Bad Disaster Novel- Calcs needed

[DPDarkPrimus]

So I am flipping through a horrible book I am about to sell at Half-Price Books, a silly disaster story called Moonfall, by Jack McDevitt, which is about a monster asteroid that's going to strike the moon... and shatter it, raining the pieces down upon the Earth in a lethal firestorm.

So, my question is... what's the actual damage potential of this thing? It says in the book that the asteroid ((Tomiko) is 180 kilometers in diameter, and is moving at 480 kilometers per second. I don't have the time, nor the inclination, to read through it and see if the author ever says what it's composition is- assume it's the norm for an asteroid.

What happens? Is the Moon toast, or does it just get a new crater?

[The Grim Squeaker]

well the kinetic energy generated by it would be:
Ek= m*v^2
m=? (What is the materila/mass of the asteroid, and would change based on how it would be affetced by the sun's gravity).
v=(m/s)=480,000 m/s.

The radius of the moon is 1740 kilometers, and The mass of the moon is 7.35x10^22 kg, but I don't know how much KE would be needed to break it up, Sorry I couldn't help more, was there any info about the "weight" of the asteroid or it's composition or mass?

Frankly, I can't see an object of that size breaking up the moon and overcoming it's gravitic "field" (the generated gravity holding it together), unless the speed is truly ludicrous.
I think that the main site's DS Alderaan balsting page had the formula for "Gravitic" binding, from which you might be able to exrapolate the enrgy needed to blow up the moon.
I could try digging around for it later, if I manage to find out the energy needed for the moon rather than Alderaan

[GrandMasterTerwynn]

So I am flipping through a horrible book I am about to sell at Half-Price Books, a silly disaster story called Moonfall, by Jack McDevitt, which is about a monster asteroid that's going to strike the moon... and shatter it, raining the pieces down upon the Earth in a lethal firestorm.

So, my question is... what's the actual damage potential of this thing? It says in the book that the asteroid ((Tomiko) is 180 kilometers in diameter, and is moving at 480 kilometers per second. I don't have the time, nor the inclination, to read through it and see if the author ever says what it's composition is- assume it's the norm for an asteroid.

What happens? Is the Moon toast, or does it just get a new crater?

An asteroid of that size, if we assume nickel-iron composition, would have a mass on the order of 2.4E+19 kilograms. Moving at 480 km/sec gives it a KE of 2.765E+30 joules. The gravitational binding energy of the Moon (assuming it's made of granite) is ~6.1E+28 joules. This means that the asteroid's kinetic energy exceeds the gravitational binding energy of the Moon by two orders of magnitude.

(These values were figured with the asteroid destruction calculator on the main site.)

[Zero]

So does the asteroid win, then? Wait, I read this book a while back. Tomiko was a comet, so the composition probably isn't iron and nickel. 480 km/s seems damned fast, too...

[Surlethe]

So does the asteroid win, then?

Depends on the direction it hits the moon from. If it hits the moon towards the Earth, the center of mass of the expanding fragment cloud will probably fall into a degenerating orbit, and eventually everything will hit the Earth. If it hits the moon away from the Earth, then the center of mass of the cloud will move away from the Earth, and only a few of the fragments will hit the Earth.

[Zero]

So does the asteroid win, then?

Depends on the direction it hits the moon from. If it hits the moon towards the Earth, the center of mass of the expanding fragment cloud will probably fall into a degenerating orbit, and eventually everything will hit the Earth. If it hits the moon away from the Earth, then the center of mass of the cloud will move away from the Earth, and only a few of the fragments will hit the Earth.

Yeah, but if it's a comet, its density and composition should be considerably different.

[Surlethe]

Yeah, but if it's a comet, its density and composition should be considerably different.

If it's a comet, then its mass will be about 2.8e15 tons = 2.8e18 kg (asteroid calculator on main site). That makes its KE ~3.2e29 J, which is still a bit greater than the gravitational binding energy of the Moon (as given by GrandMasterTerwynn). So, no, the comet still wins.

