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From MESSENGER Finds New Evidence for Water Ice at Mercury's Poles :
Mercury's polar region
Credits: NASA/John Hopkins University Applied Physics Laboratory/Carnegie Institution of Washington/National Astronomy and Ionosphere Center, Arecibo Observatory

Shown in red (in the image above) are areas of Mercury's north polar region that are in shadow in all images acquired by MESSENGER to date. The polar deposits imaged by Earth-based radar are in yellow.

New observations by the MESSENGER spacecraft provide compelling support for the long-held hypothesis that Mercury harbors abundant water ice and other frozen volatile materials in its permanently shadowed craters.

Three independent lines of evidence support this conclusion: the first measurements of excess hydrogen at Mercury's north pole with MESSENGER's Neutron Spectrometer, the first measurements of the reflectance of Mercury's polar deposits at near-infrared wavelengths with the Mercury Laser Altimeter (MLA), and the first detailed models of the surface and near-surface temperatures of Mercury's north polar regions that utilize the actual topography of Mercury's surface measured by the MLA. These findings are presented in three papers published online today in Science Express.

(Emphases by me)

For obvious reasons these findings of abundant water ice and other frozen volatile materials make the poles of Mercury much more suitable to compare with Mars in general with regard to colonization.

The most striking advantage for Mercury is of course that it gets much more energy from the Sun than Mars does:

about 12 times as much energy on average

Because the axial tilt of Mercury is only about 2 arc min., there is an elevated region near the north pole where that energy could be harvested all day and year around !

Furthermore Mercury has about the same surface gravity as Mars does, and it has a magnetic field that is strong enough to deflect the solar wind around the planet.

There's only one severe disadvantage that I could think of: the delta-v required for Mercury to enter Hohmann orbit is 7.5 km/s against 2.9 km/s for Mars, but that could be reduced with many flyby's like BepiColombo will attempt to do, so that will need much scheduling ahead !

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    $\begingroup$ this isn't exactly related to your question but it's interesting: earthsky.org/space/… $\endgroup$ – uhoh Jul 12 '20 at 17:26
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    $\begingroup$ @uhoh Interesting indeed ! And the other proposal, DAVINCI+ also for Venus , too ! "...NASA will continue development of up to two missions towards flight" . Wouldn't it be nice if htey could combine the two for Venus ! $\endgroup$ – Cornelis Jul 12 '20 at 17:53
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    $\begingroup$ It should be noted that the Hohmann transfer orbit delta-V's provided from the Wikipedia article also make the assumption that the orbit of Mercury is circular, and in the plane of the ecliptic. As anyone who's tried a flight to Moho (the Mercury analog) in Kerbal Space Program knows, those numbers are a cruel, cruel lie when your target's orbit is close the the Sun, inclined, and ellpitical. :) $\endgroup$ – notovny Jul 12 '20 at 20:09
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    $\begingroup$ related (and currently unanswered) How did Arecibo make radar images of ice on Mercury's poles? $\endgroup$ – uhoh Jul 13 '20 at 10:45
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    $\begingroup$ About that "delta-v required for Mercury to enter Hohmann orbit is 7.5 km/s against 2.9 km/s for Mars".. That's the delta-v to pass near your destination. You need to add the delta-v needed to capture (mars: 0.67km/s, Mercury 6.31km/s), and then to circularize to low orbit (Mars 2.1km/s, Mercury 1.22km/s). the totals are from LEO to Low planetary orbit: Mars=5.7km/s, Mercury=15.0km/s) $\endgroup$ – PcMan Mar 29 at 7:52
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Your delta-v analysis doesn't account for the landing delta-v. On Mars, only a fraction of a km/s has to be done propulsively, on Mercury the entire landing will be propulsive.

You also don't account for transit time. BepiColombo launched in 2018 and won't be able to enter orbit until the end of 2025. MESSENGER similarly launched in 2004 and didn't enter orbit until 2011. You're looking at a 7 year trip in a solar radiation environment that peaks at 6 times as bad as that at Earth, 12 times as bad as that at Mars, and that's just one way. Worse, your Mercury spacecraft will spend much of its overall time at the high-radiation end of that range as it does its final Mercury flyby maneuvers.

