<p>It could be as little notice as a few days.</p>
<p>There is a large variation in the amount of warning we would have before an asteroid strike similar in energy to Chicxulub.  Let’s examine the case in which we would have the least warning.</p>
<p>The lower bound to the estimated energy released by the Chicxulub collision is 1.3e24 joules <a href="https://en.wikipedia.org/wiki/Chicxulub_crater#Impact_specifics" rel="noreferrer">https://en.wikipedia.org/wiki/Chicxulub_crater#Impact_specifics</a>.
We know the object ‘Oumoamoa was moving at about 50km/s at 1 au from the sun <a href="https://en.wikipedia.org/wiki/%CA%BBOumuamua" rel="noreferrer">https://en.wikipedia.org/wiki/%CA%BBOumuamua</a> , which is the distance of the earth from the sun.  The earth itself is moving at about 30 km/s with respect to the sun, so if they struck in opposition,  the difference in velocity could be 80 km/s.</p>
<p>For kinetic energy impact, we know <span class="math-container">$e=1/2*mv^2$</span>, or energy = 0.5*mass times the square of the velocity.  So <span class="math-container">$m=2e/v^2$</span>, so m is approximately 4e14kg.  If we assume an asteroid a bit denser than iron at 1e4 kg/m^3, then the volume of the asteroid is 4e10m^3.  If we assume the asteroid is a sphere, then since the volume of a sphere is <span class="math-container">$V=4/3\pi r^3$</span>, then the radius is <span class="math-container">$r^3 = (3/4*V/\pi)$</span> or <span class="math-container">$r$</span>= 2.121km.
Let’s assume the asteroid is matte black so it hardy reflects any light, give it an albedo of 0.02.  Then the absolute magnitude of the asteroid of this size and albedo at 1 AU from an observer is 17, using the tables here <a href="https://cneos.jpl.nasa.gov/tools/ast_size_est.html" rel="noreferrer">https://cneos.jpl.nasa.gov/tools/ast_size_est.html</a>.  Let’s suppose that we spot the asteroid when it is much much brighter than at one AU (or 1.5e8 km), with the same apparent magnitude as Pluto (magnitude 14).  Since magnitude is a log scale, that is <span class="math-container">$2.5^3 = 16$</span> times as bright.</p>
<p>Brightness varies as square of distance, so the asteroid will be at <span class="math-container">$1.5e8/4$</span>, or 3.75e7 km away from earth when it’s spotted.  At 80 km/s relative velocity, the time from seeing it to impact is 4.6875e5 seconds, or about 5.4 days.</p>
<p>Notes:</p>
<ol>
<li><p>The above outlined scenario is almost impossibly unlikely. A much more likely scenario is that an asteroid from our known asteroid belt strikes earth, in which case we would probably know it was going to collide years before the strike.</p>
</li>
<li><p>The object Oumoamoa was classified as an asteroid rather than a comet due to its lack of corona, but now it has an ‘I’ designation for the first Interstellar object which is neither a comet nor an asteroid.  One could argue that I shouldn’t use it as an example given that the question asks about asteroids, but I think it is perfectly reasonable.</p>
</li>
<li><p>We don’t ever know a collision with 100% accuracy.  There is always error in our orbit estimations.  We can’t even tell if satellites are going to collide in earth orbit, it’s just probabilities.</p>
</li>
<li><p>I don't have a good idea of how long a new object of a certain apparent magnitude will be in the night sky before it's discovered by someone.  If someone thinks we would notice this object sooner due to transits or some other reason, I may amend this answer.</p>
</li>
</ol>

