Think of the simple case where the difference in mass between the two bodies is so great that the effect of the smaller mass,
m, on the greater,
M, can be ignored. Let's suppose mass
m is in a circular orbit around
M.
By definition of potential energy (and taking our "zero of potential energy" as infinity),
U = - \int_{r}^{\infty} \textbf{F} \cdot \textup{d}\textbf{r} = \frac{-GMm}{r},
where
U is potential energy due to gravity,
F gravitational force,
M the greater mass,
m the smaller mass, d
r an infinitesimal change in the position of
m (the position considered as a position vector,
r, extending from the centre of mass of the system, here effectively the centre of mass of
M, to the centre of mass of
m),
r the magnitude (length) of this position vector
r,
G the gravitational constant (which relates the units). In other words, the potential energy of the satellite
m is the work that would be done by the force of gravity if it was to move
m from infinitely far away to its current position.
By definition of kinetic energy,
T = \frac{1}{2}mv^2,
where
T is the kinetic energy of the orbiting body with mass
m, and
v its speed, defined as the magnitude of its velocity with respect to the more massive body.
By definition of circular motion,
\frac{\mathrm{d} r}{\mathrm{d} t} = 0
at all points in the orbit. This just means that the distance of the satellite from the more massive body doesn't vary over time.
Less obviously,
\frac{\mathrm{d} v}{\mathrm{d} t} = 0.
That is, the speed of the satellite is constant over time. This follows from Kepler's 2nd law in the special case of a circular orbit; the radius of the orbit doesn't change, so the only way that an equal area can be swept out over an equal time is if the speed is constant.
http://en.wikipedia.org/wiki/Kepler's_laws_of_planetary_motion#Second_law
Therefore
\frac{\mathrm{d} }{\mathrm{d} t} \left(T + U) = 0.
Kinetic energy is constant, and potential energy is constant, so their sum is constant.
