<p>If the Earth was stationary and asteroids were coming at it from a uniform direction, the angular impact distribution could be expressed concisely with a simple function involving a cosine.</p>
<p><a href="https://i.sstatic.net/XQB4w6cg.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/XQB4w6cg.png" alt="enter image description here" /></a></p>
<p>The derivation is explained in this paper <a href="https://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1993JBAA..103..123H&amp;db_key=AST&amp;page_ind=0&amp;data_type=GIF&amp;type=SCREEN_VIEW&amp;classic=YES" rel="nofollow noreferrer">https://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1993JBAA..103..123H&amp;db_key=AST&amp;page_ind=0&amp;data_type=GIF&amp;type=SCREEN_VIEW&amp;classic=YES</a></p>
<p>However, the trajectory of incoming asteroids is modified by the Earth’s gravitational field which converts their preferred straight trajectories into conic sections (usually hyperbole) with closest approach (perigee) tangential to the Earth’s surface.</p>
<p><a href="https://i.sstatic.net/zOliiaI5.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/zOliiaI5.png" alt="enter image description here" /></a></p>
<p>Most asteroids which cross Earth’s orbit (a necessity for collision) will have a prograde velocity, same as the the Earth. Comets (and their remnants) as well as extra-solar objects can have extremely high relative velocity and more random directions.</p>
<p><a href="https://i.sstatic.net/oTKamzsA.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/oTKamzsA.png" alt="enter image description here" /></a></p>
<p><a href="https://i.sstatic.net/it6LOZej.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/it6LOZej.png" alt="enter image description here" /></a></p>
<p>If the relative velocity between asteroid and Earth is fairly low (by heliocentric orbital speed standards) then its trajectory can be considered a gravity assist flyby (unless the trajectory intersects with Earth atmosphere). <a href="https://en.wikipedia.org/wiki/Gravity_assist" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Gravity_assist</a></p>

CC BY-SA 4.0
2024-07-08T09:58:27.230Z
The claim that "Any asteroids which cross Earth’s orbit (a necessity for collision) will have a faster velocity (in a solar inertial frame) than the Earth" is not true. See the [Aten](https://en.wikipedia.org/wiki/Aten_asteroid) asteroids, they'd be slower. Objects crossing the Earth's orbit also aren't required to be prograde,  For example every meteor observed from the [Orionidx](https://en.wikipedia.org/wiki/Orionids) shower are in retrograde solar orbits, as they come from [Halley's Comet](https://en.wikipedia.org/wiki/Halley%27s_Comet), which itself is in a retrograde solar orbit.
notovny
CC BY-SA 4.0
2024-07-08T03:17:18.203Z
2024-07-08T18:02:19.610Z
2024-07-08T18:02:19.610Z
Woody
<p>Last Thursday, the Planetary Defense Force(!) watched not one, but TWO asteroids miss us again. The same day.</p>
<p>They were each improbable for different reasons: one was closer than the moon, and one was a mile-wide dinosaur killer (4 times as far as the Webb telescope). But even the small one (below) was 500 feet wide:</p>
<p><a href="https://i.sstatic.net/vB9F0Yo7.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/vB9F0Yo7.png" alt="dinosaur killer" /></a></p>
<p>They weren't a threat. You multiply probabilities: small x small = teeny.</p>
<p>My question is not about the danger; it relates to the angle of entry into the atmosphere.</p>
<p>in videos, I noticed that all the meteors appear to be entering almost parallel to the Earth's surface (tangent to the planet). That always seemed to me to be an amazing coincidence. Why aren't they raining down from overhead?</p>
<p>One explanation is that if it came from overhead, it wouldn't cover very much sky.</p>
<p>But it occurred to me that if you see a meteor, it will almost always shoot across the sky approximately parallel to the surface, because the percentage of the hemisphere that is a specific angle away from dead center increases with the angle. There's much more land (and population) near the earth's limb than there is directly below the Rock of Doom.</p>
<p>HOWEVER, from the point of view of the asteroid, most of the target is close to dead center. The asteroid can barely see the cities on the horizon, but Hawaii looms large.</p>
<p>This seems a contradiction.</p>
<p><strong>MY QUESTION IS: what function describes the probability of the angle at which a meteor enters the atmosphere (it would involve a cosine)<em>, given that most asteroids orbit near the ecliptic?</em></strong></p>
<p>Except, the ecliptic clustering doesn't affect the entry angle, it would only tend to make meteors more likely to come from that direction in the sky.</p>
<p>Right?</p>

CC BY-SA 4.0
2024-07-05T00:02:47.147Z
I don't that the function you are asking for exist, because the radius of Earth is too small, so when it travels through the asteroid dense area, it will be fully submerge by asteroids. So the chances of an asteroid entering Earth in a specific angle is totally random.
Polaris5744
CC BY-SA 4.0
2024-07-05T07:35:26.940Z
Note that the Earth traverses a distance on its orbit equal to its own diameter in ~7 minutes. So the angle that an impactor would make with the ground can vary rather quickly.
PM 2Ring
CC BY-SA 4.0
2024-07-06T13:41:39.737Z
I think the idea that meteors mostly travel parallel to the ground is a mistaken impression. Unless you're directly under it, a meteor looks like it's moving across your sky, but that's an optical illusion, it's actually coming down.
Darth Pseudonym
CC BY-SA 4.0
2024-07-08T13:48:00.433Z
https://www.researchgate.net/publication/373932696_Distribution_Characteristics_of_Meteor_Angle_of_Arrival_in_Mohe_and_Wuhan_China - see figures 1(e) and 2(e) "zenith angle". Looks like the registered meteors are indeed mostly close to horizontal trajectory, with zenith angle over 50 degrees. But probably here is selection bias (my guess) - "close to horizontal" meteors make longer trail and so it's easier to register them by the radar.
Heopps
<asteroids><meteor>
Last Thursday's mile-wide asteroid that missed: what angle would it have hit?
Miss Understands
# Last Thursday's mile-wide asteroid that missed: what angle would it have hit?