CC BY-SA 4.0
2020-11-15T06:55:07.957Z
@Graipher You are absolutely correct, and I was so careful, but this is one thing I didn't check.  I ran the equations back through with this correction and reduced albedo from 0.03 to 0.02 and got the same result. My post is edited above. I am sorry.  If we get a big, dense, asteroid with low albedo coming into our solar system with escape velocity in an earth impact trajectory, it isn't good for the fate of life on this planet.  And we won't have much notice. Connor
Connor Garcia
CC BY-SA 4.0
2020-11-16T13:41:53.157Z
The brightness of an asteroid doesn't vary as the inverse square of the distance to Earth.
ProfRob
CC BY-SA 4.0
2020-11-16T14:03:04.093Z
Also if you do adopt an inverse square law, then for an object to be 16 times as bright it is at 1/4 the distance!  This places it 0.25 au away and even travelling at 80 km/s, that takes more than 5 days to get here.
ProfRob
CC BY-SA 4.0
2020-11-16T14:09:48.960Z
The same wikipedia article you reference also gives a minimum diameter for the dinosaur-killer of 11 km. This makes it more than 25 times brighter and it can therefore be seen at least 5 times further away. In fact the bigger size means that it must be moving slower thn 80 km/s to give the same minimum energy (though note that assuming a density of 10 g/cc is a bit too high - nothing in the solar system is that dense) i.e. Weeks, not hours.
ProfRob
CC BY-SA 4.0
2020-11-17T22:04:59.743Z
@RobJeffries You are right about my inverse square law calcs being wrong, I corrected them.  You are also right about my density being a bit too high and asteroid brightness not varying quite as the inverse square of distance, but I don't think either affects my answer too much.  I think I'm right about the asteroid size. Crater size is good for approximating kinetic energy, but it isn't good for approximating asteroid size.  For a worst case scenario, we should approximate our own size for a dense, fast, mover.
Connor Garcia
CC BY-SA 4.0
2020-11-17T23:27:16.083Z
"The Chicxulub impactor had an estimated diameter of 11–81 kilometers (6.8–50.3 mi), and delivered an estimated energy of 21–921 billion Hiroshima A-bombs (between 1.3×1024 and 5.8×1025 joules, or 1.3–58 yottajoules)". The SAME paper that quotes the minimum energy you used also says the minimum diameter is 11km. Yes, very much a worst possible case scenario, since there aren't a population of iron asteroids on nearly parabolic orbits and Oumuamua had an albedo of 0.1.
ProfRob
CC BY-SA 4.0
2021-05-27T17:59:02.430Z
The worst case scenario would be an interstellar object travelling much faster perpendicular to the plane of the elliptic no? Clearly 80 km/s relative velocity is not the upper limit of such objects.
M. Y. Zuo
CC BY-SA 4.0
2020-11-14T18:37:54.747Z
2020-11-16T17:26:48.383Z
2020-11-16T17:26:48.383Z
Connor Garcia
<p>An object's absolute magnitude (how bright it would appear at full phase, 1 au from both Earth and Sun) is related to its size and easier to determine.
A ~1 km asteroid has absolute magnitude <em>H</em> ≈ 17.75.
The <a href="https://en.wikipedia.org/wiki/Chicxulub_impactor" rel="noreferrer">Chicxulub impactor</a>, estimated <a href="https://ui.adsabs.harvard.edu/abs/2014arXiv1403.6391D/abstract" rel="noreferrer">between 11 and 80 km</a>, was probably 9 ≤ <em>H</em> ≤ 12.</p>
<p>Comet <a href="https://ssd.jpl.nasa.gov/sbdb.cgi?sstr=K19U060" rel="noreferrer">C/2019 U6 (Lemmon)</a> is smaller (<em>H</em> = 13.3) but a good example of modern detection capabilities:</p>
<ul>
<li>2019-10-31: <a href="https://en.wikipedia.org/wiki/Mount_Lemmon_Survey" rel="noreferrer">Mt Lemmon Survey</a> observes an unknown object at apparent magnitude 20.5.</li>
<li>2019-11-08: Minor Planet Center (MPC) <a href="https://minorplanetcenter.net/mpec/K19/K19VD1.html" rel="noreferrer">publishes</a> a preliminary parabolic orbit based on 31 observations from 5 sites. The comet is 3.36 au away from the Sun.</li>
<li>2020-03-25: MPC <a href="https://minorplanetcenter.net/mpec/K20/K20FD6.html" rel="noreferrer">publishes</a> a revised orbit with eccentricity <em>e</em> = 0.9979 based on 225 observations, including a <a href="https://en.wikipedia.org/wiki/Pan-STARRS" rel="noreferrer">Pan-STARRS</a> &quot;precovery&quot; at apparent magnitude 23 in late 2018.</li>
<li>2020-07-09: The comet crosses 0.03 au inside Earth's orbit. Earth is 0.86 au away, having passed the intersection zone 53 days earlier.</li>
</ul>
<p>That's a lead time of 8 months.</p>
<p>Asteroids can sneak up on us from sunward, but those we don't already know about tend to be rather small.
The latest <em>H</em> ≤ 15 near-Earth asteroid (NEA) discovery was in 2004.
Probably 90% or more of <em>H</em> ≤ 18 NEAs are already known; the discovery rate in that size range has been in the <a href="https://cneos.jpl.nasa.gov/stats/site_km.html" rel="noreferrer">single digits per year</a> since 2014.
Current surveys are trying to find 90% of all &quot;potentially hazardous&quot; <em>H</em> ≤ 22 asteroids, i.e. ~140 m or larger.</p>

