What you're describing is closer to a spy camera in a geosynchronous orbit. A LEO nuke would not behave like this.
I was thinking of a polar orbit when I wrote that comment. You'll always be over the same longitude with a geosynchronous orbit, as opposed to passing over any longitude. You have more flexibility in targeting locations that are mostly below you, though.
>A LEO nuke would not behave like this.
Why not? The orbit stays the same, while the Earth rotates under it. A particular point on the Earth will rotate under the orbit 0-2 times a day, depending on inclination.
...and since you have to be moving faster to hit geosynchronous orbit, you have to have a lot more ΔV to get back to the earth than if you were in a lower orbit, which means more fuel, which means more weight, which means harder to get into that orbit to start with...
Oh, and since you're still moving when you're doing that ΔV maneuver, you'll end up in a lower (and not-geosynchronous) orbit before you get all the way to the planet, so the payload would probably end up doing a few orbits on its way down anyhow (at which point, why care if you're "over the target")
> You can't just fire a rocket "straight down" -- if you just point at the Earth and thrust, you end up in an elongated orbit that misses the earth entirely.
The second sentence doesn't justify the first. You can thrust in various directions even if your intended trajectory is straight down. I surmise you would simply need to reduce orbit speed to maintain the same angular velocity the whole way down, which doesn't strike me as a computationally or physically difficult problem.
Yeah, I know pointing straight down is a horribly inefficient way to deorbit; if anything, it probably requires more delta-v than you expended getting up there in the first place. I was more thinking that with something in geosynchronous orbit, small-ish changes in whatever velocity you end up in after the deorbiting burn could result in very different targets on the ground, whereas in LEO you can't cover the same amount of ground under your orbit with the same delta-v (or at least I think; haven't actually done the numbers, if I even knew how). Still requires a heck of a lot more fuel if you just want to get from LEO to the ground, as you pointed out.
> Oh, and since you're still moving when you're doing that ΔV maneuver, you'll end up in a lower (and not-geosynchronous) orbit before you get all the way to the planet, so the payload would probably end up doing a few orbits on its way down anyhow (at which point, why care if you're "over the target")
If the deorbit burn provides enough delta-v, you can fall straight from geosynchronous altitude to Earth without being stuck doing a few orbits. And the not-geosynchronous orbit is actually what I expected. You want the Earth to rotate, since that might move the point under the geosynchronous satellite approximately under the point the nuke would reenter the atmosphere. I don't know whether an orbit with the right period exists though.