Asteroid Dimorphous easier to move because it's made up of rubble?

[DaveC426913]

TL;DR Summary

Do rock pile asteroids behave differently than classic inelastic collisions?

"There’s really no cohesion between different pieces of gravel or rocks on Dimorphos. That makeup explains why DART’s impact made such a such a surprising change in Dimorphos’ orbital period, decreasing it by about 34 minutes. A collection of boulders is easier to shift than a solid object."
https://www.universetoday.com/168008/the-surface-of-dimorphos-is-surprisingly-new/

I would have thought an elastic collision with a solid object would cause more movement than an inelastic collision with a rock pile.

Is there more nuance to this?

 

[Tazerfish]

That is weird...
I am pretty lost here, but I've got some ideas.

At the speeds that DART impacted Dimorphos, it was more like a bomb and less like a bullet.
After impacting and blowing up, it must have ejected a lot of material in the direction it was coming from.

If you've ever looked into why craters are always circular, even when the meteorites come in at a really oblique angle, it's basically that. The kinetic energy scales much faster than momentum.

When it impacts some depth into the asteroid before blowing up like a nuke, you're going to have a lot of ejecta.
Due to its larger mass, that ejecta can then carry much more momentum away, than the spacecraft ever came in with.

In the extreme case, you wouldn't need any initial momentum. If you just placed a bomb with the equivalent energy somewhere inside Dimorphos, that might have done the job as well.

It kind of reminds me of nuclear fission triggered by neutron capture.
Or even electron capture followed by alpha decay...
Small thing comes in, rips the big thing apart.

Someone with more inside knowledge will have to confirm this though, I'm just brainstorming here.

 

[DaveC426913]

If you've ever looked into why craters are always circular, even when the meteorites come in at a really oblique angle, it's basically that.

Indeed. I knew they did, never quite got the rationale why.

In the extreme case, you wouldn't need any initial momentum. If you just placed a bomb with the equivalent energy somewhere inside Dimorphos, that might have done the job as well.

I see your point.

 

[FactChecker]

At the speeds that DART impacted Dimorphos, it was more like a bomb and less like a bullet.

Yes, it is the energy, not the momentum that counts. But that is also true of bullets.

After impacting and blowing up, it must have ejected a lot of material in the direction it was coming from.

An explosion will scatter in all directions. So there will also be an increase in the speed of some material heading toward Earth.

If you've ever looked into why craters are always circular, even when the meteorites come in at a really oblique angle, it's basically that. The kinetic energy scales much faster than momentum.

When it impacts some depth into the asteroid before blowing up like a nuke, you're going to have a lot of ejecta.
Due to its larger mass, that ejecta can then carry much more momentum away, than the spacecraft ever came in with.

That is an interesting point, worth some calculation. By scattering the trajectories, a lot more matter/energy may miss the Earth than hits it even though the part that hits the Earth may have more speed and a roughly equal amount of matter may have less.

 

[DaveC426913]

Thanks FactChecker. Any light to shed on the original question though?

 

[Tazerfish]

Indeed. I knew they did, never quite got the rationale why.

I see your point.

I've found a paper that seem to confirm my intuition/reasoning:

Here we report a determination of the momentum transferred to an asteroid by kinetic impact. On the basis of the change in the binary orbit period2, we find an instantaneous reduction in Dimorphos’s along-track orbital velocity component of 2.70 ± 0.10 mm s−1, indicating enhanced momentum transfer due to recoil from ejecta streams produced by the impact3,4. For a Dimorphos bulk density range of 1,500 to 3,300 kg m−3, we find that the expected value of the momentum enhancement factor, β, ranges between 2.2 and 4.9, depending on the mass of Dimorphos. If Dimorphos and Didymos are assumed to have equal densities of 2,400 kg m−3, β=3.61−0.25+0.19(1σ). These β values indicate that substantially more momentum was transferred to Dimorphos from the escaping impact ejecta than was incident with DART. Therefore, the DART kinetic impact was highly effective in deflecting the asteroid Dimorphos.

https://www.nature.com/articles/s41586-023-05878-z

 

[FactChecker]

Thanks Factie. Any light to shed on the original question though?

You may have to work on it some yourself to see how my post relates to your original question.

PS. Please don't call me "Factie".

 

[Tazerfish]

Yes, it is the energy, not the momentum that counts. But that is also true of bullets.

An explosion will scatter in all directions. So there will also be an increase in the speed of some material heading toward Earth.

Interesting point about the bullets. I guess they really do hurt in a different way than being hit by a car or bike. But thankfully, unlike meteorites, they don't usually explode upon impact ^^

In regards to the direction of of scatter, you are right that it's not a very controlled process.
The ejecta kind of fly everywhere; nevertheless, they don't fly everywhere equally since I'd assume the probe can only penetrate a few meters or so before it becomes a fireball.
If a meteorite hits our moon, it obviously can't fly off in the direction pointing below the lunar surface, you'd have to punch though the whole thing, and that's not happening!
Unless the surface is very un-isotropic or the "projectile" comes slow, I'd assume the ejecta to shoot off fairly symmetrically, with the net momentum transfer being along the surface normal of the moon.

