Gravitational potential at the center of a uniform sphere Feynman in second Messenger Lecture said that potential at the center of ball with small radius $a$ is equal to average potential on surface of ball minus $G$ times mass inside the ball divided by $2a$. You can see it on the picture. I don't understand that. I had an idea that it is from Taylor series of potential at the centre, but I can't figure it out. Can you help me?  

 A: Late answer but I'll bite. 
Feynman's talking about a ball, which means that he is talking about a solid sphere, with uniform density, which I shall call $\rho$. You can apply Gauss's law for gravity to then calculate the potential. 
Gauss's law states that: 
$$ \int FdA = -4\pi GM$$
where F is the g-field, A is a surface area and M is the mass enclosed by our Gaussian surface. 
Let's say that our ball has radius $a$. We can imagine a Gaussian sphere, of radius $r < a$ which encloses some mass $\rho V$ where V is the volume of our Gaussian sphere (we can use multiplication instead of integration here because it's a uniform sphere). Call A the Gaussian area and by symmetry, $\int FdA = FA$, 
so that we can write, for our Gaussian sphere: 
$$F = \frac {-4πG\rho V}{A} $$
For our Gaussian sphere, $V = \frac{4}{3}\pi r^3$ and $A = 4\pi r^2$ so that:
$$\frac VA = \frac r3$$
leading to: 
$$F = \frac{-4πG\rho r}{3} $$
This tells us the field at any point in the sphere. Replacing r with a tells us the field at the surface of the sphere. 
Anyway, we want the potential at the surface of the sphere, which is the negative work done per unit mass when moving from the center of the sphere to its surface i.e from $r=0$ to $r=a$ (assuming that our reference point is set at the center of the sphere, so that the center has zero potential energy). This is the same thing as the integral of the field along this distance, since the field is the force per unit mass. Thus the potential at the surface of the sphere is: 
$$\int_0^a Fdr =\int_0^a \frac{4πG\rho r dr}{3}$$
which is:
$$ \frac{4πG\rho a^2}{6} $$
The volume of the entire sphere is $ V = \frac 43 \pi a^3$, so that $\frac {4\pi a^2}{6} = \frac {V}{2a}$. Since $M$, the mass enclosed by the sphere is $\rho V$, where this time V is the volume of the entire sphere: 
$$ \frac{4πG\rho a^2}{6}  = \frac{G\rho V}{2a} = \frac {GM}{2a}$$
So the work done in moving from the center to the surface of a ball is given by $\frac {GM}{2a}$, provided the ball is uniform. To get Feynman's equation just change the reference point for the potential energy, so that the potential energy at the center is no longer $0$.  
Edit: 
The above result holds only if the field due to matter outside the sphere is the same at the center and on the surface. If not the work done will depend on where on the surface you move to, since every point on the surface is not equidistant from external matter. If the ball is small enough, a criterion which Feynman mentions, then there's not much difference between different points on the surface or at the center, so you can approximate the field as constant on the ball. 
A: Feynman's answer does not refer to a sphere filled with uniform mass density. It is a completely general statement about the potential at the center of a (small enough) sphere given an arbitrary mass distribution. It is a different way of stating Newton's law of gravitation. It can be derived from the differential formulation of Newton's law (ΔΦ = 4πG ρ) with the help of the full-space Green's function G(r,r')=-1/(4 π |r-r'|).
