I'll Worry When; "It's The Size Of Texas Mr. President".

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https://www.yahoo.com/news/nasa-warns-massive-asteroid-headed-174111371.html

First off; am I the only one that thinks it looks like a charcoal briquet?
Next; how big does a nickle/iron asteroid have to be before it becomes dangerous?
That is, survive to hit the ground/ocean.

Now a small asteroid hitting in the middle of the Pacific wouldn't be all that big a deal whereas that same rock hitting a city would be . . . bad, really, really bad.
Of course if it was "The Size of Texas" it wouldn't be so much an "impact event" as it would be a planetesimal collision.

If you averaged the north/south distance of Texas with its east/west you would have an asteroid 725 miles in diameter.
That's a big rock, it wouldn't burn through the Earth's atmospheres, which is at best 110 miles in depth, like you see in all the movies, something that big would just punch it out of the way, kind of like dropping a bowling ball into a pie pan full of whipped cream.
In doing so it would accelerate that 110 miles deep, 725 mile wide plug of air to well beyond escape velocity and compress heat it to temperatures hotter than the surface of the sun. Gonna be a hot time in the old town tonight!!

Then it would hit, say the middle of the Pacific, and do the exact same thing to a big chunk of it.

Thankfully there doesn't seem to be anything that big wandering around loose in the solar system these days.
 
Now a small asteroid hitting in the middle of the Pacific wouldn't be all that big a deal (...)

*ahem*

https://en.wikipedia.org/wiki/Tsunami
If you averaged the north/south distance of Texas with its east/west you would have an asteroid 725 miles in diameter.
(...)
Thankfully there doesn't seem to be anything that big wandering around loose in the solar system these days.

Considering that Ceres, formerly the largest asteroid and now considered a dwarf planet, is a little less than 600 miles in diameter, no, not in the inner solar system, at least. If something that big exists that isn't known by now, it's probably somewhere in the Kuiper belt.
 
I looked up the best guess as to the size of the meteor that created the crater in Arizona and apparently it wasn't all that big, 30 to 50 meters, and it left a dent 3/4ths of a mile wide and 600 feet deep.

So I guess a rock 330 meters wide could do some serious damage.
 
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The "lucky" thing in Armageddon, was how when the nuke went off, it split the asteroid into halves, and each half flew past the Earth (one on one side, and one other other side).

Well, just as planned, right? Except they blew the landing, landing way off from where they were supposed to be to drill and all that stuff. Like let's say the landing error was 1/4 the asteroid's circumference. In which case, when it split in half, the front half would hit the Earth first, and the back half would "bounce the rubble".

But then we are talking Armageddon, and you had to turn off your science brain and just enjoy the bubble gum from the beginning.
 
For smaller objects, material composition matters. For the same size object, loose rock and you get an air-burst like Tunguska whereas solid metal (iron/nickel/etc) and you get an impact like Meteor Crater in AZ. But once you're up to a certain size, it doesn't matter any more. You're gonna have "a bad day" regardless.
 
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The "lucky" thing in Armageddon, was how when the nuke went off, it split the asteroid into halves, and each half flew past the Earth (one on one side, and one other other side).
Well we're all "Rocket Scientists" here so ask yourself; how much energy would it take to accelerate those two parts of the asteroid to a velocity that would first; overcome the gravitational attraction of the two halves so they would actually draw apart from one another and not just fall back together.

And second, accelerate them to a velocity wherein they can put enough distance between them that the Earth can pass through the gap.

My guess is that bomb would have to measure in the gigaton range.
 
Well we're all "Rocket Scientists" here so ask yourself; how much energy would it take to accelerate those two parts of the asteroid to a velocity that would first; overcome the gravitational attraction of the two halves so they would actually draw apart from one another and not just fall back together.

And second, accelerate them to a velocity wherein they can put enough distance between them that the Earth can pass through the gap.

My guess is that bomb would have to measure in the gigaton range.
Given the timeframe, I doubt that the halves would be drawn together significantly by their gravity. But I agree that acceleration of the halves so that they'd miss Earth within the next few minutes is unlikely. They'd need a couple dozen truckloads of lithium deuteride and tritium in the device.
 
