<p>A carbonaceous condrite has the same reflectivity as the moon at around 7-13%.</p>
<p>If there was ice, if the tail was 10 times smaller than Hale-Bopp, it would have auspiciously covered half of the sky.  it could have made an incredible display in the 1-2 days preceding the collision, because it was as close to the sun as Hale-Bopp, the brightest astronomical apparition in history, although it was 100 times closer to the earth in the final days than Hale-Bopp. Hale-Bopp's trail was 1.5 million km long, so the KT asteroid would have needed a tail 100 times smaller, to appear as bright as Hale Bopp in the final 2 days! <a href="https://www.google.com/amp/s/www.space.com/amp/20354-dinosaur-extinction-caused-by-comet.html" rel="nofollow noreferrer">https://www.google.com/amp/s/www.space.com/amp/20354-dinosaur-extinction-caused-by-comet.html</a></p>
<p><span class="math-container">$\frac{\text{moon distance}}{\text{moon diameter}} = 107\text{ moon diameters}$</span> away, so you can cover the moon with a thumb at arms length ( ~1cm/107cm) and the impactor would be nearly thumb sized at:  <span class="math-container">$\text{comet diameter} \cdot107 = 1500\text{km}$</span> away.</p>
<p>If you were 1000km south of Chixculub, it would stay fairly constantly thumb sized for 10-20 seconds as it arrived from the South-East at 60 degrees, and in that time it would travel from 1500km above to 1000km on the horizon. The graph suggests it had a rotation period of about 2 to 10 seconds, so it would have revolved noticeably.</p>
<p><a href="https://i.sstatic.net/qRal3.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/qRal3.jpg" alt="enter image description here" /></a></p>
<p>If you were 100km away from the impact, it would grow from thumb size to hand sized for 10-20 seconds. Then it would hit the atmosphere at 50km and the sky would ignite about 5 seconds prior to landfall.</p>
<p>For most of the dinosaurs far away, it would be like holding a pea or a matchstick at arms length.</p>
<p>Perhaps it would be brighter than Venus from 75000 km away; that's about 125 minutes prior to the impact.</p>
<p>It would have been like a fire fly landing slowly on a tennis court if the dino was on the other side of the world.</p>
<p>Mammals were nocturnal previous to the impact which is why mammals have developed whiskers, awesome hearing and only bichromatic photoreceptors of G.B, and mammal skin burns from UV. Only primates have R.G.B.
Dinosaurs had R.G.B because birds have the same gene to encode red photoreceptors as tortoises who can also see red, so scientists think that dino's had vision as good as birds, and have better vision than mammals, especially used for detecting movement and for flying.</p>
<p>EDIT: On wiki, the estimated diameter of the comet has been changed to 10-80km. Hale-Bopp comet was 20-40km, and it was visible for months as one of the brightest comets in human history, with a similar aphelion of 0.914AU. That implies that the dinosaur comet may have been visible for many days prior to impact, because it would have been 5 times closer than Hale-Bopp at the end. It's possible that it was very bright in the sky for a week prior to impact.</p>