[Kuroneko]

An asteroid of that size, if we assume nickel-iron composition, would have a mass on the order of 2.4E+19 kilograms. Moving at 480 km/sec gives it a KE of 2.765E+30 joules. The gravitational binding energy of the Moon (assuming it's made of granite) is ~6.1E+28 joules.

Why do you assume it is made of granite? Since you're assuming a uniform density anyway, all the relevant data is readibility available: U = (3/5)GM²/r = 1.24e29J.

If it's a comet, then its mass will be about 2.8e15 tons = 2.8e18 kg (asteroid calculator on main site). That makes its KE ~3.2e29 J, which is still a bit greater than the gravitational binding energy of the Moon (as given by GrandMasterTerwynn). So, no, the comet still wins.

I'm not so sure. Here's a little toy model. For a radial density function ρ(r), let M(r) = 4π\int_0^r{ ρ(x)x²dx }, which is the mass included from the center to radius r. The contribution dU to the gravitational binding energy of a shell of thickness dr at radius r is determined by the mass in that shell and M(r): dU = G(4πr²ρdr)M/r. If ρ is piecewise-linear with constant A for 0<r<c (the core) and (B-Cr) for c<r<R, then M(r) = (4/3)Aπr³ for 0<r<C, and M(r) = X - 4π[C(r^4-c^4)/4 + B(r³-c³)/3] for c<r<R, where X = M(c) is the core mass. Integrating dU is not particularly challenging, although it is somewhat long.

Assumptions: c = 3.4e5m (small), A = 7.86kg/m³ (iron), B = 5.17e3kg/m³, C = 1.42e-3kg/m⁴ [1]. Plugging and chugging gives 3.1e29J. Since a significant fraction of the impact energy will go into heat, one can safely say that the moon wins the contest with the comet--although the event will still be dramatic enough to be potentially apocalyptic. If the impact occurs on the near side of the moon, then the surface of the unlucky hemisphere will be melt in the flash of the impact, and the rest won't fare well after such an influx of energy.

[1] The parameters B and C were determined by requiring total mass to be 7.3477e22kg and outer density of 2.70e3kg/m³. I don't know the density of the moon's crust, so I simply picked the density of plagioclase, which is typical of moon rocks, particularly near the surface. If someone has more detailed density gradient data, feel free to correct me.

Edit: Lost a factor of 1/3 in M(r) above when transcribing. Actual figures unaffected.

[Gil Hamilton]

So, my question is... what's the actual damage potential of this thing? It says in the book that the asteroid ((Tomiko) is 180 kilometers in diameter, and is moving at 480 kilometers per second. I don't have the time, nor the inclination, to read through it and see if the author ever says what it's composition is- assume it's the norm for an asteroid.

Wait... 480 km/sec? How the hell is that asteroid moving so damn fast? Did aliens accelerate it up to that speed or something?

[Kuroneko]

Wait... 480 km/sec? How the hell is that asteroid moving so damn fast? Did aliens accelerate it up to that speed or something?

Good point. I suppose it could come from interstellar space--after all, it's the same order of magnitude as Sol's velocity relative to the galactic center, so such a speed would not be too hard to swallow. The problem, of course, such an object could not be an actual asteroid.

[Gil Hamilton]

Good point. I suppose it could come from interstellar space--after all, it's the same order of magnitude as Sol's velocity relative to the galactic center, so such a speed would not be too hard to swallow. The problem, of course, such an object could not be an actual asteroid.

I suppose that could be true that it's an "out of town" traveller, but I'm curious what ejected that sucker from its home system at that velocity without completely pulverizing it. Maybe its a traveller from the core of the galaxy where the stars aren't going around the center as fast and Sol (and the Moon) ran into it, rather than vice versa. Bad luck too, since there is no way the Sun would capture that thing moving at that speed.

But given that it's from a pulp fiction half price disaster novel, any explaination that involves aliens throwing giant rocks at us is an inherently superior solution.

[Gil Hamilton]

EDIT: Or would the star around the core move faster? I kind of pictured a record player thing where the inner part of the record is moving alot slower than the outside, but that could be a bad analogy.

