Dropping an Asteroid

[Toilet Cobra]

So my group of mid-level players have access to a small alien ship and have discovered a number of rocks orbiting outside the atmosphere of their planet. Because they like to shortcut the plot, they have decided to make one of those rocks fall and crash-land in a certain geographical area.

Is there a practical way this can be accomplished? The party has a bit of gold and has buddied up to some modestly powerful spellcasters. Outright wishes aren't available, but other spells or magic items might be upon request. For the sake of argument, say the asteroid in question is just Colossal size. What can alter the course of something that massive, particularly if precision is important? The best I could come up with on my own is animate object but the required caster level is prohibitive.

[DrMotives]

Animate Object on that would require a caster level of 32. You're better off with Telekinesis, although the max weight object it can affect is still too little for the rock you want to push. It would probably take several castings of Telekinesis or some of the Bigby's Hand line of spells to cause a meaningful change in the trajectory of the asteroid. And with all those castings, you're going to need some good math skills (rocket science, actually) to get it to go where you want it. Proper aim should require a high DC knowledge check. Remember, you're trying to have you players do what people with PhDs spend a lot of time & money trying to do IRL, and the hard part isn't applying the push, it's applying the push correctly.

[(Un)Inspired]

Are you the DM or a member of the party?

[BWR]

Scroll of Gate. (or the 8th level Fistandantilus' Portal from DL)
Open one in front of a small rock and straight into a city. Assuming that things keep their relative speeds, that is.

If these guys are trying to drop rocks with any sort of accuracy on spots by nudging them down from orbit I would require them to develop a bunch of new math and physics to calculate trajectories and vectors and force and stuff.

[Beneath]

Well, first, the rock has to circle the planet because gravity works in a Newtonian way, rather than because it is attached to the primum mobile by an epicycle (which, in the case of a low-orbiting rock, may well be significantly faster than the primum mobile). This is not in evidence for most D&D worlds, though it is commonly assumed to be true (Pathfinder has some strong circumstantial evidence though).

Once that's settled, though...

The orbital mechanics here are fairly simple. It's mathy, but not, as DrMotives called it, rocket science. Air resistance is more complicated, especially since it can vary with local weather conditions in theory (I haven't done the math to see how much wind might deflect a meteor), and also the fragmentation of the meteor might deflect it as well. This makes precise targeting difficult.

If you slow it down or speed it up along the direction that it orbits, it changes into a new orbit with a different amount of energy that will contain the same point as it was when you hit it (at least, it will if you're at apogee or perigee. I have no idea if this result holds for anywhere else). The change in energy you deliver is equal to the force you apply times the distance you apply it over, or, equivalently, to the change in 1/2 m v^2. The energy per mass is given by the right side of the equation here (https://en.wikipedia.org/wiki/Specific_orbital_energy) (mu/2a).

Let's say it starts in a circular orbit. The orbital height doesn't vary, and so the properties of apogee and perigee for an elliptical orbit hold at every point in the orbit. If we slow it down, it will then have an elliptical orbit with its apogee at the point we hit it and its perigee opposite it.

Its orbit will retain a focus at the planet's center of mass, but its center will be off the focus (because an ellipse has two foci at equal distances from its center in opposite directions along the major axis; if the foci and the circle are the same point then you have a circle). Its semimajor axis will be given by the equation we had (it's the a term in epsilon = -mu/2a; a = -mu/2epsilon). Epsilon being the specific energy (which has a kinetic component, proportionate to the square of the velocity, and a potential component, which is proportionate to the negative reciprocal of the orbital height. it's negative because the zero reference is set to be an object that has just enough energy to not be gravitationally bound), and mu coming from the masses of the two objects.

Decreasing the energy decreases the semimajor axis, as given in that equation. The semimajor axis of a circle is the radius; because we've decreased it but kept one point on it the same, the center moves toward us (proportionate to the decrease in semimajor axis), and the opposite end moves toward us twice as fast.

Putting this in equations: if we let f be the focus-to-focus distance and f/2 be the focus-to-center distance, a our semimajor axis, r the radius of the planet you're trying to hit, and R the radius of the initial circular orbit
a + f/2 = R (this is true no matter how much you change a as long as you only make one change)
a = f/2 + r (this is your target. At this point, at perigee its orbit will exactly intersect the surface. Note that it will come in basically parallel to the horizon, which increases air resistance (and therefore air-based inaccuracy)

All you need to do, then, is slow it down enough that the opposite end of its orbit intersects the planet's surface, while opposite the place you want to hit, and then in one-half orbital period (remember that that's proportionate to the semimajor axis, which you've just changed), it'll impact the planet. Remember to account for the rotation of the planet, too. If you're in a low orbit you'll need precision timing to do this, and you can only hit things that are already in your orbital plane (you can change that too. You might do better to ask someone who plays Kerbal Space Program how)

If you want to work out an orbit where it'll hit at a steeper angle, you'll need to work out the geometry yourself (no computer. have fun).