CC BY-SA 4.0
2020-11-15T21:27:28.197Z
2020-11-16T16:12:47.707Z
2020-11-16T16:12:47.707Z
Mike G
<p>There are two classes of object.  One is &quot;asteroids&quot; and the other is &quot;comets&quot;.</p>
<p>Asteroids orbit in fat ellipses, mostly between Mars and Jupiter, but some come closer and a few can come close to Earth. They are never very far from Earth and we discover new small ones all the time. These are &quot;potentially hazardous objects&quot;.  If there was an 11km asteroid that was potentially hazardous we would, almost certainly, have seen it already.  We know all the asteroids that are of that size are not potentially hazardous because we know their orbits and we know that they don't come near Earth.  We know that none of the potentially hazardous objects will actually hit Earth in the next 200 years or so, because we can calculate their orbit and while they come close, none will actually hit us.</p>
<p>So we are pretty sure that there are no asteroids of that size that will hit us.</p>
<p>Comets are different, because comets come from the distant parts of the solar system and fall in towards the sun on very long orbits.  We often see comets first when they are a few months from the inner part of the solar system.  So we might have as little as a few months notice.</p>
<p>Fortunately, comets like this are rare, and space is big (Really big. You just won't believe how vastly hugely mind-bogglingly big it is...) the probabilty of a large 11km comet coming out of the distant solar system and colliding with Earth is very very small.</p>
<p>So Asteroid impact is a real concern, in much the same way that Earthquakes or typhoons are a real concern.  These are things that can kill lots of people. But for now, a collision that could end human life on Earth is a very very remote possibility.</p>

CC BY-SA 4.0
2020-11-14T23:02:02.107Z
Sometimes the reason that comets enter orbits bringing them into the inner solar system is that they've interacted gravitationally or physically with other objects; they get bumped. Same can be true for asteroids. While the probability is low, it's something that can't be ruled out based on what we know now because it involves a change. Let's hope that space is really really big and the probability is very very remote.
uhoh
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2020-11-15T06:55:40.097Z
+1 for "vastly, hugely mind-bogglingly big it is."
DRF
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2020-11-16T08:47:43.173Z
@uhoh For those who wonder about the colorful wording, google the phrase! :-)
Peter - Reinstate Monica
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2020-11-16T09:29:09.290Z
@Peter-ReinstateMonica Oh I've read them all, heard the radio adaptations and seen the 2005 movie, but only bits of the old TV series.
uhoh
CC BY-SA 4.0
2020-11-16T10:24:33.703Z
@Spratty I don't remember much of anything that happened to me in 2005 - *fact.* I do remember there were many enjoyable books, one time a purse was forgotten, one trime one wasn't - *fact?* Everytime I try to remember anything about this topic I start hearing banjos - *fact!*
uhoh
CC BY-SA 4.0
2020-11-14T17:29:08.200Z
2020-11-14T17:29:08.200Z
James K
<p>The minimum diameter of the dinosaur-killing impactor is 11 km (according to the wikipedia page on the Chicxulub impact) . Let's use that as our basis for calculation. Larger objects could be seen further away.</p>
<p>Objects impacting the Earth can have relative speeds of about 11 to 90 km/s. Let's assume our object is in the middle of this range (the answer will scale linearly with the assumed closing speed).</p>
<p>We could take the optimal case that the asteroid approaches whilst fully lit by the Sun (which I think precludes the minimum and maximum speed in the range quoted above) and then scale from another similar body - say the asteroid <a href="https://en.m.wikipedia.org/wiki/4_Vesta" rel="nofollow noreferrer">Vesta</a>. This has a diameter of around <span class="math-container">$a=520$</span> km, gets as close as <span class="math-container">$d=1.14$</span> au from the Earth and has a maximum brightness of about <span class="math-container">$m=5.2$</span> apparent magnitude (and hence just visible to the naked eye) and an observed flux <span class="math-container">$f = f_0 10^{-0.4m}$</span>, where <span class="math-container">$f_0$</span> is a zeropoint for the magnitude scale.</p>
<p>Thus the flux <span class="math-container">$f_a$</span> received by a near-Earth asteroid of diameter <span class="math-container">$a_a$</span>, at a distance <span class="math-container">$d_a$</span> from Earth and with the same reflectivity would be
<span class="math-container">$$ f_a = f\left(\frac{a_a}{a}\right)^2 \left(\frac{1+d}{1+d_a}\right)^2
\left(\frac{d}{d_a}\right)^2$$</span></p>
<p>The magnitude of the dinosaur killer would then be
<span class="math-container">$$m_a = m -2.5\log (f/f_a)$$</span></p>
<p>If we say something about 100 times fainter than Vesta would be spotted, <span class="math-container">$f_a \geq 0.01f$</span>. If we assume <span class="math-container">$a_a=11$</span>km, then
<span class="math-container">$$ d_a^2(1+d_a)^2 \leq 0.27\ {\rm au}^4$$</span></p>
<p>An approximate solution is obtained by assuming <span class="math-container">$d_a \ll 1$</span> and thus we find
<span class="math-container">$ d_a \sim$</span> 0.4 au or  60 million km.</p>
<p>Travelling at 50 km/s, this takes 14 days to reach the Earth.</p>
<p>Even if we were to assume it needs to be a naked eye object, we would still have 2 days. See <a href="https://astronomy.stackexchange.com/questions/24810/could-the-dinosaurs-have-seen-the-asteroid-that-killed-them?rq=1">Could the dinosaurs have seen the asteroid that killed them?</a></p>
<p>Edit: And if you assume it would be seen when as bright as Pluto, that puts <span class="math-container">$f_a \sim 10^{-4}f$</span> and it is spotted about 2.5 au out and the warning would reach about 80 days.</p>

