Embite t1_jaijci7 wrote
I remember in the headlines after the impact everybody was amazed by how much more energy the probe delivered than predicted. Why were we so pessimistic? Isn't this a basic inelastic collision problem?
zeeblecroid t1_jail3qv wrote
I think they were expecting Dimorphos to be a lot less squishy than it turned out. Instead of hitting a rock, the spacecraft struck a rubble pile, and was able to penetrate enough to dump most of its energy into Dimorphos rather than just dumping it onto Dimorphos.
12edDawn t1_jaioiza wrote
I still don't understand why those two are different.
lagavulinski t1_jaiqukb wrote
Imagine a cinder block covered in 6 inches of compacted sand. You shoot it, and the bullet dissipates most of its energy hitting the sand. Now imagine a cinder block covered in 6 inches of loose sand just floating around it and barely touching it. You shoot it, and the bullet goes through most of the sand and hits the core of the cinder block, visibly moving it with all the energy.
Kear_Bear_3747 t1_jaje1p9 wrote
That’s not how it works in space bro
lagavulinski t1_jajernq wrote
If you don't mind elaborating on what I've said incorrectly, I'd appreciate it so I can edit my comment. :)
Kear_Bear_3747 t1_jajfhid wrote
In space, objects will absorb all of the momentum of the object. That’s basic Newtonian Physics, dealing with Inertia.
On Earth it matters because there are other forces in play like Gravity and Friction so kinetic energy can dissipate in different ways, whereas in space there’s nothing to arrest that energy, it will impart itself on whatever object it collides with.
lagavulinski t1_jajh7fb wrote
Thanks for the explanation. However, I believe you and I aren't discussing the same thing. I do agree with your explanation of Newtonian physics though.
coriolis7 t1_jajr75p wrote
Not exactly. Momentum is always conserved, but the kinetic energy is not. A fully elastic collision preserves kinetic energy, while a partially inelastic collision does not.
In both cases m1 x v1(initial) + m2 x v2(initial) = m1 x v1(final) + m2 x v2(final).
However, only in the fully elastic collision does the following hold: [m1 x v1(initial) + m2 x v2(initial)] / 2 = [m1 x v1(final) + m2 x v2(final)] / 2
It doesn’t matter if it’s in space, in a lab, or wherever.
I think what the above redditor was saying is that because the outer material was more loosely held, more of the material could be ejected. That ejected material has additional momentum. Even though the probe never bounced off (ie elastic collision) the ejected material made the collision act as partially elastic.
workingdad83 t1_jaju14g wrote
Oh yeah. 2√(brdsrntreel)+2 carry the 4. See I can just push a lot of buttons too.
Joking. I know you are smarter than me, and I was lashing out. I'm sorry.
gdpoc t1_jajjufm wrote
Doesn't it also matter whether what you're hitting is an isotropic homogeneous solid and how much of that is aggregate v fill when we talk about celestial body surface composition?
[deleted] t1_jajqpyp wrote
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ResponsiblePumpkin60 t1_jajzk3m wrote
I think the only thing that matters is how much energy is lost from ejecting impact debris off into space. If none is ejected, then no energy is lost.
Tuna-Fish2 t1_jajnj3b wrote
In an inelastic collision, momentum is conserved. So Sum(momentum of all bodies) is the same before and after the collision.
If Dimorphos was a rigid body, the spacecraft would only add it's own momentum to it.
However, because the impact by the spacecraft caused just loads of material to be thrown off, the end momentum of Dimorphos is old_momentum + dart_momentum - sum(momentum of all the ejecta). Since the ejecta is going the other way, the net effect is that dimorphos was accelerated more.
(Remember, kinetic energy = ½mv^2, momentum = mv, so spreading out the energy over more, heavier objects increases efficiency.)
Kwiatkowski t1_jalbktv wrote
someone else probably will do the maths better but I’d bet because if the squishiness it transferred more of its kinetic energy into the target instead of converting it into more heat which would be the result of a more solid hit. Don’t trust me tho, just my thought on how it works
nagabalashka t1_jajdiho wrote
Explode a balloon in front of your hand, you're fine, you feel the burst but nothing more.
Explode the same balloon in your ass and you'll be in pain.
pete_ape t1_jajo7f9 wrote
Soooo... Dimorphos got creampied by NASA?
