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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.

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  • $\begingroup$ 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. $\endgroup$
    – LDC3
    Oct 5, 2014 at 3:34
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    $\begingroup$ Meteors (actually meteorites) that hit the ground are typically slowed considerably by air resistance. $\endgroup$
    – Keith Thompson
    Oct 6, 2014 at 18:32
  • $\begingroup$ 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. $\endgroup$
    – ProfRob
    Apr 23, 2016 at 20:25

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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.

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  • $\begingroup$ 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)? $\endgroup$
    – LDC3
    Oct 5, 2014 at 16:22
  • $\begingroup$ @LDC3 clarified in answer, but this was already evident from the question. Did you vote this down?? $\endgroup$
    – Walter
    Oct 6, 2014 at 17:38
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    $\begingroup$ You seem to be addressing solar escape velocity here. I'm talking about the escape velocity of the object being hit. $\endgroup$
    – Loren Pechtel
    Apr 21, 2016 at 20:04
  • $\begingroup$ Indeed, you didn't clearly state what you meant. $\endgroup$
    – Walter
    Apr 22, 2016 at 19:19
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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

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Edited. I thought about this wrong.

A slowly spiraling in orbit would hit the earth at not much more than orbital speed, which is escape velocity over the square root of 2, but not a meteor approaching from a distance, so I thought about this the wrong way. Answer is no.

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  • $\begingroup$ What causes the spiraling orbit, though? $\endgroup$
    – Loren Pechtel
    Apr 23, 2016 at 1:50
  • $\begingroup$ Tidal forces. Uneven surface gravity. Air resistance. Granted those are mostly very gradual (Air resistance can be pretty quick but only if it's in the atmosphere). Imagine if you will, a planet with no atmosphere and an object in orbit just a few feet above the planet's surface, when that object impacts it would impact at orbital velocity not escape velocity. Granted, it's a highly unlikely scenario. A non circular orbit, but not too elliptical should impact at less than escape velocity. Shoemaker Levy-9 for example, but that case, only a hair under. $\endgroup$
    – userLTK
    Apr 23, 2016 at 6:53
  • $\begingroup$ This isn't right. Suppose something approaches the Earth that is in a similar orbit to Earth and travelling in the same direction, but just a touch faster. As it gets nearer to the Earth, the Earth's gravity accelerates the object and it picks up the escape velocity. There isn't any way to avoid this without involving a third body. $\endgroup$
    – ProfRob
    Apr 24, 2016 at 20:33
  • $\begingroup$ You're right, I hadn't considered the approach from a distance inevitably reaches at least escape velocity on impact. It's possible for an object in low orbit to strike the earth at below escape velocity, but not for a meteor approaching a planet. I'll edit. $\endgroup$
    – userLTK
    Apr 25, 2016 at 1:18

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