Voyager: Is it in "orbit" and will it return? This Phys.SE question about Voyager got me thinking, and I didn't see an answer:
Gravity - what goes up must come down. 
Will Voyager, in its current trajectory ever be pulled back into our Solar System? Thus, does it have an orbit?
Is Voyager, more or less, a long-period "comet" now?
 A: "what comes(goes?) up must come down" is wrong if the thing goes up fast enough:
https://en.wikipedia.org/wiki/Escape_velocity
There is a certain velocity above which a body will not return due to gravity. For earth, this speed is about 11 km/s, Voyager 1 had a starting speed of 15 km/s (source: german wikipedia), so no, Voyager 1 will never return.
edit:
Now that I've found out that the escape velocity of the "System" Earth/Sun is about 16km/s (https://en.wikipedia.org/wiki/Escape_velocity#List_of_escape_velocities), things seem to be more complicated...
A: The Voyager probes passed escape velocity for the solar system using gravitational boosts off of Jupiter, so they're almost certainly never coming back to us. To see this, we just calculate the total energy (divided by the probe mass) using the formula
\begin{align}
    \frac{E}{m_{\mathrm{probe}}} & = \frac{1}{2} v_{\mathrm{probe/sun}}^2 - G \frac{M_{\odot}}{r_{\mathrm{probe/sun}}}\tag1
\end{align}
and if it comes out positive, the probe is not bound to the sun anymore.
NASA provides the status of Voyagers 1 and 2 and, as of this writing, we have
\begin{equation}\begin{aligned}
    \mathrm{Voyager\ 1:\ } v_{\mathrm{probe/sun}} &= 1.6999\times 10^4 \frac{\operatorname{m}}{\operatorname{s}} & r_{\mathrm{probe/sun}} & = 2.1117 \times10^{13}\operatorname{m} \\
    \mathrm{Voyager\ 2:\ } v_{\mathrm{probe/sun}} &= 1.5374\times10^4 \frac{\operatorname{m}}{\operatorname{s}} & r_{\mathrm{probe/sun}} & = 1.7452 \times10^{13}\operatorname{m}.
\end{aligned} \tag2\end{equation}
Dropping the values from (2) into (1) results in
\begin{equation}\begin{aligned}
    \mathrm{Voyager\ 1:\ } \frac{E}{m_{\mathrm{probe}}} &= 1.44\times 10^8 \frac{\operatorname{J}}{\operatorname{kg}} \\
    \mathrm{Voyager\ 2:\ } \frac{E}{m_{\mathrm{probe}}} &= 1.18\times 10^8 \frac{\operatorname{J}}{\operatorname{kg}}.
\end{aligned} \end{equation}
As you can see, those energies are very positive, so the probes are completely unbound from the solar system. For reference, the escape velocity of the solar system (the velocity where energy is $0$) at the Earth's orbit is $42,000 \operatorname{m}/\operatorname{s}$, and at the orbit of Neptune it's $7,400\operatorname{m}/\operatorname{s}$.