CC BY-SA 4.0
2020-11-16T17:01:55.783Z
2020-11-17T23:37:01.983Z
2020-11-17T23:37:01.983Z
ProfRob
<p>There are a few other aspects to this question. As already mentioned, space is big and the Earth is a relatively small moving target. The Earth's orbital speed is around 29.8 km/s, thus it covers a distance equal to its own diameter in around 7 minutes. If you want to determine if some rogue body is going to hit the Earth, your trajectory calculations need to have that level of precision.</p>
<p>It's difficult to take accurate position &amp; velocity measurements of small distant bodies, even for bodies in sedate near-circular orbits. It's harder if they're moving swiftly and heading towards us. So even if we did spot a rogue body in the outer Solar System we may not be able to accurately determine if it's a threat until it gets into the inner region of the system, where bodies move faster, and there's more light. I suppose radar could be used to get the range of a distant object, but radar to the outer system takes a lot of power, but it certainly be useful once the rogue reaches the inner system.</p>
<p>To be honest, the collision window <em>could</em> be far longer than 7 minutes. That really only applies if the collider is heading towards us on a trajectory roughly perpendicular to Earth's orbit, either in the ecliptic plane or dropping in from above or below that plane. If a body is travelling in approximately a straight line that touches the Earth's orbit tangentially, the collision window is more like 50 hours, and perhaps even longer if its trajectory has the right curvature, and it has significant gravitational interaction with the Earth-Moon system.</p>
<p>If a rogue body is on a free-fall trajectory towards the Sun, you can calculate its speed using the <a href="https://en.wikipedia.org/wiki/Escape_velocity" rel="nofollow noreferrer">escape velocity</a>  equation:</p>
<p><span class="math-container">$$v_e=\sqrt\frac{2\mu}{r}$$</span></p>
<p>where <span class="math-container">$\mu$</span> is the Sun's <a href="https://en.wikipedia.org/wiki/Standard_gravitational_parameter" rel="nofollow noreferrer">standard gravitational parameter</a> and <span class="math-container">$r$</span> is the distance of the body from the centre of the Sun. The escape velocity is <span class="math-container">$\sqrt 2$</span> times orbital speed of a circular orbit at that radius. So a free-falling rogue will be travelling at around 42 km/s when it reaches Earth orbit. Its speed at 5 AU (out near Jupiter's orbit) is around 18.8 km/s and it will take around 280 days to drop from 5 AU to 1 AU, where we are.</p>
<p>As uhoh mentions in a comment, it's not unlikely that a rogue body may have a faster speed, due to having been bumped, or gravitationally sling-shotted into our neighbourhood. So if we don't spot it before it gets inside Jupiter's orbit we may have less than 9 months before it gets here.</p>

CC BY-SA 4.0
2020-11-16T12:48:07.047Z
2020-11-16T12:48:07.047Z
PM 2Ring
<p>Would we know months, years ahead? How sophisticated is our technology's ability to detect it at 100% accuracy?</p>