(Asked by: username_0 on 2024-07-04)

Last Thursday, the Planetary Defense Force(!) watched not one, but TWO asteroids miss us again. The same day.

They were each improbable for different reasons: one was closer than the moon, and one was a mile-wide dinosaur killer (4 times as far as the Webb telescope). But even the small one (below) was 500 feet wide:

[![dinosaur killer](https://i.sstatic.net/vB9F0Yo7.png)](https://i.sstatic.net/vB9F0Yo7.png)

They weren't a threat. You multiply probabilities: small x small = teeny.

My question is not about the danger; it relates to the angle of entry into the atmosphere.

in videos, I noticed that all the meteors appear to be entering almost parallel to the Earth's surface (tangent to the planet). That always seemed to me to be an amazing coincidence. Why aren't they raining down from overhead?

One explanation is that if it came from overhead, it wouldn't cover very much sky.

But it occurred to me that if you see a meteor, it will almost always shoot across the sky approximately parallel to the surface, because the percentage of the hemisphere that is a specific angle away from dead center increases with the angle. There's much more land (and population) near the earth's limb than there is directly below the Rock of Doom.

HOWEVER, from the point of view of the asteroid, most of the target is close to dead center. The asteroid can barely see the cities on the horizon, but Hawaii looms large.

This seems a contradiction.

**MY QUESTION IS: what function describes the probability of the angle at which a meteor enters the atmosphere (it would involve a cosine)*, given that most asteroids orbit near the ecliptic?***

Except, the ecliptic clustering doesn't affect the entry angle, it would only tend to make meteors more likely to come from that direction in the sky.

Right?

**Tags:** asteroids, meteor
> **Comments:**
> **username_1**: I don't that the function you are asking for exist, because the radius of Earth is too small, so when it travels through the asteroid dense area, it will be fully submerge by asteroids. So the chances of an asteroid entering Earth in a specific angle is totally random.  
> **username_2**: Note that the Earth traverses a distance on its orbit equal to its own diameter in ~7 minutes. So the angle that an impactor would make with the ground can vary rather quickly.  
> **username_3**: I think the idea that meteors mostly travel parallel to the ground is a mistaken impression. Unless you're directly under it, a meteor looks like it's moving across your sky, but that's an optical illusion, it's actually coming down.  
> **username_4**: https://www.researchgate.net/publication/373932696_Distribution_Characteristics_of_Meteor_Angle_of_Arrival_in_Mohe_and_Wuhan_China - see figures 1(e) and 2(e) "zenith angle". Looks like the registered meteors are indeed mostly close to horizontal trajectory, with zenith angle over 50 degrees. But probably here is selection bias (my guess) - "close to horizontal" meteors make longer trail and so it's easier to register them by the radar.  

---

## 1 answers

### Answer by username_5 on 2024-07-08. Score: 2 | Accepted answer

If the Earth was stationary and asteroids were coming at it from a uniform direction, the angular impact distribution could be expressed concisely with a simple function involving a cosine.

[![enter image description here](https://i.sstatic.net/XQB4w6cg.png)](https://i.sstatic.net/XQB4w6cg.png)

The derivation is explained in this paper <https://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?bibcode=1993JBAA..103..123H&db_key=AST&page_ind=0&data_type=GIF&type=SCREEN_VIEW&classic=YES>

However, the trajectory of incoming asteroids is modified by the Earth’s gravitational field which converts their preferred straight trajectories into conic sections (usually hyperbole) with closest approach (perigee) tangential to the Earth’s surface.

[![enter image description here](https://i.sstatic.net/zOliiaI5.png)](https://i.sstatic.net/zOliiaI5.png)

Most asteroids which cross Earth’s orbit (a necessity for collision) will have a prograde velocity, same as the the Earth. Comets (and their remnants) as well as extra-solar objects can have extremely high relative velocity and more random directions.

[![enter image description here](https://i.sstatic.net/oTKamzsA.png)](https://i.sstatic.net/oTKamzsA.png)

[![enter image description here](https://i.sstatic.net/it6LOZej.png)](https://i.sstatic.net/it6LOZej.png)

If the relative velocity between asteroid and Earth is fairly low (by heliocentric orbital speed standards) then its trajectory can be considered a gravity assist flyby (unless the trajectory intersects with Earth atmosphere). <https://en.wikipedia.org/wiki/Gravity_assist>

> **Comments:**
> **username_6**: The claim that "Any asteroids which cross Earth’s orbit (a necessity for collision) will have a faster velocity (in a solar inertial frame) than the Earth" is not true. See the [Aten](https://en.wikipedia.org/wiki/Aten_asteroid) asteroids, they'd be slower. Objects crossing the Earth's orbit also aren't required to be prograde,  For example every meteor observed from the [Orionidx](https://en.wikipedia.org/wiki/Orionids) shower are in retrograde solar orbits, as they come from [Halley's Comet](https://en.wikipedia.org/wiki/Halley%27s_Comet), which itself is in a retrograde solar orbit.  

---

