# Targeting cities with meteors

So this was done in Starship Troopers and I am curious if it is realistic for interplanetary warfare. The idea is that aliens launch meteors at major metropolitan areas...and it ends poorly for those cities...like, they're gone.

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.

Given a city that meets these criteria:

• 450 sq miles of land area
• Generally flat land area, no more than 500 feet min/max elevation change

How large would a meteor have to be to wipe out a city that size and could it be done without having major regional impacts? Essentially I want a city destroyed but I don't want regional/global firestorms or cooling.

Yes or no, please show your work. The best answer will demonstrate the process and appropriate calculations for the scenario.

This question asks for hard science. All answers to this question should be backed up by equations, empirical evidence, scientific papers, other citations, etc. Answers that do not satisfy this requirement might be removed. See the tag description for more information.

• It really depends on how fast the projectile is traveling. Bumping asteroids to intersect with earth at specific points is different than throwing them directly at a target (the the throw would likely be much more accurate. Then anything at relativistic speeds can be 'relatively' small... – bowlturner Aug 17 '15 at 19:55
• @bowlturner get math-ing then. – James Aug 17 '15 at 21:32
• I tried writing an answer based on this, but I kept getting really weird numbers. – HDE 226868 Aug 17 '15 at 23:36
• I asked a followup question about the targeting part, if anyone is interested. – Bobson Aug 18 '15 at 13:10

Using the handy Earth Impact Effects Program Calculator for dense rock asteroids I arrive at a diameter of about 150m. Or a comet of about 100m.

Calculation result

The actual numbers I use:

• Distance from impact: 12miles (corresponding to the radius of a circle with 450 sq miles area)
• Projectile density: 3000kg/m^3 - Dense rock asteroid.
• Impact velocity 17km/s - Typical for asteroids.
• Impact angle 45 degrees.

Effects;

• crater diameter: 1.64 miles
• No fireball
• 2.47 cm of ejecta.
• The major damage will be from air pressure.
• Wood frame buildings will almost completely collapse.
• Glass windows will shatter.
• Up to 90 percent of trees blown down; remainder stripped of branches and leaves.

Calculations for the comet

Similar air burst damage as for the asteroid, but no crater since the comet will burn up in the athmosphere.

• @SerbanTanasa No, I meant 1.64km. – Taemyr Aug 18 '15 at 10:34
• @SerbanTanasa The ejecta number is correct. -ish. It's the average ejact thickness 12 miles from impact. But it arrives in pieces that have an mean diameter of 1.25m. So the image of a 2.5cm thick blanket is rather misleading. – Taemyr Aug 18 '15 at 10:38
• @SerbanTanasa You are right, the 1.64 figure is miles. Evidently I was asleep when I read out these numbers. – Taemyr Aug 18 '15 at 10:43
• How do you know that the calculator is correct? – HDE 226868 Aug 18 '15 at 15:46

The energy released from the meteor impact is going to come from the meteor's kinetic energy. As the meteor comes to a sudden stop as it hits the city, most of that energy is going to be released into the environment (while some will be "wasted" in annihilating the meteor).

The kinetic energy of a body is $E=\frac{mv^{2}}{2}$. So we have two variables, mass and velocity. From the equation we can see that doubling the mass of the object, doubles the energy, while doubling the velocity, quadruples the energy. (If we get into relativistic velocities then mass dilation becomes a factor too but let's ignore that for now).

So the question of how big the meteor needs to be is very dependent on the velocity of the meteor and how much energy we want out of it (with the proviso that the meteor is sufficiently large that it wouldn't completely burn up in the atmosphere before hitting the target city).

For illustration, lets assume we want a destructive force similar to the bomb dropped on Hiroshima. That had a yield of approximately 16 kilotons. In joules, that's about $6.7 \cdot 10^{13}\;\text{J}$. So that's the kinetic energy we're looking for from our example meteor.

To keep the math simple, lets assume that our meteor is moving at $50,000 \frac{\text{m}}{\text{s}}$ (which is within the velocity range of 'normal' meteors).