Embite t1_jaimb6e wrote
If Dimorphos was a solid rock, though, there wouldn't be anywhere for the energy to go but into the asteroid. The shrapnel from the impact would have to bounce backwards or sideways, which would just add to the momentum along the approach vector.
SirLauncelot t1_jajdmou wrote
If it bounces off, that means a lot of the energy was reflected back. If it takes greater time to sink into the asteroid, more energy is transferred.
Embite t1_jajvgby wrote
If shrapnel bounces off the asteroid but doesn't accelerate it that would violate conservation of momentum?
Ouatcheur t1_jalot2h wrote
Nope. The energy that is trsnsferred from the DART to the asteroid CANNOT be higher than the initial kinetic energy it has initially, no matter the way it penetrates the asteroids.
You don't get energy out of nothing!
But people often ignnore that the total enerrgy before must equal the totlal enerrgy after.
And before it is:
DART + ASTEROID.
And after is is:
now-a-bit-saller-ASTEROID + its-EJECTA
People just tend to ignore the ejecta.
so in their minds it is
DART + ASTEROID(before) = ASTEROID(after)
or even worse it is:
DART + ASTEROID(before) + magicalwaytheimpacthappened = ASTEROID(after).
Like in "Oh it hit the dense core so it transferred MORE energy". Morer energy than what exactly? The DART fully crasthe asteroid, its, so it's gonna give 100% of it's kinetic energy no matter what. No "it transfers more than 100% because it hit something more solid". Duuh huuh huuh.
Sp, always remember the ejecta. And note that the ejecta is mostly ejected in the OTHER direction.
So the *only* way for the kinetic equations to balance out is for the asteroid to move faster once accounting for its ejecta (faster than if there hadn't been any ejecta at all).
The solidity of the "central" core part is irrelevant.
As the DART experiment proved, the amount of ejecta gives a MAJOR effect to the results.
mistaekNot t1_jaj55j4 wrote
some of the energy is in the shrapnels speed. if the spacecraft gets buried into the dimorphos then all of its energy is transferred into dimorphos
Impossible-Error166 t1_jaj9pml wrote
Its also how much the asteroid breaks apart and if gravity will pull it together.
rocketsocks t1_jajtzbs wrote
They were always expecting dimorphos to be a rubble pile. That's the whole reason they did the mission, to collect this data, because there are unknowns involved.
rocketsocks t1_jajtuvr wrote
It's not about pessimism it's about the difficulty of simulation, and lack of detailed knowledge.
And no, this is not a simple inelastic collision problem, it's not simply a matter of the final momentum of the asteroid being its starting momentum plus the probe's momentum, if it were we wouldn't have sent the probe. The asteroid is a rubble pile, which means that the impact ejected a huge plume of debris out of a crater. Because it sends a debris plume backwards (and that momentum needs to be balanced) you get greater than 1:1 momentum transfer, essentially turning the crater into a rocket engine powered by the probe's kinetic energy. The details of that plume depend greatly on the compositional and structure details of the asteroid, something we have very little firm data on up until now. We have literally fewer than five data points to go on for this sort of thing. So folks put together some simulations with variations according to the knowledge we have. As it turned out in this one instance the result was on the high end of all of the simulations, well above the average.
One thing worth pointing out here is that this is still just one data point. It may be that dimorphos is actually an outlier in terms of its compositional structure. Or it may be that the particular spot we hit was unusual. The average could be lower or higher than what we achieved with this specific instance. That's why we need a lot more studies like this one to collect enough data to be actually of practical usefulness.
CocoDaPuf t1_jan8kbi wrote
This is the most accurate and clearest answer I've seen to this question. Great job with this post!
Erinalope t1_jajc7cz wrote
Scientists are typically pessimistic with unknown variables and there may have been other forces at work that added to the effect. They’ll study the results plus the ESA is sending a follow up mission to the asteroid which will add to the data. There’s still so little we know about asteroids it might not be a bad idea to do a dart 2 with a lander to examine a different type of asteroid interior more closely.
Second_Sol t1_jakpukk wrote
It's not quite so simple, Dimorphos turned out to have a lot of loose rubble, so the impact created a lot of ejecta going the opposite direction of the site of impact, which means you get a greater pushing effect.
Imagine tossing a sticky ball at metal cube in space.
The ball will stick to the cube and they'll float away at a slower velocity than what the ball originally had
If you toss a rubber bouncy ball of the same mass at the same velocity, it'll bounce away from the metal cube, and come flying back at you while the cube will float away faster than if you used the sticky ball
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