# How early could we detect an asteroid the size of the one that caused the extinction of the dinosaurs?

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

• 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
Nov 14, 2020 at 15:08
• @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? Nov 14, 2020 at 15:12
• @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
Nov 14, 2020 at 17:04
• @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. Nov 15, 2020 at 4:28
• @CupOfGreenTea It is considered a big problem for humanity, though not given high priority. Methods to redirect or break up asteroids 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. Nov 16, 2020 at 9:46

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.

• @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 Nov 15, 2020 at 6:55
• The brightness of an asteroid doesn't vary as the inverse square of the distance to Earth. Nov 16, 2020 at 13:41
• 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. Nov 16, 2020 at 14:03
• 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. Nov 16, 2020 at 14:09
• @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. Nov 17, 2020 at 22:04

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.

• 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
Nov 14, 2020 at 23:02
• +1 for "vastly, hugely mind-bogglingly big it is."
– DRF
Nov 15, 2020 at 6:55
• @uhoh For those who wonder about the colorful wording, google the phrase! :-) Nov 16, 2020 at 8:47
• @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
Nov 16, 2020 at 9:29
• @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
Nov 16, 2020 at 10:24

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, estimated between 11 and 80 km, was probably 9 ≤ H ≤ 12.

Comet C/2019 U6 (Lemmon) is smaller (H = 13.3) but a good example of modern detection capabilities:

• 2019-10-31: Mt Lemmon Survey observes an unknown object at apparent magnitude 20.5.
• 2019-11-08: Minor Planet Center (MPC) publishes 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 a revised orbit with eccentricity e = 0.9979 based on 225 observations, including a 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 since 2014. Current surveys are trying to find 90% of all "potentially hazardous" H ≤ 22 asteroids, i.e. ~140 m or larger.

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. 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?

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.

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 equation:

$$v_e=\sqrt\frac{2\mu}{r}$$

where $$\mu$$ is the Sun's 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 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.