If you averaged the north/south distance of Texas with its east/west you would have an asteroid 725 miles in diameter... Thankfully there doesn't seem to be anything that big wandering around loose in the solar system these days.
(Underline added) If you suppose an equitorial disc of the same area as Texas it comes out to 585 miles, so Ceres is bigger.
Considering that Ceres, formerly the largest asteroid and now considered a dwarf planet, is a little less than 600 miles in diameter, no, not in the inner solar system, at least.
And Ceres is in a stable solar orbit, not "wandering around loose". In the inner solar system I seriously doubt there's anything even close to that big, by an order of magnitude, I suspect.

Anyway, the name "4660 Nereus" seems appropriate, since it's coming nere us.
 
I looked up the best guess as to the size of the meteor that created the crater in Arizona and apparently it wasn't all that big, 30 to 50 meters, and it left a dent 3/4ths of a mile wide and 600 feet deep.
The explosion was equivalent to a 15 megaton nuke. It was a metal asteroid.
 
Well we're all "Rocket Scientists" here so ask yourself; how much energy would it take to accelerate those two parts of the asteroid to a velocity that would first; overcome the gravitational attraction of the two halves so they would actually draw apart from one another and not just fall back together.

And second, accelerate them to a velocity wherein they can put enough distance between them that the Earth can pass through the gap.

My guess is that bomb would have to measure in the gigaton range.

An asteroid the size of Texas (spherical, r=500km, rho=2000kg/m^3) weighs 10^21kg.
It's gravitational binding energy is 9*10^25J. This would be required to completely blow it up. Just splitting it into two halves that aren't gravitationally bound to each other anymore would "only" require about a third of that.

That's still 8 million gigatons of TNT equivalent. Based on the Taylor limit for practical nuclear weapons (6Mt yield per ton of weapon weight) the device would weigh around a billion tons. That's about 200 times heavier than the Hoover dam or more than 100 times lighter than the Mt. Everest (pick whatever is more intuitive for you...).

I haven't seen Armageddon, but if anybody makes a movie that shows how Bruce Willis shoves down 200 Hoover dams worth of nukes down a bore hole, I might watch it.

Reinhard

PS: Everything rounded to a single significant number. I already consider it a success if I found all my "off by 3 orders of magnitude" errors. ;)
 
The "lucky" thing in Armageddon, was how when the nuke went off, it split the asteroid into halves, and each half flew past the Earth (one on one side, and one other other side).
Just reading text book on the detail behind redirecting asteroids currently. The devil is in the detail. The big problem with splitting an asteroid is that you have to impart enough energy on all the pieces so they achieve escape velocity from the clump. Otherwise you just end up with a still very dangerous clump of asteroid. Composition of the body is also important, as metal asteroids are less likely to split compared to the regular rock varieties.
 
An asteroid the size of Texas (spherical, r=500km, rho=2000kg/m^3) weighs 10^21kg.
It's gravitational binding energy is 9*10^25J. This would be required to completely blow it up. Just splitting it into two halves that aren't gravitationally bound to each other anymore would "only" require about a third of that.

10^21kg = Total mass of all pizza consumed in the U.S. in a typical year.

That's still 8 million gigatons of TNT equivalent. Based on the Taylor limit for practical nuclear weapons (6Mt yield per ton of weapon weight) the device would weigh around a billion tons. That's about 200 times heavier than the Hoover dam or more than 100 times lighter than the Mt. Everest (pick whatever is more intuitive for you...).


As I mentioned; you would need an antimatter weapon of considerable size to make a dent in something that large.
Unfortunately we cannot mass-produce antiprotons, much less have a means by which we can extract them from the particle-collision event that creates them.
Then there's the matter of storing the danged things. With atomic weapons be they fission or fusion, everything has to go perfectly or they won't go bang!
With an antimatter weapon, the least little thing that goes wrong and it goes bang!
!
 
I remember reading years ago that the asteroid that wiped out the dinosaurs was around 6 miles in diameter. On one hand, if a rock that big was plunked down on top of your town, it would be ENORMOUS! But in other ways, it didn’t sound that big to me. So I wanted to get an idea of the scale.

I had a globe about 2 feet in diameter, and when I looked at it, I realized there were no individual features marked on the map that were anywhere near that small. Earth is about 8000 miles in diameter. So on a 24” globe, a mile is .003”, and a 6-mile asteroid is .018”. So at that scale, the asteroid that wiped out the dinos is a medium grain of sand.