CC BY-SA 4.0
2020-11-14T15:08:35.107Z
I think for discovering an asteroid about 6.9 miles (11 km) in diameter, it depends on its distance, the distance of its orbit, and on coincidence. If it's a near-Earth asteroid we might know it for years before a possible Earth impact. Apophis is much smaller and was thought to possibly impact the Earth in 2029, and it's being tracked since beginning of this century. On the other hand, a long-/non-periodic comet of this size might not be discovered until a few days before an impact, and that might be too late.
John
CC BY-SA 4.0
2020-11-14T15:12:41.837Z
@John Holy That's absolutely insane. I thought we already had technology to detect things like massive asteroids in like a century but appearently not all of them. Why is this not considered a big problem for humanity?
CupOfGreenTea
CC BY-SA 4.0
2020-11-14T17:04:08.153Z
@CupOfGreenTea Because humanity has been seduced into fighting imaginary dangers instead of worrying about true threats like this. But all the currently discovered asteroids and comets don't seem to pose any threat to Earth anyway in the following decades.
John
CC BY-SA 4.0
2020-11-15T04:28:56.173Z
@CupOfGreenTea - even if you detect an asteroid which is going to impact Earth there is basically NOTHING you can do about it except sell the house, go to the beach, and enjoy yourself - right up until there's two suns in the sunset, at which point it would seem that the human race is run.
Bob Jarvis - Слава Україні
CC BY-SA 4.0
2020-11-16T09:46:36.733Z
@CupOfGreenTea It *is* considered a big problem for humanity, though not given high priority. [Methods to redirect or break up asteroids](https://en.wikipedia.org/wiki/Asteroid_impact_avoidance) are being worked on. Recent missions to asteroids are, in part, to learn more about their composition and practice techniques for tracking and altering their orbit. And [Elon Musk is working on his lifeboat](https://en.wikipedia.org/wiki/SpaceX).
Schwern
CC BY-SA 4.0
2020-11-16T13:38:38.107Z
Well we can presume we could do better than the dinosaurs. https://astronomy.stackexchange.com/questions/24810/could-the-dinosaurs-have-seen-the-asteroid-that-killed-them?rq=1
ProfRob
CC BY-SA 4.0
2020-11-17T18:51:29.037Z
I think we figured this out in a Q a few years ago.   Yes, see https://astronomy.stackexchange.com/q/11120/7142
JDługosz
CC BY-SA 4.0
2020-11-17T21:26:01.863Z
@CupOfGreenTea Yes its bad if it happens but the last one to hit was however many millions of years ago (whenever the dinosaurs were). The rate at which it happens is low compared to other issues. Climate change, famine, disease, all sorts of things could wipe us out well before an asteroid comes close. That is why it isn't a priority.
Lio Elbammalf
<asteroids><chicxulub>
How early could we detect an asteroid the size of the one that caused the extinction of the dinosaurs?
CupOfGreenTea
# How early could we detect an asteroid the size of the one that caused the extinction of the dinosaurs?

(Asked by: username_0 on 2020-11-14)

Would we know months, years ahead? How sophisticated is our technology's ability to detect it at 100% accuracy?

**Tags:** asteroids, chicxulub

> **Comments:**
> **username_1**: I think for discovering an asteroid about 6.9 miles (11 km) in diameter, it depends on its distance, the distance of its orbit, and on coincidence. If it's a near-Earth asteroid we might know it for years before a possible Earth impact. Apophis is much smaller and was thought to possibly impact the Earth in 2029, and it's being tracked since beginning of this century. On the other hand, a long-/non-periodic comet of this size might not be discovered until a few days before an impact, and that might be too late.  
> **username_0**: @username_1 Holy That's absolutely insane. I thought we already had technology to detect things like massive asteroids in like a century but appearently not all of them. Why is this not considered a big problem for humanity?  
> **username_1**: @username_0 Because humanity has been seduced into fighting imaginary dangers instead of worrying about true threats like this. But all the currently discovered asteroids and comets don't seem to pose any threat to Earth anyway in the following decades.  
> **username_2**: @username_0 - even if you detect an asteroid which is going to impact Earth there is basically NOTHING you can do about it except sell the house, go to the beach, and enjoy yourself - right up until there's two suns in the sunset, at which point it would seem that the human race is run.  
> **username_3**: @username_0 It *is* considered a big problem for humanity, though not given high priority. [Methods to redirect or break up asteroids](https://en.wikipedia.org/wiki/Asteroid_impact_avoidance) are being worked on. Recent missions to asteroids are, in part, to learn more about their composition and practice techniques for tracking and altering their orbit. And [Elon Musk is working on his lifeboat](https://en.wikipedia.org/wiki/SpaceX).  
> **username_4**: Well we can presume we could do better than the dinosaurs. https://astronomy.stackexchange.com/questions/24810/could-the-dinosaurs-have-seen-the-asteroid-that-killed-them?rq=1  
> **username_5**: I think we figured this out in a Q a few years ago.   Yes, see https://astronomy.stackexchange.com/q/11120/7142  
> **username_6**: @username_0 Yes its bad if it happens but the last one to hit was however many millions of years ago (whenever the dinosaurs were). The rate at which it happens is low compared to other issues. Climate change, famine, disease, all sorts of things could wipe us out well before an asteroid comes close. That is why it isn't a priority.  

---

## 5 answers

### Answer by username_7 on 2020-11-14. Score: 32

It could be as little notice as a few days.

There is a large variation in the amount of warning we would have before an asteroid strike similar in energy to Chicxulub. Let’s examine the case in which we would have the least warning.