CC BY-SA 3.0
2018-02-02T14:19:17.667Z
Can you imagine seeing a moon sized object for 10-20 seconds prior to impact?   I like your answer, even if it needs citations, cause I like thinking about what objects would look like. Once it hits the atmosphere it's a fireball, but a 5 km object would be about moon size at 550 km (or several seconds) if you were under it.  The most fun impact would be a near Earth object, where earth's distance wouldn't significantly accelerate it at first, so it grows larger slowly and you could watch it over time, a few days perhaps a week or two as a 2nd moon, as it moves closer and more menacing.
userLTK
CC BY-SA 3.0
2018-02-02T20:00:43.363Z
A huge number of claims, no reference or calculation to back it up, and three paragraphs that are incomprehensible or irrelevant or factually wrong. The moon is a bit bigger than 15km and a bit further away than 1470km.
Nij
CC BY-SA 3.0
2018-02-03T13:38:36.337Z
The comet was 15km, as it's written in the first post, so the comet would be as big as the moon at 1470km, i'll edit to be clearer. sorry about that.
bandybabboon
CC BY-SA 4.0
2018-02-01T13:11:47.180Z
2020-11-19T23:40:07.713Z
2020-11-19T23:40:07.713Z
bandybabboon
<p>The answer is yes; for a few nights prior to the impact (assuming they had eyes with a similar sensitivity to our own and could look up!). It could be a bit longer than this if the body was larger than 10 km (it goes up roughly in proportion to the impactor's radius) and could be much longer if the object was a cometary body or had a very high albedo.</p>
<p><strong>Details:</strong></p>
<p>Impacting solar system objects would have relative closing speeds from around <a href="https://physics.stackexchange.com/questions/232541/why-is-meteor-speed-what-it-appears-to-be">11 to 72 km/s</a>.</p>
<p>We could take the optimal case that the asteroid approaches whilst fully lit by the Sun (which probably 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 is 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">$f_n$</span> received by a near-Earth asteroid of diameter <span class="math-container">$a_n$</span>, at a distance <span class="math-container">$d_n$</span> from Earth (in au) and with the same reflectivity would be
<span class="math-container">$$ f_n = f\left(\frac{a_n}{a}\right)^2 \left(\frac{1+d}{1+d_n}\right)^2
\left(\frac{d}{d_n}\right)^2\ .$$</span>
This equation assumes the asteroids are illuminated by the Sun at a distance <span class="math-container">$(1+d)$</span> au and then viewed at a distance <span class="math-container">$d$</span> au from Earth.</p>
<p>The magnitude of the dinosaur killer would then be
<span class="math-container">$$m_n = m -2.5\log (f/f_n)$$</span></p>
<p>To be an at all conspicuous naked eye object, <span class="math-container">$f_n \geq f$</span> and the dinosaur-killer becomes a naked-eye object at a distance given by
<span class="math-container">$$ d_n^2(1+d_n)^2 \leq 0.0022 \left(\frac{a_n}{10{\rm km}}\right)^2\ .$$</span></p>
<p>The only non-imaginary (i.e. square root of a positive number) solution for <span class="math-container">$a_n \sim 10$</span> km is <span class="math-container">$ d_n \leq 0.047 (a_n/10 {\rm km})$</span> au or <span class="math-container">$ \leq 7$</span> million (<span class="math-container">$a_n/10 {\rm km})$</span> km.</p>
<p>Moving at say 30 km/s, then it gets closer by 2.6 million km per day, thus <strong>hitting the Earth about 3 days after becoming a naked eye object.</strong> Obviously this would be longer for a slower approach speed or for a larger or more reflective asteroid. But shorter for a smaller, faster asteroid or if the asteroid approached from a direction not fully illuminated by the Sun or had a smaller albedo than Vesta.</p>
<p>An interesting trade-off to think about is if what is determined by measurement is the <em>kinetic energy</em> of the impactor. This would be proportional to the product of its mass and the square of its speed. The mass will be proportional to <span class="math-container">$a^3$</span>. Thus if we fix the kinetic energy and allow the radius to be bigger, then it becomes visible from further away (roughly proportional to <span class="math-container">$a$</span>) <em>and</em> will also be moving slower. i.e. A more massive, but slower impactor will likely be visible for a longer period of time and vice-versa.</p>
<p>Another possibility is that the object is of a cometary origin with an icy composition. If that were so then it could be much brighter as a result of sublimation, outgassing and having a bright cometary nucleus and tail. The answer would still be yes, but the visibility period could be weeks (comets are rather unpredictably bright).</p>
<p>It thus seems to me that there <em>is</em> a plausible range of parameters and trajectories where a dinosaur-killing asteroid could be observed and then observed to grow brighter over a <em>few</em> nights, but probably not much longer than that unless it was a comet.</p>
<p>Another possibility is of course that the impactor approaches from the sunward side of the Earth. In which case, it would be nearer the Sun, but only partially illuminated as seen from Earth. More importantly, it is likely to be in the daytime sky and would thus remain unobserved until it was, at most, hours away from impact.</p>
<p><a href="https://i.sstatic.net/lnLZZ.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/lnLZZ.jpg" alt="Dinosaur cartoon" /></a></p>

