Hypothetically if a manufacturer was considering building a factory on the Moon or on Mars, which location has the smaller gravity well to escape? Assuming final destination of the goods produced is half way between Earth and Mars.

  • $\begingroup$ Mass of mars: 6.39 × 10^23 kg, Mass of moon: 7.34767309 × 10^22 kg. Therefore, the moon is easier to escape by a factor of: 8.69663078601, it's 8.7x easier to escape from mars (excluding atmospheric calculations, that exceeds my abilities), which means, at minimum, the moon is 8.7x easier to escape from an atmosphere-less Mars. However, given delta-v charts we can approximate slightly better: 6.4km/s from LEO to moon. 10.2km/s from LEO to mars (with aerobrake). Even getting to mars is about 1.6x harder than the moon, even using the Mars atmosphere to break the fall. $\endgroup$ Commented Oct 3, 2018 at 0:47
  • $\begingroup$ Also, welcome to this SE, as a "new" user myself, it's a lot of fun :). I'd say the gravity well is the least of the problems though, we're missing a lot of the "stay alive for more than X days" components to make either feasible (in terms of manufacturing, we're pretty much in "how the heck can we live there" mode). $\endgroup$ Commented Oct 3, 2018 at 0:51
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    $\begingroup$ @MagicOctopusUrn But the Moon is in Earth's gravity well right? Don't we have to escape the gravity well of both to get half way to Mars? Yeah i might have to follow up with a couple hundred thousand questions before I'm ready to start building the factory, haha $\endgroup$ Commented Oct 3, 2018 at 1:06
  • $\begingroup$ Technically, yes, but from LEO, once you reach the required velocity to escape the gravity well of earth, due to your distance from the moon, you've already more than exceeded the escape velocity of the moon (a smaller object) as well. The only way the moon will stop you on the way to Mars is skewing your trajectory slightly on the escape (which most n-body calculations take into account) or slamming directly into it. The escape velocity of earth is directly proportional to the mass of earth, same goes for the moon. The moon is smaller than earth, so by escaping earth we escape the moon too. $\endgroup$ Commented Oct 3, 2018 at 1:08
  • $\begingroup$ For reference, the moon's escape velocity is 2.38 km/s and the earth's is 11.2km/s. If you're going exactly 11.2km/s, leaving earth, the worst the moon can do is slow you down enough to return into earth orbit, or throw you off-target based on the direction you pass (behind = faster, in front = slower, above/below = skewed inclination). If you're going 11.2km/s the moon will not be able to capture your vessel in orbit of itself regardless of how close you pass, it will simply skew your trajectory. (Once again this is very basic wording, and is wrong in more complicated situations) $\endgroup$ Commented Oct 3, 2018 at 1:12

2 Answers 2


Halfway between earth and Mars? So an orbit 1.26 A.U. in radius, (earth orbit is 1 A.U. in radius and Mars orbit is 1.52 A.U.)

About 5 kilometers/second to leave Mars surface with enough extra delta V to reach a 1.26 A.U. perihelion. Once at perihelion it would take about .3 km/s to leave the transfer orbit and match velocities with your destination. So about 5.3 km/s.

From the moon to EML2 is about 2.5 km/s. From EML2 to an near earth perigee is about .4 km/s. At this perigee you'd be moving just a hair under earth escape velocity. At this point a .25 km/s burn would inject into a transfer orbit with a 1.26 A.U. aphelion. Once at the 1.26 aphelion it would take about .83 km/s to match velocities with the destination orbit. So a total of about 4 km/s.

So 5.3 km/s for Mars vs 4 km/s for the moon.

These approximations were made assuming circular coplanar orbits. Also I ignored Mars' atmosphere which would inflict a gravity loss penalty.

  • $\begingroup$ Thanks @HopDavid, that's the kind of answer i was after! $\endgroup$ Commented Oct 3, 2018 at 4:55

Well “escaping” the moon implies escaping the cis lunar space region which has both gravity from earth and the moon. So I’d say mars definitely. And I’d say the problem posed is more of a simulation runner than thought experiment.

  • $\begingroup$ He asked "which location has the smaller gravity well to escape" and your answer is Mars? Phrasing is important to an answer, this makes it sound like mars is easier to escape than the moon, which is not correct for many factors other than gravity. $\endgroup$ Commented Oct 3, 2018 at 1:40
  • $\begingroup$ @MagicOctopusUrn Well the location of the Moon is in the orbit of Earth. So I think it's fair that we include Earths gravity in the equation. My post contents should have been clearer. But title of my post "Comparing gravity wells: Earth from the Moon vs Surface of Mars" does say Earth's gravity well from the Moon. $\endgroup$ Commented Oct 3, 2018 at 2:10
  • $\begingroup$ The barycenter of the earth-moon system is very mild compared to others we know of (see pluto-charon). Mars also has 2 moons, Phobos and Deimos, is the gravity-well of Mars considered also as the barycenter of these three bodies? It's all relative to what you're asking. Once again, if you reach escape velocity for the Earth, it's likely that the moon won't do much other than skew your path, which possibly including slowing, until you reach a distance where the effect of gravity from the moon is negligible. $\endgroup$ Commented Oct 3, 2018 at 2:28
  • $\begingroup$ @MartianTycoon: It's true that you have to take Earth's gravity into account, but at 300,000 km out, it's far weaker than Martian gravity is at the surface. $\endgroup$ Commented Oct 3, 2018 at 2:32
  • $\begingroup$ @NathanTuggy not to mention that Earth is tugging you "in the correct direction", assuming the manufacturer's goal is returning manufactured materials to earth. The tug of the earth would actually be a boon to manufacturing on the moon. $\endgroup$ Commented Oct 3, 2018 at 2:36

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