[Kuroneko]

I suppose that could be true that it's an "out of town" traveller, but I'm curious what ejected that sucker from its home system at that velocity without completely pulverizing it.

It doesn't have to be that fast. There are nearby stars with velocities nearly 290km/s relative to Sol. In an unlucky configuarion, the velocity of Earth's orbit would provide an additional 30km/s. A 350km/s object emerging from such a system doesn't seem all that implausible. Still, 480km/s is pushing it, unless of course the fictional universe contains even higher-velocity stars.

Or would the star around the core move faster?

Yes, they would.

[Gil Hamilton]

Probably right, but 480 km/sec just struck me as really freaking fast no matter what, because what I can recall from astronomy involves asteroids in our solar system moving in low double digit klicks per second, which is plenty fast if a good sized one actually hit us.

[Zero]

In the novel, the fantastic speed of the comet was actually one major curiousity of the scientists trying to figure out a thing or two about the bloody thing. But it was a comet, which makes its origin being of interstellar space a bit less odd.

[The Dude]

We're talking about a body with KE on the order of 1e30J and momentum on the order of 4e24kgm/s.

The moon's velocity could be altered by up to 60m/s (or about 6% of its mean orbital velocity) by the collision... more than enough to knock it out of its orbit. If its new trajectory leads to Earth pulling it in....

In any case a collision of this magnitude and proximity is likely to produce sizable ejecta bound for Earth.

Finally, as Kuroneko pointed to, the heat produced could deliver upwards of 1e24J to the facing side of the Earth over a matter of seconds (thousands or millions of times the intensity of normal incident solar radiation).

In short, you're looking at a dramatic rearrangement of the Earth-moon system and a mass-extinction (if not total extinction) event on Earth.

[Zero]

I fished this book out of the pile in my closet, and the energy value given in the novel is 7 * 10^29 J. The mass of the moon is between 7.34 * 10^22 kilograms and 7.475 * 10^29 kilograms. I don't know what the equation is to calculate the gravitational binding energy, so could someone figure out that calc for me?

[The Dude]

I fished this book out of the pile in my closet, and the energy value given in the novel is 7 * 10^29 J. The mass of the moon is between 7.34 * 10^22 kilograms and 7.475 * 10^29 kilograms. I don't know what the equation is to calculate the gravitational binding energy, so could someone figure out that calc for me?

Kuroneko gave the uniform-density GBE equation in his 1st post:

U = (3/5)GM^2/r

[drachefly]

PE = GMm/r

KE = mwwrr/2

with G the universal gravitational constant, m the mass of the moon, M the mass of the earth, w = 2 pi / (one month), r = the orbital radius of the moon.

Add these together to get the energy of the moon.

To get a collision course, you need to lower the KE of the moon by the difference in PE between its present orbit and an orbit low enough it hits the Earth (assuming a perfect transfer orbit, which is highly ideal)

The kinematics of this can take on a variety of values, depending on the energy lost to heat, rotational energy, or the non-dipole component of the ejecta distribution. We can take an estimate of 95% inelastic collision from these sources (the ejecta distribution will have a noticeable dipole component, accounting for the other 5%), and approximate this as 100% inelastic for starters.

That just means that the KE of the asteroid needs to be the difference in PE between the moon's present location and where it would be sitting on the Earth's surface.

The further radius is the Moon's orbital radius; the nearer radius would be the Earth's radius plus the Moon's radius.

[drachefly]

I know, this isn't exactly what the book described, but I was just trying to figure out how to get the ENTIRE moon down. Sorry, I accidentally cut that little explanatory part out.
;)

[Zero]

http://www.astronomynotes.com/solarsys/s13.htm
That site has some stuff on the interior and composition.

The Moon's density is fairly uniform throughout and is only about 3.3 times the density of water. If it has an iron core, it is less than 800 kilometers in diameter. This is a sharp contrast from planets like Mercury and the Earth that have large iron-nickel cores and overall densities more than 5 times the density of water. The Moon's mantle is made of silicate materials, like the Earth's mantle, and makes up about 90% of the Moon's volume. The temperatures do increase closer to the center and may be high enough to partially liquify the material close to the center. Its lack of a liquid iron-nickel core and slow rotation is why the Moon has no magnetic field.