Substituting the known values, we get $6.7 \cdot 10^{13}\;\text{J} = \frac{1}{2}m \cdot (5 \cdot 10^{4}\;\frac{\text{m}}{\text{s}})^{2}$.

Solving for $m = 2\cdot \frac{6.7 \cdot 10^{13}\;\text{J}}{(5 \cdot 10^{4}\;\frac{\text{m}}{\text{s}})^{2}} = \frac{1.34 \cdot 10^{14}}{2.5 \cdot 10^{9}}\;\frac{\text{J}}{\frac{\text{m}^{2}}{\text{s}^{2}}} = 53,600\;\text{kg}$ or $53.6$ metric tonnes.

The actual volume of the meteor will depend on its density. For example, Iron has a density of $7850\;\frac{\text{kg}}{\text{m}^{3}}$, so a $53.6$ tonne iron meteor would have a volume of just $6.8\;\text{m}^{3}$. Thats equivalent to a sphere just $2.35\;\text{m}$ in diameter, which is tiny in cosmic terms.

If you want a larger amount of damage to cover your theoretical target city, you simply increase the size of the object or its velocity or both. Or you simply hit the city with a number of smaller meteors.

• Note that this needs to be the size and mass on impact. To know how large the meteor needs to be when you nudge them along an appropriate transfer orbit, you need to also take atmospheric ablation into account. – a CVn Aug 18 '15 at 7:24
• Uh, Hiroshima blast was 15kT, and could be expected to level about 9 sq. km. The OP specifies an area of 1165 sq. km. So your blast would level 0.7% of the city. Bit like building a highway. – Serban Tanasa Aug 18 '15 at 10:33
• Good science, but not quite a big enough bang. – ArtOfCode Aug 18 '15 at 15:02
• I'd used the Hiroshima example simply as a tool to show how the calculations would work and because it's a concrete example that people are generally familiar with. As I said in the answer, there are two variables both of which can be manipulated to get the desired result so there's no single right answer to the question (so I didn't try to provide one). – Steve Bird Aug 18 '15 at 20:33

Using asteroids (or better yet, comets) would seem to be ideal at first because of the vast amounts of kinetic energy they can deliver. Comets are much better than asteroids because they are generally moving much faster in the inner solar system, in fact, a comet could theoretically be moving at something like 70Km/sec before it is no longer bound by the Sun's gravity. Even a small chunk of ice will deliver a pretty whopping blow at that speed.

The problem is that asteroids and comets are quite a long way away, and you would need a lot of energy to change their orbits. Since the calculations will be all over the place depending on which particular object you are choosing and when you start the project (if the object is in opposition to the Earth [i.e. on the other side of the sun] when you begin your thrust manoeuvre, then you have the option of using a minimum energy orbit), we can just hand wave at this point and say "a lot" of deltaV would be required. As a military project, you want the strike to be as soon as possible after the order is given, but the reality is that using current and projected technology, the travel time could range from months to years. A minimum energy trajectory from Earth to Mars takes about 6 months to complete, and asteroids are even further away...

Of course, making your energy budget even larger is the fact you are not accelerating a small object, but an object ranging in size from a hundred metres to many kilometres in diameter. You would need a ridiculously large rocket engine, or some sort of substitute like mile long mass drivers or improbably large solar sails. The energy rerelease or signature of such a power plant would be highly visible from Earth, so everyone on the planet would quickly figure out that something was up and start making plans to counter your asteroid strike. They would also have many months to years to prepare, so your asteroid would be met by a hail of kinetic energy impactors or nuclear weapons, leaving Dr Evil contemplating a rapidly expanding cloud of gas and rubble where the asteroid used to be.

Frankly, it would be mush faster, simpler and even cheaper to send a fleet of nuclear tipped missiles towards the Earth rather than try pushing an asteroid or comet into an intersecting orbit. At least you know nukes work.

Nuke them from orbit. It's the only way to be sure....

• I have the impression that this does not answer the question? Plus, it's not hard-science. – Burki Aug 18 '15 at 7:35
• Since there is no mass, orbital periods or any other data, only a general answer can be made at this point. – Thucydides Aug 18 '15 at 22:57