What should be the escape velocity for our galaxy and can we calculate it?
For instance, if we assume that we don't know the mass of our galaxy, you may consider it as small 'm'.
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2$\begingroup$ Such questions are more fitting for astronomy.stackexchange.com $\endgroup$– PhilippMar 9, 2014 at 23:55
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4$\begingroup$ @Philipp Why? If it's not strictly off-topic here, then we shouldn't push for migration. We have questions about Hill sphere, escape velocity of a binary star, and so on. And they're all relevant to space exploration too, not merely movement of stars in a galaxy. Say, you want to send a probe into the intergalactic space, you'd want to know what's the escape velocity it should reach first. But we probably do need to discuss demarcation line between Astronomy and Space Exploration to avoid such issues, so I suggest opening a new thread in out Space Exploration Meta where our members could state their opinion. $\endgroup$– TildalWaveMar 10, 2014 at 8:54
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$\begingroup$ a.k.a. Fourth Cosmic Velocity - 330km/s $\endgroup$– SF.Apr 25, 2019 at 8:52
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This is actually pretty difficult to do, because it depends on where from due to uneven distribution of matter (local parameters), how far from the galactic center due to radial velocity, the direction in which you want to reach escape velocity (how much of the radial velocity can be used), and that it's hard to estimate mass of the Milky Way (global parameters) and there will always be significant uncertainty with methods that are currently available to us.
We can probably expect a much more reliable overview of mass distribution, stellar velocities and total mass estimates once the Gaia observatory completes its billion stars stellar kinematics survey, but so far the best we have is probably the results of the Radial Velocity Experiment (RAVE) survey;
RAVE took a sample of 90 high-velocity Milky Way stars (some moving at over 300 km/s) for which their position and velocity was determined sufficiently precise, and then compared their movement to models of other, similar spiral galaxies, to reach an estimate of Milky Way's total mass at about 1.6 trillion solar masses. Once they had that, they could calculate estimated escape velocities for our stellar neighbourhood, for which mass distribution and radial velocity with respect to the galactic barycenter are of course most well understood.
Solar system's orbital velocity is estimated at roughly 220 km/s, and galactic escape velocity for our vicinity at about 537 km/s. So in the direction of Solar system's velocity vector, velocity required to escape Milky Way is ~ 317 km/s. And much more, if this Solar system's own orbital momentum cannot be used to full extent and a launch in other directions is required. This is of course assuming you can launch on a trajectory that avoids getting too close to gravitational influence of other solar systems.
Exact methods used to calculate this are a fair bit too complex to even attempt describing them here, so I'll refer you to some interesting sources:
- The RAVE survey: the Galactic escape speed and the mass of the Milky Way, Piffl et al., 17 Sep 2013 (Complete PDF)
- The RAVE Survey: Constraining the Local Galactic Escape Speed, Smith et al., 9 May 2007 (Complete PDF)
And a bit lighter reading:
answered Mar 9, 2014 at 20:11
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9$\begingroup$ To get that 317 km/s, you "only" need about 77 km/s in a single (fast) burn near the surface of our Sun. $\endgroup$ Mar 12, 2014 at 4:38
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1$\begingroup$ This paper arxiv.org/abs/1408.1787 is slightly more recent than the references you gave, and it estimates escape velocity to be 551 km/s, with error bars of about 20 or 30 km/s. I haven't studied it carefully enough to understand how the method or data differ from what was used in your references. $\endgroup$– user687Jun 29, 2016 at 22:50
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1$\begingroup$ Escape velocity is independent of direction. It does not matter what direction you are heading. $\endgroup$– ErikJun 30, 2016 at 18:34
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3$\begingroup$ @Erik: The answer's 317 km/s refers to the velocity relative to the solar system. If you add a delta vee vector of a fixed magnitude to the solar system's velocity vector, the magnitude of the result does depend on the direction of the delta vee. $\endgroup$– user687Jul 2, 2016 at 1:36
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1$\begingroup$ @BenCrowell which means you are answering a different question than the one asked. Escape velocity is relative to the body in question - the barycenter of the galaxy in this case. $\endgroup$– ErikJul 2, 2016 at 1:40