Lunar samples brought back by the Apollo astronauts show that compared to the Earth, the Moon is deficient in iron and nickel and volatiles (elements and compounds that turn into gas at relatively low temperatures) such as water and lead. The Moon is richer in elements and compounds that vaporize at very high temperatures. The Moon's material is like the Earth's mantle material but was heated to very high temperatures so that the volatiles escaped to space.

[Kuroneko]

Just a slight update for The Dude's figures considering the new information...

I fished this book out of the pile in my closet, and the energy value given in the novel is 7 * 10^29 J.

For this kinetic energy, the mass must be around m = 6e18kg, so that the density is 2e3kg/m³ assuming rough sphericity. That's a bit strange for a comet, but I suppose one should be grateful that the numbers are on about the same order of magnitude (most sci-fi situations do worse). The momentum would be around 3e24kgm/s, thus altering the moon's orbit by at least 40m/s, although realistically more since launching fragments back into space on the side of the impact will add to this.

I know, this isn't exactly what the book described, but I was just trying to figure out how to get the ENTIRE moon down. Sorry, I accidentally cut that little explanatory part out.

The minimum energy is, of course, G(5.974e24kg)(7.348e22kg)[1/(3.844e9m) - 1/(6.371e6m+1.371m)] = 3.54e30J.

[Surlethe]

If it's a comet, then its mass will be about 2.8e15 tons = 2.8e18 kg (asteroid calculator on main site). That makes its KE ~3.2e29 J, which is still a bit greater than the gravitational binding energy of the Moon (as given by GrandMasterTerwynn). So, no, the comet still wins.

I'm not so sure. Here's a little toy model. For a radial density function ρ(r), let M(r) = 4π\int_0^r{ ρ(x)x²dx }, which is the mass included from the center to radius r. The contribution dU to the gravitational binding energy of a shell of thickness dr at radius r is determined by the mass in that shell and M(r): dU = G(4πr²ρdr)M/r. If ρ is piecewise-linear with constant A for 0<r<c (the core) and (B-Cr) for c<r<R, then M(r) = (4/3)Aπr³ for 0<r<C, and M(r) = X - 4π[C(r^4-c^4)/4 + B(r³-c³)/3] for c<r<R, where X = M(c) is the core mass. Integrating dU is not particularly challenging, although it is somewhat long.

Assumptions: c = 3.4e5m (small), A = 7.86kg/m³ (iron), B = 5.17e3kg/m³, C = 1.42e-3kg/m⁴ [1]. Plugging and chugging gives 3.1e29J. Since a significant fraction of the impact energy will go into heat, one can safely say that the moon wins the contest with the comet--although the event will still be dramatic enough to be potentially apocalyptic. If the impact occurs on the near side of the moon, then the surface of the unlucky hemisphere will be melt in the flash of the impact, and the rest won't fare well after such an influx of energy.

[1] The parameters B and C were determined by requiring total mass to be 7.3477e22kg and outer density of 2.70e3kg/m³. I don't know the density of the moon's crust, so I simply picked the density of plagioclase, which is typical of moon rocks, particularly near the surface. If someone has more detailed density gradient data, feel free to correct me.

Edit: Lost a factor of 1/3 in M(r) above when transcribing. Actual figures unaffected.

I follow this. My mistake was implicitly assuming all of the energy would be transferred to the Moon directly, rather than some being converted into heat, correct?

[Kuroneko]

I follow this. My mistake was implicitly assuming all of the energy would be transferred to the Moon directly, rather than some being converted into heat, correct?

Well, that too, but my main concern was the figure of 3.2e29J of the comet impact (although now known to be more than doubled, which makes a strange comet) was too close to the estimate of the gravitational binding energy to make a straightforward comparison as to which will 'win'. The assumption of uniform density is an understimate, after all. Hence, I tried making a better estimate and would expect the majority of the moon to still be there post-impact at that energy level, although molten and rearranged.

[Zero]

So wait... I may just be dumb. I read through stuff, but I still can't quite tell. The comet still loses, right?