A medium grain of sand hits a 2’ globe at 9 feet per hour and digs out a 1/4” wide crater, killing most life on the globe.

I‘m pretty sure anything the size of Texas would split the planet into pieces that might form some kind of ring system that would rain asteroids down on the remnants of the Earth for millennia or maybe consolidate into something in orbit like the moon.
 
I‘m pretty sure anything the size of Texas would split the planet into pieces that might form some kind of ring system that would rain asteroids down on the remnants of the Earth for millennia or maybe consolidate into something in orbit like the moon.
Have you read about the Impact Theory as to the origin of the Moon?
 
I‘m pretty sure anything the size of Texas would split the planet into pieces that might form some kind of ring system that would rain asteroids down on the remnants of the Earth for millennia or maybe consolidate into something in orbit like the moon.

Theia, the theoretical protoplanet whose collision with Earth is believed to have resulted in the formation of the moon, is believed to be the size of Mars.
 
I had a globe about 2 feet in diameter .... Earth is about 8000 miles in diameter.
Hi TRF colleagues,

This reply is not addressing @ThirstyBarbarian's argument. His position may be correct or incorrect.

I am citing his post just to show another reason why the metric system is better to use than the foot-pound system.

@ThirstyBarbarian makes the ratio of his 2-foot globe to the Earth's diameter of approximately 8000 miles. Fine. However, visualizing that ratio cannot easily be done without converting units.

So instead, let's use metres. Then the ratio becomes a globe measuring 0.6096 metres compared to the Earth's diameter of approximately 12,874,752 metres. And that's it. The ratio is immediately clear.

I thank everyone for listening.

Stanley
 
Good sirs.

I wish to express my heartfelt gratitude for the well-reasoned replies to my post regarding the relative sizes of the Chicxulub asteroid compared to our beloved home planet, the Earth, and my analogy utilizing a 2-foot globe and grain of sand for scale.

I wholeheartedly agree the Metric System of weights and measures is far superior to the Imperial System (the one named for the British Empire, and not to be confused with the system utilized by the Galactic Empire, which is also pretty good, except in regard to parsecs). The Metric System requires far less conversion between units than the Imperial System, making calculations simpler, faster, and less prone to error, whilst fostering a more genteel, polite, and in all ways more proper society.

However, in my defense, I feel it is fair to point out that I am in fact a Barbsrian who is at all times Thirsty, and as such, I measure my beverages in butts and flagons. In short, the Metric System does not come naturally to me, but I shall endeavor to improve in that regard.

Returning to my analogy, the aforementioned 2-foot globe was not measured with precision. The 2-foot figure was in fact pulled out of my butt (not the one that contains my mead, the other one). Therefore, the 2-foot approximation cannot properly be converted to the precise metric measurement of 0.6096 metres. Can we agree it would be more “accurate” in a sense to say the globe measures ’bout in the neighborhood of half a meter, more or less?

Thank you for considering my correction on this matter.

Yours truly,

Thirsty A Barbarian
 
Hi @ThirstyBarbarian and everyone else,

Thank you for your nice reply.

Returning to my analogy, the aforementioned 2-foot globe was not measured with precision. The 2-foot figure was in fact pulled out of my butt (not the one that contains my mead, the other one). Therefore, the 2-foot approximation cannot properly be converted to the precise metric measurement of 0.6096 metres. Can we agree it would be more “accurate” in a sense to say the globe measures ’bout in the neighborhood of half a meter, more or less?

Right. I do agree with that. So the ratio of your globe’s diameter to planet Earth’s diameter is roughly half-a-metre to 13 million metres, or approximately 1:26,000,000.

A good way to begin to get used to metric-denominated lengths is to learn one’s own height in centimetres (for example, 175 cm which equals 1.75 m). Then, as you go about your day’s business, you can compare your height to objects nearby. Or as rocketeers, we always have our rockets in mind. So for the person who is 1.75 m tall, a two-metre rocket is a bit taller than they are, and a rocket that measures one-and-a-half metres in length is a bit shorter than they are.

Incidentally, you taught me about two measurement units that I did not know existed — the butt and the flagon. Since I am highly interested in weights and measures, I looked up those terms and was genuinely glad to learn about them.

Stanley
 
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