# Is there any way a meteor can hit at less than escape velocity?

Note: I have figured out one case: The meteor is splash from an impact on a close-in moon. I'm looking for any path from deep space.

My understanding is that anything from deep space will inherently pick up the escape velocity in it's fall and thus there is no possibility in the simple case. I'm not so sure about what could happen after an encounter with one or more moons, though.

• There might be a possibility, but it would be very unlikely. The meteor would need to have its orbit deflected into an orbit very similar to Earth and then have the moon slow its speed when it's about to collide. – LDC3 Oct 5 '14 at 3:34
• Meteors (actually meteorites) that hit the ground are typically slowed considerably by air resistance. – Keith Thompson Oct 6 '14 at 18:32
• Meteorites hit the Earth's atmosphere at 11-72 km/s. The minimum is basically the Earth's escape velocity, since none originate from close to the Earth. The maximum is for an object hitting the Earth head on in a parabolic orbit around the Sun. – Rob Jeffries Apr 23 '16 at 20:25

Edited. No if you talk about the escape velocity form Earth. This follows simply from the fact that energy $E$ is conserved. An object that is not gravitationally bound to Earth has $E>0$ and hence $v>v_{\mathrm{esc}}$ when hitting ground.

Yes, if you meant the escape velocity from the Solar system, because the Earth moves with $v_{\rm Earth}=v_{\rm escape}/\sqrt{2}$ relative to (but not towards) the Sun. Here $v_{\rm escape}=\sqrt{2GM_{\odot}/1{\rm AU}}$ is the local escape speed from the Sun, while $v_{\rm Earth}=\sqrt{GM_{\odot}/1{\rm AU}}$ is the speed of the local circular orbit. An object at 1AU form the Sun and bound to the Sun cannot have speed greater than $v_{\rm escape}$.

Now, the impact speed of an object that moves at $v_{\rm escape}$ can be as low $v_{\rm escape}-v_{\rm Earth}=v_{\rm escape}(1-1/\sqrt{2})$ if it hits Earth "from behind", i.e. moving in the same direction as Earth at the time of impact.

Note also that meterors typically move not faster than $v_{\rm escape}$, for they don't come from outer space, but from the Solar system.

• How do you know that the Earth's velocity is $1/\sqrt {2}$ times the escape velocity? Also, which escape velocity are you referring to (Earth's escape velocity or the solar system escape velocity)? – LDC3 Oct 5 '14 at 16:22
• @LDC3 clarified in answer, but this was already evident from the question. Did you vote this down?? – Walter Oct 6 '14 at 17:38
• You seem to be addressing solar escape velocity here. I'm talking about the escape velocity of the object being hit. – Loren Pechtel Apr 21 '16 at 20:04
• Indeed, you didn't clearly state what you meant. – Walter Apr 22 '16 at 19:19

If you mean the escape velocity of the Earth, then no. Any meteor that enters the earth's atmosphere and burns up or collides with the earth must have had higher than the escape velocity, purely because they were not from that one body.

Unless of course as you say they have ricocheted of another object in orbit sufficiently reducing the kinetic energy to below escape levels