CC BY-SA 3.0
2018-02-01T12:14:15.857Z
Interesting to calculate how long it would have been visible during the day.
Nathaniel Bubis
CC BY-SA 3.0
2018-02-01T12:22:32.017Z
An unstated assumption is that dinosaur eyes are as good at night vision as human eyes, no better or worse :)
gerrit
CC BY-SA 3.0
2018-02-01T13:45:49.987Z
@gerrit Question says "human" naked eye. No unstated assumptions there.
ProfRob
CC BY-SA 3.0
2018-02-01T13:57:43.063Z
it would be brighter than venus for 20 minutes and brighter than the moon for about 3 minutes if you were on the same 2000 miles.
bandybabboon
CC BY-SA 3.0
2018-02-01T19:53:51.987Z
@gerrit Most extant dinosaur species have better eyesight than your average mammal, so the same may well be true of extinct species.
Mike Scott
CC BY-SA 3.0
2018-02-02T17:32:38.317Z
3 days? 72 hours remain
BlueBuddy
CC BY-SA 3.0
2018-02-03T19:39:44.387Z
@MikeScott While it's not an entirely unreasonable suggestion, it does bear mentioning that extant dinosaurs are a) in aggregate smaller, so more dependent on fine vision, b) almost universally fly, so more dependent on gross vision, and more dependent on discerning detail at range, and c) have 65 million years of dinosauring it up on their extinct brethren.
Williham Totland
CC BY-SA 3.0
2018-02-04T13:01:09.873Z
Days were shorter back then, maybe 23 hours?   Still about 3 days.
userLTK
CC BY-SA 4.0
2018-01-31T23:54:49.133Z
2023-03-27T15:56:33.517Z
2023-03-27T15:56:33.517Z
ProfRob
<p>Wikipedia says the <a href="https://en.wikipedia.org/wiki/Chicxulub_impactor" rel="noreferrer">Chicxulub impactor</a> is thought to have been a 10-15 km diameter object. Would it have been visible to a (human<sup>*</sup>) naked eye before impact? And if so, would it have appeared like a star that grew brighter and brighter each night?</p>

<p><sup>* I know, there were no humans at the time.</sup></p>

CC BY-SA 3.0
2018-02-01T09:33:31.193Z
I'm assuming if you were *right under it*, you'd definitely be able to see it, y'know, before it hit you.
Valorum
CC BY-SA 3.0
2018-02-01T12:49:19.870Z
Chances are if they did not see it, they did feel it.
Neil Meyer
CC BY-SA 3.0
2018-02-01T13:08:49.010Z
A lot of dinosaurs seem to have had ridges over their eyes and inflexible necks so whether they could look up at all is in question.
hwerwerh
CC BY-SA 3.0
2018-02-01T14:00:54.677Z
No, they only looked down at the food they ate.  That's what killed them.
Mikhail Katz
CC BY-SA 3.0
2018-02-01T23:00:59.373Z
That and an asteroid :-)
Laurence
CC BY-SA 3.0
2018-02-02T02:41:13.963Z
https://s-media-cache-ak0.pinimg.com/originals/37/87/b0/3787b0625764e1f51f961397cf247aa0.jpg
Hot Licks
CC BY-SA 3.0
2018-02-03T21:57:58.527Z
if they could have looked up and had seen the asteroid they could have built shelters or something.
Natsfan
CC BY-SA 3.0
2018-02-04T03:13:12.557Z
Yeah, if they hadn't had such tiny hands.
Dawood ibn Kareem
CC BY-SA 3.0
2018-02-04T14:43:05.420Z
Why has this particular QA attracted so many (lame) jokes? :)
Fattie
CC BY-SA 4.0
2022-07-09T02:46:08.160Z
They hadn't even enough time to kiss their asteroid goodbye.
WPWPWP
<asteroids><history><comets><naked-eye><chicxulub>
Could the dinosaurs have seen the asteroid that killed them?
Robert
# Could the dinosaurs have seen the asteroid that killed them?

(Asked by: username_0 on 2018-01-31)

Wikipedia says the [Chicxulub impactor](https://en.wikipedia.org/wiki/Chicxulub_impactor) is thought to have been a 10-15 km diameter object. Would it have been visible to a (human\*) naked eye before impact? And if so, would it have appeared like a star that grew brighter and brighter each night?

\* I know, there were no humans at the time.

