Gravitational Potential

Potential energy belongs to a pair of bodies. Divide out the test mass and you get a property of the place itself: the gravitational potential, a scalar map of how deep each point lies in the well.

Definition#

Equivalently, V at a point is the work per kilogram an external agent must do to bring a mass slowly from the reference point to that point. Because V is a scalar, contributions from different masses simply add, with no directions to worry about.

Calculating V#

A point mass

Using U = −GMm/r with zero at infinity:

A ring, on its axis

Every bit dm of a ring of radius a is the same distance $z = \sqrt{a^2 + r^2}$ from a point P on the axis a distance r from the centre. So each contributes $-G\,dm/z$ and the sum is easy:

A thin spherical shell

Slice a shell of mass M, radius a, into rings by angle θ from the line OP (O the centre, OP = r). The ring between θ and θ + dθ has mass $dm = \tfrac M2\sin\theta\,d\theta$, and all of it is at distance z from P, where $z^2 = a^2 + r^2 - 2ar\cos\theta$. Differentiating, $z\,dz = ar\sin\theta\,d\theta$, so $dm = \dfrac{M}{2ar}z\,dz$, and the ring’s potential at P is

Now integrate over z. The limits are where the cases differ:

• P outside (r > a): z runs from r − a to r + a, a range of 2a. $V = -\dfrac{GM}{2ar}(2a) = -\dfrac{GM}{r}$ (11.9), exactly as if all the mass were at the centre.
• P inside (r < a): z runs from a − r to a + r, a range of 2r. $V = -\dfrac{GM}{2ar}(2r) = -\dfrac{GM}{a}$ (11.10): the same at every point inside.

A solid sphere

Outside, treat it as a stack of concentric shells, each acting as a point at the centre: $V = -GM/r$ (11.11). Inside, at distance r from the centre, split the sphere in two. The inner ball (mass $Mr^3/a^3$) acts like a point at the centre; the outer shells each give their own constant interior potential $-G\,dm/x$, with $dm = 3Mx^2dx/a^3$. Adding up:

At the centre this is $-\tfrac32 GM/a$, one and a half times as deep as at the surface.