)\,dr+g(\theta,\phi) \tag 5$$
where $g(\theta,\phi)$ is an integration constant.  Using $(5)$ in $(3)$ we find that 
$$\frac{\partial g(\theta,\phi)}{ \partial \theta}=0\implies g(\theta,\phi)=h(\phi) \tag 6$$
where $h(\phi)$ is an integration constant.
Finally, using $(6)$ in $(4)$ reveals
$$\frac{\partial h(\phi)}{ \partial \phi}=0\implies h=C \tag 7$$
where $C$ is an integration constant.
Putting it all together yields
$$\psi=\int f(r)\,dr+C$$
For $f(r)=-\frac{GM}{r^2}$, we have
$$\bbox[5px,border:2px solid #C0A000]{\psi=\frac{GM}{r}+C}$$
as was to be shown!  If we require $\lim_{r\to \infty}\psi=0$, then $C=0$.