**Tags:** asteroids, history, comets, naked-eye, chicxulub

> **Comments:**
> **username_1**: I'm assuming if you were *right under it*, you'd definitely be able to see it, y'know, before it hit you.  
> **username_2**: Chances are if they did not see it, they did feel it.  
> **username_3**: A lot of dinosaurs seem to have had ridges over their eyes and inflexible necks so whether they could look up at all is in question.  
> **username_4**: No, they only looked down at the food they ate.  That's what killed them.  
> **username_5**: That and an asteroid :-)  
> **username_6**: https://s-media-cache-ak0.pinimg.com/originals/37/87/b0/3787b0625764e1f51f961397cf247aa0.jpg  
> **username_7**: if they could have looked up and had seen the asteroid they could have built shelters or something.  
> **username_8**: Yeah, if they hadn't had such tiny hands.  
> **username_9**: Why has this particular QA attracted so many (lame) jokes? :)  
> **username_10**: They hadn't even enough time to kiss their asteroid goodbye.  

---

## 2 answers

### Answer by username_14 on 2018-01-31. Score: 101 | Accepted answer

The answer is yes; for a few nights prior to the impact (assuming they had eyes with a similar sensitivity to our own and could look up!). It could be a bit longer than this if the body was larger than 10 km (it goes up roughly in proportion to the impactor's radius) and could be much longer if the object was a cometary body or had a very high albedo.

**Details:**

Impacting solar system objects would have relative closing speeds from around [11 to 72 km/s](https://physics.stackexchange.com/questions/232541/why-is-meteor-speed-what-it-appears-to-be).

We could take the optimal case that the asteroid approaches whilst fully lit by the Sun (which probably 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 is 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 $f\_n$ received by a near-Earth asteroid of diameter $a\_n$, at a distance $d\_n$ from Earth (in au) and with the same reflectivity would be
$$ f\_n = f\left(\frac{a\_n}{a}\right)^2 \left(\frac{1+d}{1+d\_n}\right)^2
\left(\frac{d}{d\_n}\right)^2\ .$$
This equation assumes the asteroids are illuminated by the Sun at a distance $(1+d)$ au and then viewed at a distance $d$ au from Earth.

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

To be an at all conspicuous naked eye object, $f\_n \geq f$ and the dinosaur-killer becomes a naked-eye object at a distance given by
$$ d\_n^2(1+d\_n)^2 \leq 0.0022 \left(\frac{a\_n}{10{\rm km}}\right)^2\ .$$

The only non-imaginary (i.e. square root of a positive number) solution for $a\_n \sim 10$ km is $ d\_n \leq 0.047 (a\_n/10 {\rm km})$ au or $ \leq 7$ million ($a\_n/10 {\rm km})$ km.

Moving at say 30 km/s, then it gets closer by 2.6 million km per day, thus **hitting the Earth about 3 days after becoming a naked eye object.** Obviously this would be longer for a slower approach speed or for a larger or more reflective asteroid. But shorter for a smaller, faster asteroid or if the asteroid approached from a direction not fully illuminated by the Sun or had a smaller albedo than Vesta.

An interesting trade-off to think about is if what is determined by measurement is the *kinetic energy* of the impactor. This would be proportional to the product of its mass and the square of its speed. The mass will be proportional to $a^3$. Thus if we fix the kinetic energy and allow the radius to be bigger, then it becomes visible from further away (roughly proportional to $a$) *and* will also be moving slower. i.e. A more massive, but slower impactor will likely be visible for a longer period of time and vice-versa.

Another possibility is that the object is of a cometary origin with an icy composition. If that were so then it could be much brighter as a result of sublimation, outgassing and having a bright cometary nucleus and tail. The answer would still be yes, but the visibility period could be weeks (comets are rather unpredictably bright).

It thus seems to me that there *is* a plausible range of parameters and trajectories where a dinosaur-killing asteroid could be observed and then observed to grow brighter over a *few* nights, but probably not much longer than that unless it was a comet.

Another possibility is of course that the impactor approaches from the sunward side of the Earth. In which case, it would be nearer the Sun, but only partially illuminated as seen from Earth. More importantly, it is likely to be in the daytime sky and would thus remain unobserved until it was, at most, hours away from impact.

