Gradient of Newtonian Potential

Theorem
Let $R$ be a region of space.

Let $S$ be a Newtonian potential over $R$ defined as:
 * $\forall \mathbf r = x \mathbf i + y \mathbf j + z \mathbf k \in R: \map S {\mathbf r} = \dfrac k r$

where:
 * $\tuple {\mathbf i, \mathbf j, \mathbf k}$ is the standard ordered basis on $R$
 * $\mathbf r = x \mathbf i + y \mathbf j + z \mathbf k$ is the position vector of an arbitrary point in $R$ with respect to the origin
 * $r = \norm {\mathbf r}$ is the magnitude of $\mathbf r$
 * $k$ is some predetermined constant.

Then:
 * $\grad S = -\dfrac {k \mathbf r} {r^3} = -\dfrac {k \mathbf {\hat r} } {r^2}$

where:
 * $\grad$ denotes the gradient operator
 * $\mathbf {\hat r}$ denotes the unit vector in the direction of $\mathbf r$.

The fact that the gradient of $S$ is negative indicates that direction of the vector quantities that compose the vector field that is $\grad S$ all point towards the origin.

Proof
From the geometry of the sphere, the equal surfaces of $S$ are concentric spheres whose centers are at the origin.

As the origin is approached, the scalar potential is unbounded above.

We have: