I saw one on a blackboard and I think that some introductory physics books should have that.
Anyway, I can derive it here, too:
With Newtonian gravity, each pair of objects i,j has a potential V_{ij}=m_i m_j \frac{G}{|r_{ij}|} where rij is the distance between the objects.
The total energy of the system is then given by
E=\frac{1}{2} \sum_k m_k\, v_k^2 + \frac{1}{2} \sum_{i,j,i \ne j} V_{ij}
where the second 1/2 is required as the sum counts each pair twice. The time-derivative is then given by
\frac{dE}{dt}=\sum_i m_i\, v_i \cdot a_i + \frac{1}{2} \sum_{i,j,i \ne j} m_i\, m_j\, G \frac{d}{dt}\frac{1}{|r_{ij}|}
The dot indicates a scalar product. The chain rules now gives
\frac{d}{dt}\frac{1}{|r_{ij}|}<br />
= \frac{1}{|r_{ij}|^2} \frac{d}{dt} |r_{ij}|<br />
= \frac{1}{|r_{ij}|^3} (x_i-x_j) \cdot (v_i-v_j)
The force on an object is now given by the gravitational forces of all other objects, which is the negative gradient of the potential.
m_i a_i = - \sum_{j \ne i} m_i\, m_j\, G\; \nabla \left(\frac{1}{|r_{ij}|}\right) = - \sum_{j \ne i} m_i\, m_j\, G \frac{x_i-x_j}{|r_{ij}|^3}
If you insert both expressions in the expression for dE/dt, you will see that both parts just cancel and you get 0.
This is not restricted to potentials which are proportional to 1/r. In the general case of V(|r|), you get \frac{d}{dt}V(r)=\left(\frac{d}{d|r|}V(|r|)\right) \frac{1}{|r|} (x_i-x_j)\cdot(v_i-v_j) and m_i a_i = - \sum_{j \ne i} m_i\, m_j\, G \left(\frac{d}{d|r|}V(|r|)\right) \frac{x_i-x_j}{|r_|}
