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Personally, I would rather live near Mercury's north pole than anywhere on Mars mainly because:

But it's been stated many times that it's so hard to get there because of the proximity of the Sun.

From the BBC news article "What is Elon Musk's Starship?":

"You could conceivably have five or six people per cabin, if you really wanted to crowd people in. But I think mostly we would expect to see two or three people per cabin, and so nominally about 100 people per flight to Mars," Musk said. The payload bay would also host common areas, storage space, a galley and a shelter where people could gather to shield from solar storms, where the Sun belches out harmful charged particles into space.

I really would like to know how many people could be transported to the surface of Mercury's north pole with such a Starship, taking into account the longer trajectory to the aphelion of Mercury's orbit than that to an orbit around Mars.

Please with the necessary calculations to be able to control the assertions in your answer !

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    $\begingroup$ Read up on the route taken by the Messenger probe. It took 6½ years to get to Mercury. It's about as hard as getting to Saturn. $\endgroup$
    – JDługosz
    Jan 14 at 15:51
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    $\begingroup$ Also, the "100 people on a starship" is an Elon Number, i.e. the same as a Trump number with twice the inaccuracy. A quick back-of-the-envelope calculation shows it is beyond sane, see e.g. youtube.com/watch?v=SummGvYrHqw $\endgroup$
    – Jens
    Jan 15 at 20:08
  • $\begingroup$ Getting to Mercury directly is ridiculously hard - A landing mission needs about the same $\Delta v$ as landing on Pluto! $\endgroup$
    – asdfex
    Jan 18 at 19:45
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I take "after one Hohmann-like transfer" to mean a direct Earth to Mercury interplanetary transfer (i.e., no Venus shenanigans), please let me know if that is not the case.


Unless serious lithobraking is possible, no humans will go to Mercury.

Starship, even with no payload, doesn't have the juice to do anything other than fly by Mercury:

I did a trajectory search from 2025-2030 from Earth to Mercury direct transfers. The cheapest launch $\Delta V$ (from a 250 km LEO) is 4.87 km/s (red dot, $C3=$~$40$ $km^2/s^2$, L: 27-Nov-2026, A: 23-Mar-2027):

min launch dV (personal work)

Sounds nice, but the arrival speed (Mercury relative speed @ 0 altitude) for these specific dates is a cool 13.65 km/s (!yikes). In fact, the minimum arrival speed for the period examined is 8.91 km/s, only slightly less (0.3%) than the total $\Delta V$ the payload-less Starship has available.

So if you could throw a fully fueled, yet still empty, Starship at Mercury on that trajectory (L: 09-May-2029, A: 31-Mar-2030, $C3=$~$104$ $km^2/s^2$!) you might have a shot at landing safely. Good luck!


Edit: A premise of the question is incorrect:

taking into account the longer trajectory to the aphelion of Mercury's orbit than that to an orbit around Mars

The transfer time to Mercury (for the launch example) is only ~120 days, less than half of a typical transfer to Mars. The minimum arrival speed example uses a longer type III/IV trajectory (I didn't check which & it doesn't really matter), yet is still comparable to a typical Mars transfer at ~330 days.

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    $\begingroup$ Could you mention the source of the diagram in the answer ? $\endgroup$
    – Cornelis
    Jan 14 at 11:02
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    $\begingroup$ @Cornelis source is personal work. That's not how the question reads, but if I have some time I may consider it. $\endgroup$ Jan 14 at 12:13
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    $\begingroup$ @Cornelis those are different questions that have mostly been answered on this site: here, here, and here for starters $\endgroup$ Jan 14 at 16:25
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    $\begingroup$ @Cornelis if you'd like an explanation on how to produce such a plot and how to interpret it (get values like $\Delta V$ etc.), you would have to ask those questions $\endgroup$ Jan 14 at 16:26
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    $\begingroup$ Very interesting. Presumably a modified Starship with stretched tanks as now seems likely and with other modifications might be in with a chance? $\endgroup$
    – Slarty
    Jan 15 at 9:52
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The delta-v to Mercury is 2.5x greater than to Mars; given the dry mass of Starship I doubt that it could land any payload at all without some heliocentric fuel depot, which would be quite tricky and expensive to set up. This is without taking into consideration that Starship just isn't designed to survive the heat it would absorb by traveling so close to the sun; Mercury-bound spacecraft have to be specially designed to handle the temperatures.

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    $\begingroup$ Heliocentric fuel depots won't be helpful unless you throw them along the exact same transfer orbit as the spacecraft is on, in which case, you might as well attach them to the spacecraft itself, and they become extra stages or drop tanks. $\endgroup$
    – notovny
    Jan 13 at 17:19
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    $\begingroup$ Not to mention that you can aerobrake at Mars, there's no aerobrake at Mercury. $\endgroup$ Jan 16 at 21:34
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Landing on Mercury is supremely difficult due to how deep Mercury lies in the Suns gravitational well. In addition it has no significant atmosphere so retro propulsive landings would be required making it harder still.

It is very unlikely that Starship would be a good choice for landing on Mercury owing to it's high mass, it can't be ruled out entirely, but it would certainly need considerable modification. Stretched tanks, lowered payload capacity and more.

It would also be necessary to re-tank in a very highly elliptical Earth orbit to maximise it's capability and use a fleet of disposable tankers. But even that might not be enough. Another option would be a series of gravitational assists from the Earth and Venus. That might just do it but at the cost of a very extended mission duration with subsequent problems of crew exposure to radiation and zero gravity, crew life support requirements and the practicality of retaining cryogenic propellants for many years.

In summary. It's not going to happen.

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  • $\begingroup$ A light-weight heat shield could be attached on one half of the spacecraft to protect against both radiation and high temperature exposure. $\endgroup$
    – Cornelis
    Jan 13 at 18:41
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    $\begingroup$ A light weight heat shield will do very little to protect against radiation. You will either need a great deal of mass, or a magnetic shield which at present is purely theoretical. $\endgroup$ Jan 13 at 19:31
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    $\begingroup$ @Cornelis this kind of radiation $\endgroup$ Jan 13 at 22:21
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    $\begingroup$ @Vikki GCRs aren't really "the worst of the ionizing radiation", they're just the kind with the highest energy per-particle. Increasing the lower-energy, but still dangerous solar radiation by an order of magnitude to somewhat reduce a component of the radiation that's mainly relevant to long-term exposure is a poor trade. $\endgroup$ Jan 14 at 3:53
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    $\begingroup$ @RockPaperLz look at this vid: youtube.com/watch?v=T0uwtmHt0v4 The hole in the middle is the Sun, the coins the planets. Much faster at the bottom of the funnel A vast difference in orbital speed that must be taken into account when changing orbits. The orbital speed of the planets can be found here: en.wikipedia.org/wiki/Orbital_speed#Planets Any kind of image of the solar system based on A road map is hopeless. Better to imagine this: youtube.com/watch?v=J3HXVfz8r4A Imagine playing catch with someone doing 60mph around the middle of the spinner $\endgroup$
    – Slarty
    Jan 16 at 16:00

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