How come we talk about gravitational potential energy and not gravitational potential? With regards to gravity the equation learned is $$U=-\frac{GMm}{r}$$
And the relationship to force is $$F=-\frac{dU}{dr}$$
In electrostatics we instead talk about electric field and electric potential:
$$V=\frac{kQ}{r}$$
$$E=-\frac{dV}{dr}$$
Why do we do this if the situation between gravity fields and electric fields is analogous?
 A: I think you are reading a lot into what is a minor distinction.
Strictly speaking I suppose the gravitational potential is the energy per unit mass, i.e. $m=1$ in your first equation, while the gravitational potential energy is the potential times the mass. In practice no-one I know has ever bothered to make the distinction because it's usually obvious what is meant.
In electrostatics particles can be opposite signs or uncharged, so maybe this is why there is more of a distinction.
A: I too was confused by this difference between gravity and electromagnetism. Hopefully the following clears things up.
The gravitational potential a distance $r$ from a mass $M$ is
$$
\phi_g=-\frac{GM}{r},
$$
the gravitational field is
$$
{\bf g} = - \nabla \phi_g,
$$
and the gravitational potential energy (of two masses $M$ and $m$ separated by a distance $r$) is
$$
U_g=m \phi_g =-\frac{GMm}{r}.
$$
The force on the mass $m$ is
$$
{\bf F}_g = - \nabla U_g = -m \nabla \phi_g = m {\bf g}.
$$
The electric potential a distance $r$ from a charge $Q$ is
$$
\phi_e=\frac{kQ}{r},
$$
the electric field is
$$
{\bf E} = - \nabla \phi_e,
$$
and the electric potential energy (of two charges $Q$ and $q$ separated by a distance $r$) is
$$
U_e= q \phi_e =\frac{kQq}{r}.
$$
The force on the charge $q$ is
$$
{\bf F}_e = - \nabla U_e = -q \nabla \phi_e = q {\bf E}.
$$
As an aside, to make matters worse, the standard symbol used for the potential energy term in the Schrödinger equation is $V$, and it is referred to as the potential.
A: Well, the electric field $\vec E$ is different from the force field $\vec F$ a test charge will feel. That difference is exactly the charge of the test particle. That force field is given by the gradient of a function, too
$$ q \vec E = \vec F = - \frac{\mathrm d}{\mathrm d r} W$$
where I use the letter $W$ in order not to have confusing notation.
The relation between $W$ and your $V$ would just be
$$ W = qV = \frac{kQq}{r}.$$
Exactly analogous to the gravitational potential energy of a test mass, we found the "electric potential energy" of a test charge.
The difference might come from the fact that  gravitational fields are usually assumed to exist and the question becomes of a test mass' motion in those given fields. Note that we can not create gravitational fields at will but have to do with what nature hands us. 
On the other hand, the range of interesting questions for electric fields goes beyond the motion of a test charge, thereby triggering interest in notation that is independent of the actual charge that will be subject to the forces in the end.
