The Kepler virial theorem fixes mean energies

For a bound orbit in $U(r)=-k/r$, use $G=\mathbf p\cdot\mathbf r$ to derive the time-averaged virial theorem without assuming a circular orbit. Express $\langle T\rangle$ and $\langle U\rangle$ in terms of total energy $E$ and semimajor axis $a$. For a unit mass with $\mu=k/m=3.986\times10^{14}\,\mathrm{m^3/s^2}$ and $a=1.00\times10^7\,\mathrm m$, compute the mean specific kinetic and potential energies.

$d(\mathbf p\cdot\mathbf r)/dt=2T+\mathbf r\cdot\mathbf F$.