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You have to leave Mars's surface at 6300 m/s to get to Earth, ignoring slowing down at the end which is accomplished by the atmosphere. To get to Earth from th

You have to leave Mars's surface at 6300 m/s to get to Earth, ignoring slowing down at the end which is accomplished by the atmosphere. To get to Earth from the Moon takes 5670 ms/s. The Moon is much closer in Cartesian space, but that just means that the delay from an object leaving the Moon's surface to getting to the Earth's is much smaller, it affects latency but not throughput and on geologic timescales both latencies are small. EDIT: See this map of the solar system in delta-v terms: http://i.imgur.com/AAGJvD1.png http://i.imgur.com/AAGJvD1.png EDIT2: Argh, I messed up my calculations and the distance from the moon is just 2410 m/s. You have to factor in how often Mars gets hit by stuff and the speed at which stuff is going given Mars's gravity well.

bostonpete 9y ago
It seems like your comment ignores the relevance of proximity in the likelihood that an ejected object with the necessary velocity ends up at Earth. Wouldn't the probability of such an object hitting Earth from Mars would be many orders of magnitude smaller than one from the Moon (like trying to hit a target from 100x further away)? I always understood the large numbers of meteorites from Mars to be related to the fact that Mars is closer to the asteroid belt and therefore has had a lot more impacts historically...
- Symmetry 9y ago
  The levels of delta-v we're talking about here aren't nearly enough to knock these objects out of the solar system. When they leave the Moon or Mars they're still in orbit and will remain in orbit until they collide with something. In the case of a moon rock that means either hitting the Earth or returning to the Moon in what is likely a single orbit so 2 weeks. In the case of Mars it will either hit Mars again or the Earth. Or possibly the Moon, one Mars's moons, or an asteroid but probably the first two. The odds of this collision taking place on any given orbit around the Sun are low but the object isn't going anywhere and eventually, possibly millions of orbits later, it will end up hitting something. So no, the probability that it will eventually hit the Earth doesn't really go down with the distance in this case since we're only talking millions of years for a collision to occur rather than billions.

bostonpete 9y ago

It seems like your comment ignores the relevance of proximity in the likelihood that an ejected object with the necessary velocity ends up at Earth. Wouldn't the probability of such an object hitting Earth from Mars would be many orders of magnitude smaller than one from the Moon (like trying to hit a target from 100x further away)? I always understood the large numbers of meteorites from Mars to be related to the fact that Mars is closer to the asteroid belt and therefore has had a lot more impacts historically...

Symmetry 9y ago
The levels of delta-v we're talking about here aren't nearly enough to knock these objects out of the solar system. When they leave the Moon or Mars they're still in orbit and will remain in orbit until they collide with something. In the case of a moon rock that means either hitting the Earth or returning to the Moon in what is likely a single orbit so 2 weeks. In the case of Mars it will either hit Mars again or the Earth. Or possibly the Moon, one Mars's moons, or an asteroid but probably the first two. The odds of this collision taking place on any given orbit around the Sun are low but the object isn't going anywhere and eventually, possibly millions of orbits later, it will end up hitting something. So no, the probability that it will eventually hit the Earth doesn't really go down with the distance in this case since we're only talking millions of years for a collision to occur rather than billions.