The lower bound to the estimated energy released by the Chicxulub collision is 1.3e24 joules <https://en.wikipedia.org/wiki/Chicxulub_crater#Impact_specifics>.
We know the object ‘Oumoamoa was moving at about 50km/s at 1 au from the sun <https://en.wikipedia.org/wiki/%CA%BBOumuamua> , which is the distance of the earth from the sun. The earth itself is moving at about 30 km/s with respect to the sun, so if they struck in opposition, the difference in velocity could be 80 km/s.

For kinetic energy impact, we know $e=1/2\*mv^2$, or energy = 0.5\*mass times the square of the velocity. So $m=2e/v^2$, so m is approximately 4e14kg. If we assume an asteroid a bit denser than iron at 1e4 kg/m^3, then the volume of the asteroid is 4e10m^3. If we assume the asteroid is a sphere, then since the volume of a sphere is $V=4/3\pi r^3$, then the radius is $r^3 = (3/4\*V/\pi)$ or $r$= 2.121km.
Let’s assume the asteroid is matte black so it hardy reflects any light, give it an albedo of 0.02. Then the absolute magnitude of the asteroid of this size and albedo at 1 AU from an observer is 17, using the tables here <https://cneos.jpl.nasa.gov/tools/ast_size_est.html>. Let’s suppose that we spot the asteroid when it is much much brighter than at one AU (or 1.5e8 km), with the same apparent magnitude as Pluto (magnitude 14). Since magnitude is a log scale, that is $2.5^3 = 16$ times as bright.

Brightness varies as square of distance, so the asteroid will be at $1.5e8/4$, or 3.75e7 km away from earth when it’s spotted. At 80 km/s relative velocity, the time from seeing it to impact is 4.6875e5 seconds, or about 5.4 days.

Notes:

1. The above outlined scenario is almost impossibly unlikely. A much more likely scenario is that an asteroid from our known asteroid belt strikes earth, in which case we would probably know it was going to collide years before the strike.
2. The object Oumoamoa was classified as an asteroid rather than a comet due to its lack of corona, but now it has an ‘I’ designation for the first Interstellar object which is neither a comet nor an asteroid. One could argue that I shouldn’t use it as an example given that the question asks about asteroids, but I think it is perfectly reasonable.
3. We don’t ever know a collision with 100% accuracy. There is always error in our orbit estimations. We can’t even tell if satellites are going to collide in earth orbit, it’s just probabilities.
4. I don't have a good idea of how long a new object of a certain apparent magnitude will be in the night sky before it's discovered by someone. If someone thinks we would notice this object sooner due to transits or some other reason, I may amend this answer.

> **Comments:**
> **username_7**: @Graipher You are absolutely correct, and I was so careful, but this is one thing I didn't check.  I ran the equations back through with this correction and reduced albedo from 0.03 to 0.02 and got the same result. My post is edited above. I am sorry.  If we get a big, dense, asteroid with low albedo coming into our solar system with escape velocity in an earth impact trajectory, it isn't good for the fate of life on this planet.  And we won't have much notice. username_7  
> **username_4**: The brightness of an asteroid doesn't vary as the inverse square of the distance to Earth.  
> **username_4**: Also if you do adopt an inverse square law, then for an object to be 16 times as bright it is at 1/4 the distance!  This places it 0.25 au away and even travelling at 80 km/s, that takes more than 5 days to get here.  
> **username_4**: The same wikipedia article you reference also gives a minimum diameter for the dinosaur-killer of 11 km. This makes it more than 25 times brighter and it can therefore be seen at least 5 times further away. In fact the bigger size means that it must be moving slower thn 80 km/s to give the same minimum energy (though note that assuming a density of 10 g/cc is a bit too high - nothing in the solar system is that dense) i.e. Weeks, not hours.  
> **username_7**: @RobJeffries You are right about my inverse square law calcs being wrong, I corrected them.  You are also right about my density being a bit too high and asteroid brightness not varying quite as the inverse square of distance, but I don't think either affects my answer too much.  I think I'm right about the asteroid size. Crater size is good for approximating kinetic energy, but it isn't good for approximating asteroid size.  For a worst case scenario, we should approximate our own size for a dense, fast, mover.  
> **username_4**: "The Chicxulub impactor had an estimated diameter of 11–81 kilometers (6.8–50.3 mi), and delivered an estimated energy of 21–921 billion Hiroshima A-bombs (between 1.3×1024 and 5.8×1025 joules, or 1.3–58 yottajoules)". The SAME paper that quotes the minimum energy you used also says the minimum diameter is 11km. Yes, very much a worst possible case scenario, since there aren't a population of iron asteroids on nearly parabolic orbits and Oumuamua had an albedo of 0.1.  
> **username_8**: The worst case scenario would be an interstellar object travelling much faster perpendicular to the plane of the elliptic no? Clearly 80 km/s relative velocity is not the upper limit of such objects.  

