<p>Depending on how something is launched from its original location, potentially an interstellar object could be travelling significantly faster than in-system velocities.</p>
<p>If anything flies into our system at relativistic speeds, though, it's a weapon, most likely: and we're toast.</p>

CC BY-SA 4.0
2020-12-15T02:41:22.877Z
Welcome to Stack Exchange! You've used an answer post to leave what is really just a casual comment. Once you reach 50 reputation points you'll be able to leave comments on other people's posts.
uhoh
CC BY-SA 4.0
2020-12-15T03:47:36.763Z
FWIW, we do get single atoms, mostly hydrogen, and some helium, from outside the Solar System (ok, mostly nuclei not whole atoms). Some of them have considerable speed, eg https://en.wikipedia.org/wiki/Oh-My-God_particle
PM 2Ring
CC BY-SA 4.0
2020-12-15T02:16:48.820Z
2020-12-15T02:16:48.820Z
Marc Wilson
<p>Although most objects orbit the Sun in the same direction — having emerged out of the same rotating gas cloud that spawned the Solar System — some asteroids and other <a href="https://en.wikipedia.org/wiki/Minor_planet" rel="noreferrer">minor planets</a> do move in opposite, or <em>retrograde</em>, orbits (see <a href="https://en.wikipedia.org/wiki/List_of_exceptional_asteroids#Retrograde" rel="noreferrer">this Wikipedia article</a> for a list of such objects).</p>

<p>The minimum speed for an asteroid is achieved if it has more or less the same velocity around the Sun as the Earth. In this case the gravitational attraction of Earth will accelerate the object to the escape velocity of Earth, i.e.
$$
v_\mathrm{min} = v_{\mathrm{esc,}\oplus} = \sqrt{\frac{2GM_\oplus}{R_\oplus}} \simeq 11\,\mathrm{km}\,\mathrm{s}^{-1}.
$$
Here, $G$, $M_\oplus$, and $R_\oplus$ are the gravitational constant and the mass and radius of Earth, respectively.</p>

<p>The maximum speed is achieved at a "head-on" collision. Earth's speed around the Sun is
$$
v_{\mathrm{orb,}\oplus} = \sqrt{\frac{GM_\odot}{d}} \simeq 30\,\mathrm{km}\,\mathrm{s}^{-1},
$$
where $M_\odot$ and $d$ are the mass of the Sun and the distance from Sun to Earth (1 AU).</p>

<p>If the asteroid travels on the same orbit, but in the opposite direction, the impact will then be at 60 km/s. However, if the asteroid comes from far away (e.g. the Oort Cloud), it will be accelerated by the Sun and achieve a velocity equal to <em>the escape velocity from the Sun at the location of Earth</em>. As is seen from the two equations above, the orbital speed and the escape velocity differ by a factor of $\sqrt{2}$. That is, an object falling from infinity toward the Sun, will have a speed equal to $30\,\mathrm{km}\,\mathrm{s}^{-1}\times\sqrt{2}=42\,\mathrm{km}\,\mathrm{s}^{-1}$ when it reaches Earth.</p>

<p>Hence, the maximum impact velocity is
$$
v_\mathrm{max} = 30\,\mathrm{km}\,\mathrm{s}^{-1} + 42\,\mathrm{km}\,\mathrm{s}^{-1} = 72\,\mathrm{km}\,\mathrm{s}^{-1}.
$$</p>

