# 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?

• It will return in 2271: en.wikipedia.org/wiki/Star_Trek:_The_Motion_Picture#Plot – Hot Licks Feb 6 '18 at 22:03
• @HotLicks that made me laugh... and think of the ever powerful V-Giny - although I haven't found a date for its return. – WernerCD Feb 6 '18 at 23:33
• @HotLicks that's Voyager 6 (no one has even proposed Voyager 3 yet) but still worth the mention ;) – honeste_vivere Feb 7 '18 at 15:54

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{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 Dropping the values from (2) into (1) results in \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}

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 $7,400\operatorname{m}/\operatorname{s}$.

• Might we meet it later after some turns around the galactic center? – Paŭlo Ebermann Feb 6 '18 at 20:31
• @PaŭloEbermann I do not know. I doubt that orbits around the Milky Way galaxy are closed, so I think they're unlikely to meet again. It's possible, though. – Sean E. Lake Feb 6 '18 at 20:39
• Galactic orbits tend to be box orbits but an actual intercept is possible. Anyway, Voyager will orbit the galaxy as it's something like 492 km/s short of escape. – Joshua Feb 6 '18 at 21:29

"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...

• The escape velocity for earth? Or the solar system? ... using those terms (specifically escape velocity), I found this article that says that it will continue toward the constellation Ophiuchus - so, not coming back. – WernerCD Feb 6 '18 at 17:02
• @WernerCD correct, it is not coming back. – zeta-band Feb 6 '18 at 17:11
• Voyager picked up additional speed from its slingshot passages, so it ended up going faster than you'd expect from the launch velocity – PhillS Feb 6 '18 at 17:12
• Note: Escape velocity from a body is not a constant: It is a function of distance from the body's center of mass. The further you are from the body, the less velocity you need to escape from it. 11km/s is approximately equal to escape velocity from the Earth's surface. – Solomon Slow Feb 6 '18 at 17:27