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### Answer by username_10 on 2020-11-14. Score: 17

There are two classes of object. One is "asteroids" and the other is "comets".

Asteroids orbit in fat ellipses, mostly between Mars and Jupiter, but some come closer and a few can come close to Earth. They are never very far from Earth and we discover new small ones all the time. These are "potentially hazardous objects". If there was an 11km asteroid that was potentially hazardous we would, almost certainly, have seen it already. We know all the asteroids that are of that size are not potentially hazardous because we know their orbits and we know that they don't come near Earth. We know that none of the potentially hazardous objects will actually hit Earth in the next 200 years or so, because we can calculate their orbit and while they come close, none will actually hit us.

So we are pretty sure that there are no asteroids of that size that will hit us.

Comets are different, because comets come from the distant parts of the solar system and fall in towards the sun on very long orbits. We often see comets first when they are a few months from the inner part of the solar system. So we might have as little as a few months notice.

Fortunately, comets like this are rare, and space is big (Really big. You just won't believe how vastly hugely mind-bogglingly big it is...) the probabilty of a large 11km comet coming out of the distant solar system and colliding with Earth is very very small.

So Asteroid impact is a real concern, in much the same way that Earthquakes or typhoons are a real concern. These are things that can kill lots of people. But for now, a collision that could end human life on Earth is a very very remote possibility.

> **Comments:**
> **username_11**: Sometimes the reason that comets enter orbits bringing them into the inner solar system is that they've interacted gravitationally or physically with other objects; they get bumped. Same can be true for asteroids. While the probability is low, it's something that can't be ruled out based on what we know now because it involves a change. Let's hope that space is really really big and the probability is very very remote.  
> **username_12**: +1 for "vastly, hugely mind-bogglingly big it is."  
> **username_13**: @username_11 For those who wonder about the colorful wording, google the phrase! :-)  
> **username_11**: @username_13-ReinstateMonica Oh I've read them all, heard the radio adaptations and seen the 2005 movie, but only bits of the old TV series.  
> **username_11**: @Spratty I don't remember much of anything that happened to me in 2005 - *fact.* I do remember there were many enjoyable books, one time a purse was forgotten, one trime one wasn't - *fact?* Everytime I try to remember anything about this topic I start hearing banjos - *fact!*  

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### Answer by username_9 on 2020-11-15. Score: 12

An object's absolute magnitude (how bright it would appear at full phase, 1 au from both Earth and Sun) is related to its size and easier to determine.
A ~1 km asteroid has absolute magnitude *H* ≈ 17.75.
The [Chicxulub impactor](https://en.wikipedia.org/wiki/Chicxulub_impactor), estimated [between 11 and 80 km](https://ui.adsabs.harvard.edu/abs/2014arXiv1403.6391D/abstract), was probably 9 ≤ *H* ≤ 12.

Comet [C/2019 U6 (Lemmon)](https://ssd.jpl.nasa.gov/sbdb.cgi?sstr=K19U060) is smaller (*H* = 13.3) but a good example of modern detection capabilities:

* 2019-10-31: [Mt Lemmon Survey](https://en.wikipedia.org/wiki/Mount_Lemmon_Survey) observes an unknown object at apparent magnitude 20.5.
* 2019-11-08: Minor Planet Center (MPC) [publishes](https://minorplanetcenter.net/mpec/K19/K19VD1.html) a preliminary parabolic orbit based on 31 observations from 5 sites. The comet is 3.36 au away from the Sun.
* 2020-03-25: MPC [publishes](https://minorplanetcenter.net/mpec/K20/K20FD6.html) a revised orbit with eccentricity *e* = 0.9979 based on 225 observations, including a [Pan-STARRS](https://en.wikipedia.org/wiki/Pan-STARRS) "precovery" at apparent magnitude 23 in late 2018.
* 2020-07-09: The comet crosses 0.03 au inside Earth's orbit. Earth is 0.86 au away, having passed the intersection zone 53 days earlier.

That's a lead time of 8 months.

Asteroids can sneak up on us from sunward, but those we don't already know about tend to be rather small.
The latest *H* ≤ 15 near-Earth asteroid (NEA) discovery was in 2004.
Probably 90% or more of *H* ≤ 18 NEAs are already known; the discovery rate in that size range has been in the [single digits per year](https://cneos.jpl.nasa.gov/stats/site_km.html) since 2014.
Current surveys are trying to find 90% of all "potentially hazardous" *H* ≤ 22 asteroids, i.e. ~140 m or larger.