Then there's this handy calculator (http://impact.ese.ic.ac.uk/ImpactEffects/) to work out how much damage your rock does.

Oh, and remember, their accuracy is bounded by the number of sig figs in their calculations and the precision with which they can do these maneuvers. A 90-minute orbital period crosses 4 degrees in its orbit (equivalent to, like, 240 nautical miles I think?) in a six-second combat round.

Of course, to even begin the calculations, their characters would have to have a better grasp of physics than some modern-day engineers (I'd expect an engineer who doesn't work on orbits would have to look up the energy equation I used on wikipedia, like I did. Their characters don't have wikipedia), in a world where a Newtonian understanding of physics would be anachronistic (same for a Newtonian understanding of calculus, which many of these results can't be derived without). So it depends also on whether their stolen spaceship has physics and calculus textbooks they can understand on it and whether they can teach themselves from them.

If they're not using, like, precision-accurate computers to guide all of this, you're entirely justified in saying that they can't do it with any accuracy. Provided a means of decelerating or accelerating a space rock they can do some pretty immense damage to some part of the world, but they can't pick where, which should deter most PCs from attempting it and should prevent them from using it to derail plots.

[tiercel]

Well, this is D&D, so space itself is optional and physics is generally about as relevant as in a James Bond film.

The most obvious way to move around a rock in space would be with their magical spaceship; if it can move itself and its occupants around, presumably it can nudge around a big rock.

In terms of accuracy? It seems unlikely. If you WANT, as DM, the PCs to succeed, then, yay? Otherwise your PCs need to make a trained-only Knowledge (you have no ranks in this) check, fail, and you get to send the rock someplace amusing, causing maximally inconvenient-for-the-PCs destruction (if you're feeling mean), or plunging toward a treasured city/location, evoking horror in the PCs until the big rock smashes into Cobblestone #13428, dealing it 20d6 damage. (See comment above about D&D and physics.)

[nedz]

All you have to do is find some way of moving the smallest rock.

Then have the Cleric cast Divine Insight and Guidance of the Avatar to boost their Perform (Billiards).

[BWR]

You might want to check out Malhavoc Press' "When the sky falls". While primarily concerned with mystical events and a campaign plot based on a magical meteorite, it does have some stuff on mundane meteorites and their effects.

[Toilet Cobra]

Thank you all for your input! I've got a lot to consider before Sunday's game. Special thanks to Beneath for the super detailed and educational post.

For the record, I am the DM, and I don't mind at all if they wanna drop rocks on villains (plenty more where that came from). I just want to make sure that, within the squidgy rules of our Pathfinder universe, their plan actually makes sense. After all, I'm the guy who gave them a ship, so I knew the risks.

Points you guys gave me I will probably bring up to them: Inventing new math, asteroid fragmentation, gate spells. The players actually do have an advanced computer, but no real idea how to use it; it has the power to run their equations if they ever figure them out. More likely they will try to use some kind of divination to predict their impact site.

I think the best bet is to show them the realistic drawbacks with their plan, and let them spend a level or two working towards overcoming them so the rock drops like they intend. Maybe they can go interplanar and hire some axiomites to help with their calculations.

[Calimehter]

One other obstacle the players might run into is that anyone can dodge a dropped object with a DC 15 Reflex save. I don't have the page number, but it is in Heros of Battle in the section describing dropping rocks from flying creatures.

You as the DM are of course free to ignore that, as I'm sure that the authors were not thinking of orbital bombardment when they wrote those rules. It is there if you want it, though. :)

[Inevitability]

One other obstacle the players might run into is that anyone can dodge a dropped object with a DC 15 Reflex save. I don't have the page number, but it is in Heros of Battle in the section describing dropping rocks from flying creatures.

You as the DM are of course free to ignore that, as I'm sure that the authors were not thinking of orbital bombardment when they wrote those rules. It is there if you want it, though. :)

I'm fairly sure 'dodging' won't do any good. Imagine being hit by a ten-ton asteroid. Now imagine standing next to said asteroid as it hits the ground. Do you really think the second situation will end better for you?

[ben-zayb]

Wouldn't the Rockburst (Shining South) spell be perfect in this scenario? It's low level enough that finding scrolls or NPC casters shouldn't be a problem, and you need not worry about the collateral damage because that is contained within a 20ft radius area.

[nedz]

One other obstacle the players might run into is that anyone can dodge a dropped object with a DC 15 Reflex save. I don't have the page number, but it is in Heros of Battle in the section describing dropping rocks from flying creatures.

So what Reflex save does your city have again ?

[The Grue]

So what Reflex save does your city have again ?

Depends if the city is considered an attended object or not. :smallbiggrin:

As is usual in the cases where one is attempting to apply real-world science to a D&D universe, the answer to OP's question is essentially "Whatever the DM wants to happen, happens". An asteroid impact is a sufficiently large-scale event that you can throw rules and dice out the window and resolve it as pure narrative. And then you can play Numenaria or Dark Sun or something similarly post-apocalyptic in the ruins of your campaign setting.