[![Dinosaur cartoon](https://i.sstatic.net/lnLZZ.jpg)](https://i.sstatic.net/lnLZZ.jpg)

> **Comments:**
> **username_15**: Interesting to calculate how long it would have been visible during the day.  
> **username_16**: An unstated assumption is that dinosaur eyes are as good at night vision as human eyes, no better or worse :)  
> **username_14**: @username_16 Question says "human" naked eye. No unstated assumptions there.  
> **username_11**: it would be brighter than venus for 20 minutes and brighter than the moon for about 3 minutes if you were on the same 2000 miles.  
> **username_17**: @username_16 Most extant dinosaur species have better eyesight than your average mammal, so the same may well be true of extinct species.  
> **username_18**: 3 days? 72 hours remain  
> **username_19**: @MikeScott While it's not an entirely unreasonable suggestion, it does bear mentioning that extant dinosaurs are a) in aggregate smaller, so more dependent on fine vision, b) almost universally fly, so more dependent on gross vision, and more dependent on discerning detail at range, and c) have 65 million years of dinosauring it up on their extinct brethren.  
> **username_12**: Days were shorter back then, maybe 23 hours?   Still about 3 days.  

---

### Answer by username_11 on 2018-02-01. Score: 13

A carbonaceous condrite has the same reflectivity as the moon at around 7-13%.

If there was ice, if the tail was 10 times smaller than Hale-Bopp, it would have auspiciously covered half of the sky. it could have made an incredible display in the 1-2 days preceding the collision, because it was as close to the sun as Hale-Bopp, the brightest astronomical apparition in history, although it was 100 times closer to the earth in the final days than Hale-Bopp. Hale-Bopp's trail was 1.5 million km long, so the KT asteroid would have needed a tail 100 times smaller, to appear as bright as Hale Bopp in the final 2 days! <https://www.google.com/amp/s/www.space.com/amp/20354-dinosaur-extinction-caused-by-comet.html>

$\frac{\text{moon distance}}{\text{moon diameter}} = 107\text{ moon diameters}$ away, so you can cover the moon with a thumb at arms length ( ~1cm/107cm) and the impactor would be nearly thumb sized at: $\text{comet diameter} \cdot107 = 1500\text{km}$ away.

If you were 1000km south of Chixculub, it would stay fairly constantly thumb sized for 10-20 seconds as it arrived from the South-East at 60 degrees, and in that time it would travel from 1500km above to 1000km on the horizon. The graph suggests it had a rotation period of about 2 to 10 seconds, so it would have revolved noticeably.

[![enter image description here](https://i.sstatic.net/qRal3.jpg)](https://i.sstatic.net/qRal3.jpg)

If you were 100km away from the impact, it would grow from thumb size to hand sized for 10-20 seconds. Then it would hit the atmosphere at 50km and the sky would ignite about 5 seconds prior to landfall.

For most of the dinosaurs far away, it would be like holding a pea or a matchstick at arms length.

Perhaps it would be brighter than Venus from 75000 km away; that's about 125 minutes prior to the impact.

It would have been like a fire fly landing slowly on a tennis court if the dino was on the other side of the world.

Mammals were nocturnal previous to the impact which is why mammals have developed whiskers, awesome hearing and only bichromatic photoreceptors of G.B, and mammal skin burns from UV. Only primates have R.G.B.
Dinosaurs had R.G.B because birds have the same gene to encode red photoreceptors as tortoises who can also see red, so scientists think that dino's had vision as good as birds, and have better vision than mammals, especially used for detecting movement and for flying.

EDIT: On wiki, the estimated diameter of the comet has been changed to 10-80km. Hale-Bopp comet was 20-40km, and it was visible for months as one of the brightest comets in human history, with a similar aphelion of 0.914AU. That implies that the dinosaur comet may have been visible for many days prior to impact, because it would have been 5 times closer than Hale-Bopp at the end. It's possible that it was very bright in the sky for a week prior to impact.

> **Comments:**
> **username_12**: Can you imagine seeing a moon sized object for 10-20 seconds prior to impact?   I like your answer, even if it needs citations, cause I like thinking about what objects would look like. Once it hits the atmosphere it's a fireball, but a 5 km object would be about moon size at 550 km (or several seconds) if you were under it.  The most fun impact would be a near Earth object, where earth's distance wouldn't significantly accelerate it at first, so it grows larger slowly and you could watch it over time, a few days perhaps a week or two as a 2nd moon, as it moves closer and more menacing.  
> **username_13**: A huge number of claims, no reference or calculation to back it up, and three paragraphs that are incomprehensible or irrelevant or factually wrong. The moon is a bit bigger than 15km and a bit further away than 1470km.  
> **username_11**: The comet was 15km, as it's written in the first post, so the comet would be as big as the moon at 1470km, i'll edit to be clearer. sorry about that.  

---