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### Answer by username_4 on 2020-11-16. Score: 4

The minimum diameter of the dinosaur-killing impactor is 11 km (according to the wikipedia page on the Chicxulub impact) . Let's use that as our basis for calculation. Larger objects could be seen further away.

Objects impacting the Earth can have relative speeds of about 11 to 90 km/s. Let's assume our object is in the middle of this range (the answer will scale linearly with the assumed closing speed).

We could take the optimal case that the asteroid approaches whilst fully lit by the Sun (which I think precludes the minimum and maximum speed in the range quoted above) and then scale from another similar body - say the asteroid [Vesta](https://en.m.wikipedia.org/wiki/4_Vesta). This has a diameter of around $a=520$ km, gets as close as $d=1.14$ au from the Earth and has a maximum brightness of about $m=5.2$ apparent magnitude (and hence just visible to the naked eye) and an observed flux $f = f\_0 10^{-0.4m}$, where $f\_0$ is a zeropoint for the magnitude scale.

Thus the flux $f\_a$ received by a near-Earth asteroid of diameter $a\_a$, at a distance $d\_a$ from Earth and with the same reflectivity would be
$$ f\_a = f\left(\frac{a\_a}{a}\right)^2 \left(\frac{1+d}{1+d\_a}\right)^2
\left(\frac{d}{d\_a}\right)^2$$

The magnitude of the dinosaur killer would then be
$$m\_a = m -2.5\log (f/f\_a)$$

If we say something about 100 times fainter than Vesta would be spotted, $f\_a \geq 0.01f$. If we assume $a\_a=11$km, then
$$ d\_a^2(1+d\_a)^2 \leq 0.27\ {\rm au}^4$$

An approximate solution is obtained by assuming $d\_a \ll 1$ and thus we find
$ d\_a \sim$ 0.4 au or 60 million km.

Travelling at 50 km/s, this takes 14 days to reach the Earth.

Even if we were to assume it needs to be a naked eye object, we would still have 2 days. See [Could the dinosaurs have seen the asteroid that killed them?](https://astronomy.stackexchange.com/questions/24810/could-the-dinosaurs-have-seen-the-asteroid-that-killed-them?rq=1)

Edit: And if you assume it would be seen when as bright as Pluto, that puts $f\_a \sim 10^{-4}f$ and it is spotted about 2.5 au out and the warning would reach about 80 days.


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### Answer by username_14 on 2020-11-16. Score: 3

There are a few other aspects to this question. As already mentioned, space is big and the Earth is a relatively small moving target. The Earth's orbital speed is around 29.8 km/s, thus it covers a distance equal to its own diameter in around 7 minutes. If you want to determine if some rogue body is going to hit the Earth, your trajectory calculations need to have that level of precision.

It's difficult to take accurate position & velocity measurements of small distant bodies, even for bodies in sedate near-circular orbits. It's harder if they're moving swiftly and heading towards us. So even if we did spot a rogue body in the outer Solar System we may not be able to accurately determine if it's a threat until it gets into the inner region of the system, where bodies move faster, and there's more light. I suppose radar could be used to get the range of a distant object, but radar to the outer system takes a lot of power, but it certainly be useful once the rogue reaches the inner system.

To be honest, the collision window *could* be far longer than 7 minutes. That really only applies if the collider is heading towards us on a trajectory roughly perpendicular to Earth's orbit, either in the ecliptic plane or dropping in from above or below that plane. If a body is travelling in approximately a straight line that touches the Earth's orbit tangentially, the collision window is more like 50 hours, and perhaps even longer if its trajectory has the right curvature, and it has significant gravitational interaction with the Earth-Moon system.

If a rogue body is on a free-fall trajectory towards the Sun, you can calculate its speed using the [escape velocity](https://en.wikipedia.org/wiki/Escape_velocity) equation:

$$v\_e=\sqrt\frac{2\mu}{r}$$

where $\mu$ is the Sun's [standard gravitational parameter](https://en.wikipedia.org/wiki/Standard_gravitational_parameter) and $r$ is the distance of the body from the centre of the Sun. The escape velocity is $\sqrt 2$ times orbital speed of a circular orbit at that radius. So a free-falling rogue will be travelling at around 42 km/s when it reaches Earth orbit. Its speed at 5 AU (out near Jupiter's orbit) is around 18.8 km/s and it will take around 280 days to drop from 5 AU to 1 AU, where we are.

As username_11 mentions in a comment, it's not unlikely that a rogue body may have a faster speed, due to having been bumped, or gravitationally sling-shotted into our neighbourhood. So if we don't spot it before it gets inside Jupiter's orbit we may have less than 9 months before it gets here.


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