Also, consider that space is very big and very empty.
I really mean that. If you've ever calculated the volumetric density of space debris or an asteroid field, you'll know what I'm talking about.
There's nothing out there. If you pick a random point in the sky, there's a 99.999% chance it'll shoot off into the void, never hitting anything like a planet, moon or star.
5 parts per million for the sun, 5 parts per million for the moon. Much less for all the planets combined and an rounding error on a rounding error for other stars.
I'll repeat myself: space is empty!


Only a miniscule fraction of phase space will lead to a collision with earth. Any random deviation off a collision course is likely to produce a "not collision–course"
Plus, if we're unlucky and some of the ejecta hits the earth—it's still much better to be hit by a hundred small things than one huge one.
The atmosphere will make short work of it and hey, maybe we'll get a nice meteor shower out of it, haha.

 

[DaveC426913]

If a meteorite hits our moon, it obviously can't fly off in the direction pointing below the lunar surface,

Sure it can. It transfers its kinetic energy into the Moon's bulk in the form of shock waves.

If the masses of the two objects were more similar, we would see the second object actually change course, like a red ball in billiards. Which is the case with Dimorphos.

But how does any of this illuminate the thread question?

 

[DaveC426913]

You may have to work on it some yourself to see how my post relates to your original question.

Alas, my math-fu is not up to it.

PS. Please don't call me "Factie".

Fixed.

 

[FactChecker]

Alas, my math-fu is not up to it.

Neither is mine. And math is my field. It seems very complicated.

Fixed.

Thanks.

 

[Ibix]

Consider a simple model. The asteroid has mass $M$ and is initially at rest. The probe has mass $m$ and initial speed $u$ in the $+x$ direction. Momentum and energy are therefore$$\begin{eqnarray}
p&=&mu\\
E&=&\frac 12mu^2
\end{eqnarray}$$The asteroid and probe collide. Some fraction $\phi$ (where $0<\phi<1$) of the kinetic energy is lost in damaging the asteroid and probe. Some fraction $f$ (where $0<f<1$) of the total mass, $m+M$, flies back the way the probe came at velocity $v$ and the rest continues on at $V$ (both $v$ and $V$ are positive in the $+x$ direction). Conserving momentum and conserving energy (accounting for the losses in the latter):$$\begin{eqnarray}
p&=&(1-f)(M+m)V+f(M+m)v\\
(1-\phi)E&=&\frac 12(1-f)(M+m)V^2+\frac 12f(M+m)v^2
\end{eqnarray}$$Eliminating $u$, $E$, and $v$ by substituting 1, 2 and 3 into 4 and solving the result for $V$ gives us$$
V=\frac{p}{m+M}\left[1+\sqrt{\frac{f}{1-f}}\sqrt{\frac Mm-\phi\left(1+\frac Mm\right)}\right]$$This is rather messy, but there are a couple of points of interest.

First, if we assume $f$ is very small (i.e., only a relatively small amount of matter is kicked up by the impact), then increasing it increases $V$. So the more matter is kicked free of the asteroid the faster the asteroid recoils. Second, if we decrease $\phi$, the fraction of energy lost in breaking off the chunk of matter, then $V$ increases. So the easier it is to kick up matter, the faster the asteroid recoils.

That is consistent with what the article is claiming: a loosely bound asteroid is easier to break up (not much energy lost, so a low $\phi$) and that probably allows more matter to escape (higher $f$, although still very small), so the asteroid recoils with a higher $V$ than a more tightly bound rock.

 

[Tazerfish]

But how does any of this illuminate the thread question?

I thought we had basically solved the thread question.
What in particular do you still want to know?

It mostly fits with my intuition. For a more comprehensive understanding, there are simulations of it. I'm satisfied with where we're at.

While I appreciate the effort to put formal equations to it by Ibix (and maybe that's exactly what you were looking for) it doesn't really do much for me.

Real life is messy.
I imagine the craft impacted, mixing with a bit of the asteroid, as the kinetic energy got transferred into thermal energy via shockwaves and compression.
During this process, it may have penetrated a bit into the surface of the asteroid.
This extremely hot cloud of gas/plasma then exploded, potentially taking a few chunks with it as the gas rushed outwards.


500kg of mass, 6.6km/s impact speed or so.
If we convert that to thermal energy of the craft, we come out at about 20,000K, significantly hotter than the surface of the sun.
Even if we mix two parts asteroid with one part probe, the temperature is still 7,000K.
The gas will then expand more or less equally in all directions, shooting off into space. I'd expect it to go several hundred meters a second, if not a few kilometres a second, but I've not done any fancy simulations of explosions in space, so who knows.
The nature paper I've linked below saw nothing faster than 500m/s.
If something is in the way of this shockwave, it may get ripped outwards with the blast.
This is where we can get a decent amount of momentum transfer, beyond the outrushing gas.

