<p>Edited.
<strong>No</strong> 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 $E&gt;0$ and hence $v&gt;v_{\mathrm{esc}}$ when hitting ground.</p>

<hr>

<p><strong>Yes</strong>, 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 <em>local escape speed</em> 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 $v_{\rm escape}$.</p>

<p>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.</p>

<p>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.</p>

CC BY-SA 3.0
2016-04-21T20:04:49.173Z
You seem to be addressing solar escape velocity here.  I'm talking about the escape velocity of the object being  hit.
Loren Pechtel
CC BY-SA 3.0
2016-04-22T19:19:19.757Z
Indeed, you didn't clearly state what you meant.
Walter
CC BY-SA 3.0
2014-10-05T16:22:14.607Z
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
CC BY-SA 3.0
2014-10-06T17:38:19.230Z
@LDC3 clarified in answer, but this was already evident from the question. Did you vote this down??
Walter
CC BY-SA 3.0
2014-10-05T16:00:55.960Z
2016-04-22T19:22:12.610Z
2016-04-22T19:22:12.610Z
Walter
<p>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.</p>

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

CC BY-SA 3.0
2016-04-21T14:37:26.540Z
2016-04-21T14:37:26.540Z
Tanenthor
<p>Edited.  I thought about this wrong.</p>

<p>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.</p>

CC BY-SA 3.0
2016-04-23T01:50:54.180Z
What causes the spiraling orbit, though?
Loren Pechtel
CC BY-SA 3.0
2016-04-23T06:53:01.993Z
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.
userLTK
CC BY-SA 3.0
2016-04-24T20:33:31.730Z
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.
ProfRob
CC BY-SA 3.0
2016-04-25T01:18:31.973Z
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.
userLTK
CC BY-SA 3.0
2016-04-23T00:31:45.510Z
2016-04-25T01:20:07.747Z
2016-04-25T01:20:07.747Z
userLTK
<p>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.</p>

<p>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.</p>

CC BY-SA 3.0
2016-04-23T20:25:50.387Z
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.
ProfRob
CC BY-SA 3.0
2014-10-05T03:34:24.100Z
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
CC BY-SA 3.0
2014-10-06T18:32:29.797Z
Meteors (actually meteorites) that hit the ground are typically slowed considerably by air resistance.
Keith Thompson
<gravity><speed><meteor>
Is there any way a meteor can hit at less than escape velocity?
Loren Pechtel
# Is there any way a meteor can hit at less than escape velocity?

(Asked by: username_0 on 2014-10-05)

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.

**Tags:** gravity, speed, meteor
> **Comments:**
> **username_2**: 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.  
> **username_3**: Meteors (actually meteorites) that hit the ground are typically slowed considerably by air resistance.  
> **username_1**: 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.  

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## 3 answers

### Answer by username_4 on 2014-10-05. Score: 5

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.

> **Comments:**
> **username_2**: 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)?  
> **username_4**: @LDC3 clarified in answer, but this was already evident from the question. Did you vote this down??  
> **username_0**: You seem to be addressing solar escape velocity here.  I'm talking about the escape velocity of the object being  hit.  
> **username_4**: Indeed, you didn't clearly state what you meant.  

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### Answer by username_5 on 2016-04-21. Score: 2

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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### Answer by username_6 on 2016-04-23. Score: 0

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.

> **Comments:**
> **username_0**: What causes the spiraling orbit, though?  
> **username_6**: 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.  
> **username_1**: 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.  
> **username_6**: 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.  

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