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Can somebody please tell me what this is?

Piece of cake. Start with a small moon in orbit around Ceres. A big enough meteorite smacks into the front of the moon, reducing its momentum enough to make its new orbit intersect Ceres. So the (ex)moon hits with orbital velocity, which is well below escape velocity.

Nope, that won't work, the moon will be accelerated by Ceres gravity and will impact at or above escape velocity.
Of course it will work. The moon will be accelerated by Ceres' gravity but will not be accelerated enough to impact at or above escape velocity. This follows from the conservation of energy. "Escape velocity" is simply the velocity at which the moon's kinetic energy is equal to the increase in potential energy it would take for it to climb out of Ceres' gravity well. The fact that it starts out in orbit proves it has less kinetic energy than that. Being slowed down by a collision reduces its kinetic energy further, without changing its potential energy. In its subsequent trajectory the sum of kinetic and potential energy doesn't change. So if hasn't got enough kinetic energy to climb out of the gravity well right after the collision, which of course it does't, then it still hasn't got enough when it hits Ceres.
 
If the ejecta left the satellite (if it had one) of Ceres and got to the joint center of mass (barycenter) with just a smidge of speed and then landed on Ceres...

I guess that unless the satellite was very close the velocity would be almost the same as the escape velocity of Ceres alone. I can't see it being that close in a stable orbit.

Actually, the fact that it was moving at a crawl at the barycenter instead of at the "reverse escape velocity" at that distance would probably make more of a difference than drag from the satellite.
 
Wow, just reading about 90 Antiope, a true binary asteroid system:

http://en.wikipedia.org/wiki/90_Antiope

I will do some math for the scenario in my previous post.

Datum:
Assume both radii are 44 km
Distance between centers is 171 km
Density is 1.3 g/cc

Calculate escape velocity for A assuming that impactor comes in such that asteroid B is directly behind it. Will have to add both potential energies up and then square root to get velocity.

Compare that to the impactor coming from at rest at the barycenter.

Before I do the math I will guess that not above a 5-10% difference between both velocities.
 
Ok, this is interesting. For having just one asteroid alone, I get the escape velocity at ~38 m/s. For an impactor coming in parallel to both it will be about 9.8% faster, or ~41.7 m/s

For impact debris that barely made it past the barycenter (85.5 km from each center) it will impact the second asteroid at ~ 21.2 m/s

Oh my god that was way more work than I though it would be...

Once it gets past the barycenter at a crawl it would hit the second asteroid at 26.2 m/s if the first asteroid didn't slow it down to 21.2 m/s.
 
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If the ejecta left the satellite (if it had one) of Ceres and got to the joint center of mass (barycenter) with just a smidge of speed and then landed on Ceres...
I think you're mixing up the barycenter with the L1 point. Ejecta from a secondary can't get to a barycenter with just a smidge of speed: by that point it would be pulled much harder by the primary than by the secondary so it would have been falling toward the primary for a while. (Never mind that the barycenter would certainly be inside Ceres, since the upper bound on possible satellite size is about a kilometer.)

I guess that unless the satellite was very close the velocity would be almost the same as the escape velocity of Ceres alone. I can't see it being that close in a stable orbit.
It's speculative since we don't know the orbit of the hypothetical satellite; but we can take the Earth-Moon system as a model. If we imagine ejecta from an impact on the Moon reaching the L1 point with just a smidge of speed and then falling to the Earth, back of the envelope I have it reaching the Earth at about 99.7% of escape velocity. So I think your guess is spot on.

ETA: If instead we take Phobos as a model, similar ejecta would hit Mars at about 92% of escape velocity. So that gives us the error bars on our guesstimates.
 
I wasn't disagreeing with your Antiope calculations; I was only referring to post #22. Sorry if that wasn't clear. For a normal system with a large primary and a small secondary it's unlikely for the fraction of escape velocity to be below 90% and impossible for it to be below 70%; but yes, as you've shown, it can hit a lot slower than that in a system like Antiope where the two bodies are about the same size.
 
The Dawn probe is powered by an ion thruster that can run continuously for years, accelerating the craft over time.

In September, the thruster failed when a highly energetic particle slammed into its electronics, forcing the probe into safe mode for four days. The loss of power means that, while Dawn will arrive at Ceres on time, it will take weeks to correct its orientation and point its cameras at the surface.

“We’ll be pointing in the wrong direction when we get there,” said Russell. “We have a period coming up where we won’t be taking much data, but we will get the next set of good pictures at the end of April.”

I bet the electronics has a thicker wiring of circuitry (as in higher nanometers) for the CPU and so on compared to new consumer electronics. Otherwise the particle (and weaker ones) would have totally fried the circuitry.

Too sleepy to investigate now.
 
Obviously, it's the staging area for the alien invasion that I imagine will happen any month now.
 
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