Calculating specific orbital energy, semi-major axis, and orbital period of an orbiting body Is it possible to calculate the specific orbital energy $ϵ$, the semi-major axis $a$, and the orbital period $T$ (or $P$) without any of them being available to you? The values I do have available to me are the velocity of the orbiting body relative to the center of gravity, its current position (also relative to the center of gravity), and the central mass that is providing the source of gravity $M$. I also have the mass of the orbiting body, but it is negligible.
So, given all of these things and no outside factors, is it possible to calculate any of the values listed above? According to Kepler's Third Law, the orbital period is given in the proportion $4π^2/T^2 = GM/R^3$  where $T$ is the orbital period, $G$ is Newton's Universal Gravitational Constant, $M$ is the mass of the larger body (given the orbiting body's mass is negligible), and $R$ is the distance between the center of gravity and the orbiting body. This doesn't help very much simply because it is a proportion and cannot be worked around with algebra to get an real value for $T$ (I think?).
Anyway, I have scoured the internet and Wikipedia trying to find a way to calculate these values, but I am at a loss. I am trying to see if there is a way to calculate these things for a small programming project/simulator. Otherwise, it would be necessary to simulate the  program for a period to determine one of these values to calculate the others.
 A: Yes, you can derive all of these quantities. The specific orbital energy $E$ is 
$$
\begin{align}
E &= \frac{1}{2}v^2 - \frac{\mu}{r}= -\frac{\mu}{2a},
\end{align}
$$
where $ \mu = GM^3/(M+m)^2 $, and $a$ is the semi-major axis. The orbital period follows from Kepler's Third Law:
$$
T^2 = (2\pi)^2\frac{a^3}{\mu}.
$$
If you also know the radial velocity $v_r$ and the tangential velocity $v_T$ separately at $r$, then you can also calculate the specific relative angular momentum $h$ and the orbital eccentricity $e$:
$$
\begin{align}
h^2 &= r^2\,v^2_{T} = \mu a(1-e^2).
\end{align}
$$

Edit
Several people have tried to change $\mu$ into $\mu = G(M+m)$. This is wrong, because that is the formula for relative motion instead of motion with respect to the centre of mass. The equations of motion of the two-body problem are
$$
\begin{align}
m\ddot{\boldsymbol{r}}_m &= - \frac{GmM}{|\boldsymbol{r}_m - \boldsymbol{r}_M|^3}\left(\boldsymbol{r}_m - \boldsymbol{r}_M\right),\\
M\ddot{\boldsymbol{r}}_M &= \frac{GmM}{|\boldsymbol{r}_m - \boldsymbol{r}_M|^3}\left(\boldsymbol{r}_m - \boldsymbol{r}_M\right),\\
\end{align}
$$
where $\boldsymbol{r}_m$ and $\boldsymbol{r}_M$ are the positions of the small and large body with respect to the centre of mass. What we want is to express the motion of the small body in terms of $\boldsymbol{r}_m$. By definition, the position of the centre of mass remains constant,
$$
m\boldsymbol{r}_m + M\boldsymbol{r}_M = \boldsymbol{0},
$$
so that
$$
\boldsymbol{r}_m - \boldsymbol{r}_M = \frac{M+m}{M}\boldsymbol{r}_m.
$$
Therefore,
$$
m\ddot{\boldsymbol{r}}_m = -GmM\frac{M^3}{(M+m)^3r^3_m}\left(\frac{M+m}{M}\boldsymbol{r}_m\right),
$$
or
$$
\ddot{\boldsymbol{r}}_m = -\frac{\mu}{r^3_m}\boldsymbol{r}_m,
$$
with $\mu = GM^3/(M+m)^2$. In my answer, $r = r_m$. I hope this clears things up.