CC BY-SA 3.0
2018-01-22T16:01:41.257Z
@BSeven Why straight up?  Pela adressed that: "the more inclined your path is, the more atmosphere you'd also have to go through".  More atmosphere is, quite literally, a drag.  The sooner that you get out of the atmosphere, the better.
badjohn
CC BY-SA 3.0
2017-04-21T22:07:04.703Z
The head on collision should be close to perpendicular which would require vector addition at close to a 90 degree angle.   Also, you should add the 11 km/s for Earth's gravity.   The highest velocity impact might work out to be pure retrograde not high eccentricity.
userLTK
CC BY-SA 3.0
2017-04-21T22:31:45.957Z
@userLTK: Actually, it doesn't have to be perpendicular. You can think of it in reverse: If you launch an object tangentially to Earth's object, it will escape if the speed is ≥42 km/s (similarly, if you ride your bike along the bike lane at ≥11 km/s, you will leave Earth; you don't have to leave Earth perpendicularly). Also, I think that at a head-on collision, the increase in speed due to Earth's gravity can be neglected.
pela
CC BY-SA 3.0
2017-04-21T22:36:18.203Z
Good points.   I think you're right about that.   Should I delete my comment?
userLTK
CC BY-SA 3.0
2017-04-21T22:51:10.200Z
@userLTK: No, I think it's a good comment because you're definitely not the only one who thinks this. Intuitively, it seems easier to leave Earth "straight up" than at an angle, but that's not the case. Although in reality, the more inclined your path is, the more atmosphere you'd also have to go through. But if you leave Earth at an angle in the eastward direction, you gain some speed due to Earth's spin, so in practice I think they use some compromise between speed gain and not-too-much atmosphere (but this is spaceflight, of which I know next to nothing).
pela
CC BY-SA 3.0
2017-04-22T17:53:44.830Z
@pela - Thanks for your awesome answer. Your comment "it seems easier to leave Earth "straight up" than at an angle, but that's not the case" didn't make sense to me. If this is not true, then why are rockets launched straight up?
B Seven
CC BY-SA 3.0
2017-04-24T07:31:04.543Z
@BSeven: First the disclaimer: I'm an theoretical astronomer, not a rocket scientist, so my knowledge about the real world is limited :) But I think that rockets are launched straight up, but then turn their path to an angle in the same direction as the rotation of Earth, benefitting from the rotation. Hmm… I wrote some more here, but I think I shouldn't, so perhaps go to Space Exploration SE instead.
pela
CC BY-SA 4.0
2019-09-24T21:40:27.860Z
Do those limits hold up for interstellar objects (are they still called asteroids, or something else?) that might impact Earth? or is it conceivable that an object originating outside our solar system could arrive at a relative speed, and from a relative direction, that could change those limits somehow?
Harthag
CC BY-SA 4.0
2019-09-25T12:40:27.143Z
@Dalila It pretty much does hold for interstellar asteroids, like 'Oumuamua. Things in the Solar neighborhood don't have wildly different velocities, so an extra-Solar object entering the Solar System will reach velocities of the order of the escape speed. But if an asteroid from another solar system were slingshot out of that system by coming very close to a planet, it may escape with a larger velocity, and if that star had a large velocity wrt. us, it may enter ours with a higher velocity. But relative velocities between stars close to each other are a few 10 km/s, so not a big difference.
pela
CC BY-SA 3.0
2017-04-21T21:54:39.140Z
2017-04-22T15:26:54.617Z
2017-04-22T15:26:54.617Z
pela
<p>I gather that the impact speed depends on the radius of the orbit of the asteroid.</p>

<p>Is the orbit of asteroids in the same direction around the sun, or can they move in the "opposite" direction?</p>

CC BY-SA 3.0
2017-04-22T07:10:54.873Z
see https://physics.stackexchange.com/questions/162386/how-gently-could-a-comet-asteroid-meteorite-hit-earth/162576#162576
ProfRob
<asteroids>
What are the maximum and minimum impact speeds for an asteroid that would strike the Earth?
B Seven
# What are the maximum and minimum impact speeds for an asteroid that would strike the Earth?

(Asked by: username_0 on 2017-04-21)

I gather that the impact speed depends on the radius of the orbit of the asteroid.

Is the orbit of asteroids in the same direction around the sun, or can they move in the "opposite" direction?

**Tags:** asteroids

> **Comments:**
> **username_1**: see https://physics.stackexchange.com/questions/162386/how-gently-could-a-comet-asteroid-meteorite-hit-earth/162576#162576  

---

## 2 answers

### Answer by username_5 on 2017-04-21. Score: 9 | Accepted answer

Although most objects orbit the Sun in the same direction — having emerged out of the same rotating gas cloud that spawned the Solar System — some asteroids and other [minor planets](https://en.wikipedia.org/wiki/Minor_planet) do move in opposite, or *retrograde*, orbits (see [this Wikipedia article](https://en.wikipedia.org/wiki/List_of_exceptional_asteroids#Retrograde) for a list of such objects).

The minimum speed for an asteroid is achieved if it has more or less the same velocity around the Sun as the Earth. In this case the gravitational attraction of Earth will accelerate the object to the escape velocity of Earth, i.e.
$$
v\_\mathrm{min} = v\_{\mathrm{esc,}\oplus} = \sqrt{\frac{2GM\_\oplus}{R\_\oplus}} \simeq 11\,\mathrm{km}\,\mathrm{s}^{-1}.
$$
Here, $G$, $M\_\oplus$, and $R\_\oplus$ are the gravitational constant and the mass and radius of Earth, respectively.

The maximum speed is achieved at a "head-on" collision. Earth's speed around the Sun is
$$
v\_{\mathrm{orb,}\oplus} = \sqrt{\frac{GM\_\odot}{d}} \simeq 30\,\mathrm{km}\,\mathrm{s}^{-1},
$$
where $M\_\odot$ and $d$ are the mass of the Sun and the distance from Sun to Earth (1 AU).

