# Targeting cities with meteors (part 2)

This question asked about aliens launching meteors at major metropolitan areas on Earth, but explicitly excluded the accuracy of the meteors from the question.

In this specific scenario aliens are using asteroids from our asteroid belt and launching them at earth to target specific cities. Never mind how they target (unless it is relevant to your answer) so accurately.

I'm interested in the targeting part of this question. How accurately could you hit a city on Earth with an asteroid from the belt?

Assumptions about the city are the same as the previous question:

• 450 sq miles of land area
• Generally flat land area, no more than 500 feet min/max elevation change
• Essentially I want a city destroyed but I don't want regional/global firestorms or cooling.

Another assumption: Once the asteroid is initially accelerated, no further course corrections can be made. This is an unguided, unpowered projectile.

• If a species has the means to freely manipulate the path of a small celestial body, they can probably aim it with at least as good precision as we can do today with missiles. I'll let someone better versed write up an answer about it, though. – Frostfyre Aug 18 '15 at 13:23
• Seconding Frostfyre's comment, but not enough for an answer, I imagine the problem will be to apply sufficient delta-v to the asteroid to slam it into the Earth at all, rather than once you have that capability how to aim accurately enough to hit an approximately 20x20 miles area on the ground. Consider how accurately we were able to send Rosetta and Philae to 67P, or New Horizons to Pluto. It's all about accurate delta-v and orbital calculations; I know the saying is cliché, but that's hardly rocket science. – user Aug 18 '15 at 13:49
• @Frostfyre - Missiles are self-guided, though. Rockets aren't, and they're definitely less accurate (although still generally good). – Bobson Aug 18 '15 at 13:58
• @MichaelKjörling - As with the missile comparison, those were both powered spacecraft. An asteroid wouldn't be able to make course corrections. I'll edit that into the question. – Bobson Aug 18 '15 at 14:01
• With a decent model of the original orbit, and a well-enough known imparted delta-v (change in velocity), midcourse corrections should be avoidable. Most often in practice the problem is the accuracy of orbit and delta-v, but I think if a civilization is capable of moving asteroids around at will, they will likely have good enough models that imprecision in those are not going to pose major problems. Unpowered cruise spaceflight is, relatively speaking, trivially predictable. Not really paper-and-pencil trivial, but even early computers could make the calculations quickly enough to be useful. – user Aug 18 '15 at 14:14

$\Large{Yes.}$

Just throw it on a Hohmann transfer orbit, which is the delta-v minimizing transfer between wherever it is and Los Angeles. The energy formula is trivial:

Let:

$v \,\!$ be the speed of an orbiting body,
$\mu = GM\,\!$ be the standard gravitational parameter of the primary body, assuming M+m is not significantly bigger than M (which makes $v_M \ll v$),
$r \,\!$ is the distance of the orbiting body from the primary focus, and,
$a \,\!$ is the semi-major axis of the body's orbit.

Then: $$\Delta v_1 = \sqrt{\frac{\mu}{r_1}} \left( \sqrt{\frac{2 r_2}{r_1+r_2}} - 1 \right)$$

Since you can also calculate the timing to any arbitrary precision using:

$$t_H = \begin{matrix}\frac12\end{matrix} \sqrt{\frac{4\pi^2 a^3_H}{\mu}} = \pi \sqrt{\frac {(r_1 + r_2)^3}{8\mu}}$$

...you can most definitely time it with the known orbital and rotational parameters of Earth's motion to target Los Angeles.

After all, primitive earth-people were able to target one of their probes into a 100x100 km window as part of a slingshot maneuver around Jupiter with 21st century technology.

Since the object will likely be in the megaton range to raze anything the size of a metropolis, you will likely need a fusion-powered mass driver on the impactor surface. Given that the energy will not be instantaneous, the actual calculations will require integral calculus, but are still rather trivial.

Good luck!