Mars, for comparison, can be reached with transits of just a few months. In Musk's 2016 IAC talk, he proposed high-energy transits lasting 80-150 days. These start out in the near-Earth radiation environment and move rapidly outward, spending the bulk of the transit closer to the near-Mars radiation environment, with about half the solar radiation experienced at the distance of Earth. And once you land, the atmosphere provides radiation protection similar to that provided in LEO by Earth's magnetosphere: https://www.jpl.nasa.gov/spaceimages/details.php?id=PIA03480

You need a major advantage to compensate for the delta-v issue, and there really isn't one. Volatile ices are available on Mercury, but they really don't compare to the large bodies of pure water ice and the CO2/N2/Ar atmosphere of Mars. Launch windows are more frequent, but given a certain size of vehicle, you're going to need a lot more launches to move the same amount of material due to the lower payload fractions. And Mercury does have 12 times the power for a solar panel facing the sun...but you'll be at the poles, where you'll have to space panels widely apart or erect them on towers so they don't shade each other, or have them at such a steep angle that they don't actually catch much sunlight. On Mars, you can land at mid-latitudes and just unroll/unfold solar panels in fields.

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    $\begingroup$ @Cornelisinspace Because even the minimum energy transit with nothing left for entering orbit or landing takes 7.5 km/s. That's more delta-v than a fully loaded Starship can manage. $\endgroup$ – Christopher James Huff Jul 12 '20 at 16:57
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    $\begingroup$ Completely emptying the Starship wouldn't provide enough extra delta-v to enter orbit around Mercury, let alone land there...only around 8 km/s total, depending on how heavy Starship turns out to be. That's a vehicle that easily has enough for a high energy transit to Mars with a full load. The delta-v difference is not a minor issue, in practice it means you'd need something exotic like nuclear or solar thermal propulsion just to get humans there, while Mars is easily reachable with chemical propulsion...not even the highest performance chemical propulsion, at that. $\endgroup$ – Christopher James Huff Jul 12 '20 at 17:59
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    $\begingroup$ @JohnDvorak Not for quickly getting people there and back. $\endgroup$ – Christopher James Huff Jul 13 '20 at 12:02
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    $\begingroup$ Plus the whole point of starship was to be big enough to be able to quickly bring a lot of stuff (things and people) to the Mars landing site , not to be stripped down to near nothing and slowly limp to mercury. And colonization requires literal tons and tons of stuff, so how much you can bring in one go becomes incredibly important. $\endgroup$ – eps Jul 13 '20 at 15:20
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    $\begingroup$ @eps Exactly. SpaceX's Mars colonization plan has them landing at a site with several hundred tons of supplies already there waiting for them in other Starships. It also means the advantage of the more frequent launch windows is greatly reduced: the higher delta-v requirement means a lower payload fraction, so a vehicle of the same size will need more trips to deliver the same payload. $\endgroup$ – Christopher James Huff Jul 13 '20 at 23:39
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You can't land on the day side of Mercury nor on a peak of perpetual light because it's far too hot (800 deg F / 430 deg C), even if not as hot as on Venus. A crewed mission must land either on the night side or in a crater of long darkness. This crater can and probably will be on the north or south pole if there's perpetual darkness.

Also, a crewed flight can't afford multiple planet flybys like BepiColombo. Their goal would be to get to Mercury (and back to Earth if they wanna) as fast as possible unless they have a really large spacecraft and one that can provide centripetal gravity.

Permanent colonization of the entire planet is impossible with current and near-future technology because Mercury is not in synchronous rotation with the Sun, therefore all of Mercury (except craters of perpetual darkness) will eventually be enlightened by the Sun.

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    $\begingroup$ The day side very close to the poles would not be so hot. The angle between the Sun and the horizon of Mercury should be small, 5.8 ° would reduce the solar intensity to a horizontal plane to 10 % of the intensity at the equator. $\endgroup$ – Uwe Jul 12 '20 at 15:52
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    $\begingroup$ The question is explicitly about the poles of Mercury versus Mars ! $\endgroup$ – Cornelis Jul 12 '20 at 16:44
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    $\begingroup$ @Cornelisinspace Peaks of perpetual light and craters of perpetual darkness can only be found near/on the poles of Mercury. Mars is tilted and doesn't have such. You're also invited to look onto my question: space.stackexchange.com/questions/44566/… $\endgroup$ – LoveForChrist Jul 12 '20 at 17:14
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    $\begingroup$ How large is the largest such known polar crater of perpetual darkness? Is it Prokofiev at 85.77°N 62.92°E, with 112km diameter? How much would that limit a hypothetical Mercury colony? (population/raw materials). Also, how much distance separation would there need to be between the colony and the return launch site? $\endgroup$ – smci Jul 14 '20 at 8:12
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    $\begingroup$ @Uwe: since Mercury has no atmosphere, I suspect the unshielded UV and gamma radiation levels would be lethal, even at 86°N. (Anyone got any rough numbers?) So, the colonists would have to hide below their crater rim almost all the time. $\endgroup$ – smci Jul 14 '20 at 9:28

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