It's a little confusing that it moves to the left. My first reaction was to think it got hit from the right, but I think that's incorrect. I expect most of the debris to eject in the direction the probe came in on so I think the probe impacted from the left. (It's also possible that the viewing direction is more parallel to the trajectory than I expect. I'm saying left and right, assuming the probe came in somewhere in the plane of the image, but perhaps it went more "front to back")


https://www.nature.com/articles/s41586-023-06998-2
This paper has some really amazing images of the aftermath and it mentions having tracked at least two bits that moved off with somewhere between 30 and 100 m/s.
I suspect these are basically boulders/rocks that were ejected by "the bomb".
The speeds are actually pretty pityful, considering it was coming in more than 60 times faster...
It also says that the ejects plume had an "aperture angle" of 140°, meaning that the most oblique ejection if material happened at 140/2, so 70° angle with respect to the surface normal. Most of it shoots back into direction that the probe came from.
And it seemed to form filaments. I'm not sure if that material was liquid, solid or gaseous though.

 

[Tazerfish]

I read through the OG post again and I think I came upon a misconception.

I would have thought an elastic collision with a solid object would cause more movement than an inelastic collision with a rock pile.

Is there more nuance to this?

How in the world are you supposed to elastically collide at 6.6km/s ?

That's friggin faster than the speed of sound in even iron! The impact is supersonic *in metal*.
Take a moment to appreciate this absurdity.

As you increase the velocity, it becomes ever more difficult for things to elastically rebound.
Shoot a tennis ball at a wall at the speed of sound and that thing isn't coming back.
Even metal bullets don't really ricochet back with significant speed when impacting mostly normal to a surface. At these forces, they just deform plastically.
When you shoot a bullet at 6.6km/s, it might actually act more "elastic" again, I'm the sense that the expanding cloud of gas that used to be your projectile explodes away from the target, back in your direction.
But the speed won't be very close to the impact speed.

At these velocities, things behave *nothing* like the classical collision problems we did back in school/uni.
Sure, momentum is conserved.
However, the objects don't stay distinct. They certainly don't act like solid bodies and energy is nowhere close to being conserved.

The only way you get an "elastic collision" at those speeds is as a gravity assist/slingshot maneuver around a planet.

Elastic rebound is just impossible at those speeds.

 

[Tazerfish]

The Wikipedia article is very insightful.
https://en.m.wikipedia.org/wiki/Double_Asteroid_Redirection_Test

Look at the "Course of the Impact" part.
Apparently, more than 1000 tons of dusty ejecta have been displaced into space. At 2000 times the spacecraft weight, that is a lot more than I thought.

 

[DaveC426913]

That is consistent with what the article is claiming: a loosely bound asteroid is easier to break up (not much energy lost, so a low $\phi$) and that probably allows more matter to escape (higher $f$, although still very small), so the asteroid recoils with a higher $V$ than a more tightly bound rock.

Ah! Now I get it!

 

[DaveC426913]

I read through the OG post again and I think I came upon a misconception.

How in the world are you supposed to elastically collide at 6.6km/s ?

The only way you get an "elastic collision" at those speeds is as a gravity assist/slingshot maneuver around a planet.

Elastic rebound is just impossible at those speeds.

Hang on. We're not talking billiard ball level elastic collision here.

Elastic versus inelastic collison is characterized simply (or at least simpistically) by the amount of permanent deformation of the objects. Obviously the smaller one is pulverized in our scenario. But the larger one's resultant shape will be largely determined by its initial rigidity. A solid rock will come away with less deformity than a rock pile. So the solid rock collision is more elastic than the rock pile's.

 

[Tazerfish]

Hang on. We're not talking billiard ball level elastic collision here.

Elastic versus inelastic collison is characterized simply (or at least simpistically) by the amount of permanent deformation of the objects. Obviously the smaller one is pulverized in our scenario. But the larger one's resultant shape will be largely determined by its initial rigidity. A solid rock will come away with less deformity than a rock pile. So the solid rock collision is more elastic than the rock pile's.

I'm glad that Ibix's equations solved your problem!

I guess we just think quite differently about the elasticity of a collision.
In my mind, the elasticity of a collision is what fraction of the initial kinetic energy, as seen from the center of mass frame of the entire system, gets conserved.
The concept is similar to the coefficient of restitution, except that the coefficient of restitution looks at velocities, not energy.

This view of elasticity is largely decoupled from the deformation of an object—hell, a rock skipping over water is quite elastic in this view, even though one of the colliders is a liquid—so it's unsurprising we'd miscommunicate about it.

Thanks for clarifying :)