Okay.
So in order to destroy the moon, we need to smash it with a huge rock.
Well no, not really. That would be one of the most inefficient uses of energy, i just used an asteroid as an example since that was the largest easily imaginable explosion i had at hand.
Trying to destroy a moon by smashing it with a physical object would be very difficult if not impossible. There's a lot that can go wrong.
If the density of both object is too similar they'll smash together and quickly form a homogenous blob (remember: they behave like liquids). Much of the energy will be lost in deformation, friction etc. ultimately: heat.
If the projectile is too dense it will pass the moon.
If it is not dense enough it will "explode" on contact, i.e. the impact will resort in mostly heat energy.
If it is too slow, the same happens.
If it is too fast, again it would pass right through.
Obviously, the more massive the rock you throw, the bigger the combined gravity you'll have to overcome.
Now, i've not dony any calculations (hah, you'd need a real simulation for that anyway) but i only see two likely scenarios: One, a fast, dense object impacts, passes through, takes a huge spray of moon material with it and leaves the rest behind. Think of a bullet hitting a watermelon style impact.
Ok, *technically* it's an apple.
The other, a slow object impacts, smashing the moon to bits, producing a cone shaped spray of rocks across the solar system that are nicely sorted by size. (small splinters go faster). No handy picture here, Just imagine how that apple would look with a bullet five times the size.
Slowly, or somewhat slowly, breaking apart, i can only see happening if there's some kind of explosion directly inside the moon pushing outwards, not with an impact from the side.
Say one 50k times the size of the "dinosaur" rock. One that is 1.1X10^17 cubic meters, or about 1,300,000 meters in diameter. The moon is 3,480,000 meters in diameter--our cue-ball is about a third as big. This is based on the cue-ball being made of rock the same density as the moon and moving at the above mentioned 20 km/s velocity. I assume a faster moving body, or one more dense might be smaller.
The (newtonian) kinetic energy is 0.5mv², so yes, velocity is very important. However as mentioned above, simply increasing the velocity won't necessarily do the trick.
As for my mention of "slowly" coming apart, I didn't mean leisurely. I meant that I'd like it to come apart, just not immediately become a cloud of dust.
Well the problem with this is that gravity acts as a constant force of acceleration, and even if the acceleration is slow, if it's constant objects pick up sizeable speeds quickly. The number best for vizualising this is the escape velocity. For the moon it's 2,4km/s. Anything slower than that will fall back. unfortunately, 2,4km/s is pretty much immediate cloud of dust velocity.
If the cue-ball moves at 20km/s, then it would take half an hour to cross the diameter of the moon.
Slight miscalculation. The Moon's radius is about 1700km, so crossing the diameter would take less than 3 minutes.
If it hit the moon dead center, then much of that velocity would be absorbed in the collision. But as has been pointed out, if it sufficiently large (or massive) then there is enough energy to squirt moon stuff all over the vicinity.
Well velocity means energy, and energy is never lost, only converted. Depending on a multitude of factors this can either end up as kinetic energy, the kind we want, or heat, the kind we don't want. Unfortunately, heat is by far nature's favorite kind of energy.
But say the thing isn't as massive, but makes up for it with increased velocity. The total energy is the same, isn't it?
Can be. the formula is 0.5mv², so it's directly proportional to mass and squared to velocity.
Let's increase velocity ten-fold.
Then we'll have a hundred times the energy. If we decrease the mass by a factor of a hundred we quickly run into the watermelon-bullet scenario because we have something very small and very fast.
Now the cue-ball is moving at near escape velocity for the moon's gravity well.
Actually, 20km/s already was eight time the escape velocity.
Just btw, that number (for the dinosaur killer) isn't totally arbitrary, it's kind of an average of what to expect from an asteroid in our solar system. But this can vary of course, i don't think the story would even need an explanation for that. Maybe the rock is from deepspace. Extremely unlikely but possible.
Wham. Instant melted rock, I suppose,
Yes, lots of it. Vaporized rock even. The kinetic energy of an asteroid of that speed is a lot more than is nessary to melt it.
which, in the coldness of space congeal rapidly enough, becoming meteoroids.
Actually space is not all that cold. Or rather it is cold, but also a very good insulator, being a vacuum after all. Things in space will cool down much much slower than in, say, the sibirian winter chill. However once they did, they'll be a lot colder.
How many of them move perpendicular to the vector of the cue-ball, percentage wise? How many of them follow the cue-ball?
No that's a complicated question i have no idea how to answer truthfully. The majority will follow the bullet like in the apple picture. However how much material flies in which direction depends on pretty much everything, from density to velocity of both objects.
Now, increase the mass of the cue-ball as well. What if it isn't silica rock, or even iron, but is solid uranium? Is there enough gravity in the cue-ball to pull more of the moon's bits with it?
Unless the mass of the projectile is a sizeable fraction of the moons mass, the gravitational pull won't matter. The bits fly in the same direction because that's the direction they got shoved in when they were hit.
What if the thing strikes the moon tangental to the moon's orbit? Do the remnants of the explosion zing away into space, or do some continue to orbit Earth? What if it strikes perpendicular to the plane of that orbit?
The orbit doesn't really matter. You could define the orbital vector as 0 and go from there. I.e. the only rocks that will continue to orbit will be the ones virtually unmoved by the impact. Now, "unmoved" has some leeway since earths gravity will catch even some of the stuff that gets moved, however most will either fall down (on earth) or zing away. Depending on the individual trajectory of course. Since such an impact is very chaotic you'd see some rocks of every kind.
The direction they get moved in doesn't matter either, to orbit, something needs a certain height, vector and velocity. If you change the vector too much they'll no longer orbit. If you slow them down, they fall. If you push them in a different direction they'll fly (spiral) away.
And, lastly before I shut up, how much of the initial energy of the cue-ball is lost to the collision?
Well lost can't happen but it can be converted to heat. And will. How much i can't possibly say. In asbolute terms there will be lots and lots of heat, especially in the place of the original impact, but also everywhere else the moon gets deformed or broken apart. Enough heat to melt and vaporize rocks.
However what percentage of the initial kinetic energy ends up that way i have no idea, and google wasn't informative. I suppose you'd need to ask someone who works with asteroid impact simulations to get an informed answer to that.
Does the impact slow it enough for it to be deflected from its original trajectory by the Earth, and if so, is it a significant amount?
Not really, liquid, remember? Real deflection will not happen, however any parts of the moon will of course have the combined vector of the original orbit and the one imparted by the impact, and bullet would get a push in the direction of the moons orbit as well. How big that effect is would mainly depend on it's speed. The faster it passes the moon the less effect on its vector of course.
Well and i'd think the spread from the impact would be far greater than the effect of the moons orbital vector.