If the asteroid travels on the same orbit, but in the opposite direction, the impact will then be at 60 km/s. However, if the asteroid comes from far away (e.g. the Oort Cloud), it will be accelerated by the Sun and achieve a velocity equal to *the escape velocity from the Sun at the location of Earth*. As is seen from the two equations above, the orbital speed and the escape velocity differ by a factor of $\sqrt{2}$. That is, an object falling from infinity toward the Sun, will have a speed equal to $30\,\mathrm{km}\,\mathrm{s}^{-1}\times\sqrt{2}=42\,\mathrm{km}\,\mathrm{s}^{-1}$ when it reaches Earth.

Hence, the maximum impact velocity is
$$
v\_\mathrm{max} = 30\,\mathrm{km}\,\mathrm{s}^{-1} + 42\,\mathrm{km}\,\mathrm{s}^{-1} = 72\,\mathrm{km}\,\mathrm{s}^{-1}.
$$

> **Comments:**
> **username_7**: The head on collision should be close to perpendicular which would require vector addition at close to a 90 degree angle.   Also, you should add the 11 km/s for Earth's gravity.   The highest velocity impact might work out to be pure retrograde not high eccentricity.  
> **username_5**: @username_7: Actually, it doesn't have to be perpendicular. You can think of it in reverse: If you launch an object tangentially to Earth's object, it will escape if the speed is ≥42 km/s (similarly, if you ride your bike along the bike lane at ≥11 km/s, you will leave Earth; you don't have to leave Earth perpendicularly). Also, I think that at a head-on collision, the increase in speed due to Earth's gravity can be neglected.  
> **username_7**: Good points.   I think you're right about that.   Should I delete my comment?  
> **username_5**: @username_7: No, I think it's a good comment because you're definitely not the only one who thinks this. Intuitively, it seems easier to leave Earth "straight up" than at an angle, but that's not the case. Although in reality, the more inclined your path is, the more atmosphere you'd also have to go through. But if you leave Earth at an angle in the eastward direction, you gain some speed due to Earth's spin, so in practice I think they use some compromise between speed gain and not-too-much atmosphere (but this is spaceflight, of which I know next to nothing).  
> **username_0**: @username_5 - Thanks for your awesome answer. Your comment "it seems easier to leave Earth "straight up" than at an angle, but that's not the case" didn't make sense to me. If this is not true, then why are rockets launched straight up?  
> **username_5**: @BSeven: First the disclaimer: I'm an theoretical astronomer, not a rocket scientist, so my knowledge about the real world is limited :) But I think that rockets are launched straight up, but then turn their path to an angle in the same direction as the rotation of Earth, benefitting from the rotation. Hmm… I wrote some more here, but I think I shouldn't, so perhaps go to Space Exploration SE instead.  
> **username_6**: @BSeven Why straight up?  Pela adressed that: "the more inclined your path is, the more atmosphere you'd also have to go through".  More atmosphere is, quite literally, a drag.  The sooner that you get out of the atmosphere, the better.  
> **username_8**: Do those limits hold up for interstellar objects (are they still called asteroids, or something else?) that might impact Earth? or is it conceivable that an object originating outside our solar system could arrive at a relative speed, and from a relative direction, that could change those limits somehow?  
> **username_5**: @Dalila It pretty much does hold for interstellar asteroids, like 'Oumuamua. Things in the Solar neighborhood don't have wildly different velocities, so an extra-Solar object entering the Solar System will reach velocities of the order of the escape speed. But if an asteroid from another solar system were slingshot out of that system by coming very close to a planet, it may escape with a larger velocity, and if that star had a large velocity wrt. us, it may enter ours with a higher velocity. But relative velocities between stars close to each other are a few 10 km/s, so not a big difference.  

---

### Answer by username_2 on 2020-12-15. Score: 1

Depending on how something is launched from its original location, potentially an interstellar object could be travelling significantly faster than in-system velocities.

If anything flies into our system at relativistic speeds, though, it's a weapon, most likely: and we're toast.

> **Comments:**
> **username_3**: Welcome to Stack Exchange! You've used an answer post to leave what is really just a casual comment. Once you reach 50 reputation points you'll be able to leave comments on other people's posts.  
> **username_4**: FWIW, we do get single atoms, mostly hydrogen, and some helium, from outside the Solar System (ok, mostly nuclei not whole atoms). Some of them have considerable speed, eg https://en.wikipedia.org/wiki/Oh-My-God_particle  

---

