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12edDawn t1_jaioiza wrote

Reply to comment by zeeblecroid in NASA’s DART data validates kinetic impact as planetary defense method | DART altered the orbit of the asteroid moonlet Dimorphos by 33 minutes by mepper

I still don't understand why those two are different.

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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.

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Kear_Bear_3747 t1_jaje1p9 wrote

That’s not how it works in space bro

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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. :)

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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.

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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.

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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.

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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.

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coriolis7 t1_jak3dle wrote

Sorry about the formatting, it makes a lot more sense when you see it properly formatted. The unformatted above is kinda dry to the eye.

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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?

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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.

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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.)

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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

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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.

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