Gravitational Potential V(r)

Definition. The gravitational potential $V(r)$ at a distance $r$ from a mass is defined as the amount of work required to bring a unit mass from that point out to infinity -- equivalently, it is the gravitational potential energy per unit mass at that point:

$V$ is a scalar quantity, with SI unit joule per kilogram ($\text{J kg}^{-1}$).

Why bother with a scalar version of potential energy? The gravitational field $\vec{E}$ and gravitational force $\vec{F}$ are both vectors, needing components and directions to add correctly. Gravitational potential $V$ and potential energy $U$, being scalars, add up as simple algebra -- which makes many multi-mass problems dramatically easier (this is the same trick used later for electric potential).

Near the Earth's surface. At height $h$ above the surface, $V(r)=-\dfrac{GM_e}{R_e+h}$, and on the surface itself $V(R_e)=-\dfrac{GM_e}{R_e}$; since $R_e+h>R_e$, we always have $V(R_e+h) > V(R_e)$ in the algebraic sense that the potential becomes less negative (i.e. increases) as you move away from the Earth. Equivalently, near the surface, $V(h) = U(h)/m = gh$, and the potential is zero right at the surface ($h=0$). This is why "water always falls downhill": the top of a hill is at a higher gravitational potential than the ground below it, and any freely-moving mass slides from a region of higher potential to a region of lower potential.

Worked example (four equal masses on a circle). Four equal masses $M$ sit equally spaced (i.e. at $90^\circ$ intervals) on a circle of radius $R$ around a centre point $O$. Since potential is a scalar, the net potential at $O$ is just the algebraic sum of the four individual potentials:

$V_O=-\frac{GM}{R}-\frac{GM}{R}-\frac{GM}{R}-\frac{GM}{R}=-\frac{4GM}{R}.$ …