Where on Earth is statistically safer if an asteroid was to break up in the Roche limit and rain meteorites effectively resetting the Earth? I would imagine that a ring of falling meteors would form around the equator making the poles more safe?

  • 7
    $\begingroup$ You seem to suppose that an asteroid will get captured into an Earth orbit. The orbit will decay to the point that the roche limit is passed and the asteroid breaks up. But that is the most unlikely scenario. Far more likely is that the asteroid just hits the Earth. And it could hit anywhere. $\endgroup$
    – James K
    Feb 13 at 7:16
  • 6
    $\begingroup$ Most asteroids are not held together by gravity, but are simply solid chunks of rock. They won't be pulled apart at the Roche limit. $\endgroup$ Feb 13 at 13:42
  • 9
    $\begingroup$ @NuclearHoagie I'm not sure we can say with clarity what the makeup of "most" asteroids is. Recent missions suggest that more asteroids are rubble-piles than previously thought. $\endgroup$ Feb 13 at 15:34
  • 1
    $\begingroup$ The asteroid can travel many Roche radii in the timescale on which it might fall apart. Therefore tidal break up can be neglected (see edit in my answer), even for rubble piles. $\endgroup$
    – ProfRob
    Feb 13 at 17:09
  • 4
    $\begingroup$ What does "effectively resetting the Earth" mean? $\endgroup$
    – PM 2Ring
    Feb 14 at 1:11

1 Answer 1


Robertson et al. (2021) (note the erratum) provide a very comprehensive answer to your question.

Surprisingly (to me) the asteroidal flux (impacts per unit area) is slightly higher at the poles (by 8%) than at the equator.

However, the impact angles tend to be slightly shallower at the equator, which leads to airbursts that are more dangerous for smaller asteroids. Larger asteroids reach the ground whatever and so the lower flux at the equator means slightly less risk. Asteroid velocities are similar at all latitudes.

Note that asteroids would hit the Earth at greater than the Earth's escape velocity and have a similar density to the Earth. They are unlikely to be captured and unlikely to be broken up by tidal forces before entering the atmosphere unless they were so large they would end all life on Earth.

Your best bet is to always stay at the point on Earth that is antipodal to the direction of motion in its orbit around the Sun, since the flux will be higher from that direction and the incoming relative velocities larger. Of course, the Earth spins, so that is difficult...

EDIT: To address some comments on the question. The break up timescale of a gravitationally bound asteroid is going to be a small multiple of the freefall timescale of $\tau_{ff} \simeq (G\rho)^{-1/2}$, where $\rho \sim 3000$ kg/m$^3$ for a typical asteroid density and $\tau_{ff} \sim 2000$ s.

At a minimum closing relative velocty of $\sim 11$ km/s, the asteroid will travel $>20,000$ km during $\tau_{ff}$, but given that the Roche radius will not be much larger than the Earth's radius, it will already have impacted before it has any chance to respond to tidal forces.

  • 1
    $\begingroup$ Re your last paragraph: This suggests flying in a jet that keeps you at the antipode? But if you're already in a jet you should be able to just fly away from the impact area (I assume astronomers can predict it in time). $\endgroup$
    – Barmar
    Feb 13 at 15:52
  • $\begingroup$ And the thicker atmosphere at the equater would also likely help. $\endgroup$ Feb 13 at 16:54
  • 1
    $\begingroup$ @Barmar you might find that tickets are quite expensive... $\endgroup$
    – ProfRob
    Feb 13 at 17:00
  • $\begingroup$ Just keep that airplane flying into the evening sunset at all times at because the sunrise and the dawn is where the asteroids hit, it's like the front side and the back of a car driving into the rain, the dawn is the front side and the dusk is the back. $\endgroup$ Feb 13 at 20:06
  • 1
    $\begingroup$ @PierrePaquette thanks. I made an edit. The answer is qualitatively the same, but "slightly" becomes even more slightly. $\endgroup$
    – ProfRob
    Feb 14 at 12:29

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .