# Ballistics 102: Can North Korean missiles really hit a major US city?

On July 28, 2017, North Korea launched a ballistic missile on a high trajectory that reached 3700km altitude and traveled 1000km downrange. We don’t know whether the missile had a full payload, however assuming it did we can use the parameters of the flight to get an upper bound on the maximum range of the missile. The Union of Concerned Scientists claims that this missile would have a maximum range of 10,800km, not accounting for Earth’s rotation, and somewhat more if rotation is included. As this is a calculation with such grave implications, let’s check it for ourselves and learn some physics along the way!

The flight of a ballistic missile can be broadly lumped into three parts:

1. The rocket fires, and the missile accelerates up to the desired burnout speed $v_b$ and desired loft angle $\theta$. This takes it out of the atmosphere and into space.
2. The missile coasts through space on an elliptical orbit.
3. The warhead re-enters the atmosphere and descends upon the target.

The initial boost phase is generally much shorter than the total flight time: the rocket burns for a couple minutes at most, and a 10,000km flight takes around 25 minutes. This means that it’s OK to treat this problem as that of a projectile being launched with an initial speed $v_b$ and angle $\theta$. Except we’re not going to get the good old Physics 101 projectile motion solutions because we’re dealing with a trajectory that takes us around the circumference of the Earth. Both centrifugal force and the variation of Earth’s gravitational field will come into play, and the beloved parabolic arc turns to out be just the tip of an elliptical orbit.

## The Actual Range of a Projectile

The Hamiltonian of a projectile of mass $m$ orbiting at distance $r$ from the centre of the Earth is:

$\mathcal{H} = \frac{\mathbf{p}^2}{2m} - \frac{G M_\oplus m}{r},$

where $\mathbf{p}$ is the projectile’s momentum, $G$ is the gravitational constant, and $M_\oplus$ is the mass of the Earth.

This Hamiltonian gives us everything we need, but it warrants simplification. First, we can divide out $m$ and the equations of motion will not change because both the inertia and the gravitational force are proportional to $m$. Second, we can adopt a convenient system of units where $G M_\oplus=R_\oplus=1$. Then all distances we get will be in units of the Earth radius (6,371km) and all velocities will be expressed as a fraction of $v_{circ}=\sqrt{\frac{G M_\oplus}{R_\oplus}}=7.9\mathrm{km\,s^{-1}}$, which is the velocity needed to achieve a circular orbit around the Earth.

Once we adopt these simplifications, the dimensionless Hamiltonian becomes

$\hat{\mathcal{H}} = \frac{1}{2}v^2-\frac{1}{r}$

The Hamiltonian is time-independent and spherically symmetric, so the specific energy $\mathcal{E} = \frac{1}{2}v^2 - \frac{1}{r}$ and the specific angular momentum $h = v_\phi r$ are conserved. Their values are, from our initial conditions,

$\mathcal{E} = \frac{1}{2}v_b^2 - 1,\,h = v_b \cos \theta.$

With the constants of motion, we obtain an equation for the radial velocity as a function of radius:

$\frac{1}{2}v_r^2 + \frac{h^2}{2r^2} - \frac{1}{r} = \mathcal{E}$

To find the apocentric and pericentric radii of the elliptical orbit, we set $v_r=0$ and solve:

$r_{a/p} = -\frac{1}{2\mathcal{E}}\left(1\pm\sqrt{1+h^2 \mathcal{E}}\right) = a\left(1\pm e\right)$

The quantity $a=-\frac{1}{2\mathcal{E}}$ is known as the semimajor axis, and the quantity $e=\sqrt{1+h^2 \mathcal{E}}$ is the orbital eccentricity. It would be at this point that one should show explicitly that the orbit is an ellipse, but I don’t find it particularly illuminating; it is done well enough here. Instead, we’ll just write down the radial equation of the ellipse that has those radii we got, as a function of the angle from apogee $\phi$:

$r\left(\phi\right) = \frac{a \left(1-e^2\right)}{1-e\cos\phi}.$

To find the range, we just need to find what $\phi$ has to be so that $r=R_\oplus$, which is just equal to 1 in our units. Then, doubling that $\phi$ and multipying by the Earth’s radius gives the arc length between the launch site and the target. Thus, the range is:

$d\left(v_b,\theta\right) = 2 \cos ^{-1}\left(\frac{1-v_b^2 \cos ^2(\theta)}{\sqrt{\left(v_b^2-2\right) v_b^2 \cos^2(\theta )+1}}\right).$

As a sanity check, we can Taylor expand this expression about $v_b=0$ to get:

$d\left(v_b,\theta\right) = v_b^2 \sin (2 \theta )+2 v_b^4 \sin (\theta ) \cos ^3(\theta )+\mathcal{O}\left(v_b^6\right).$

The leading term is just the solution we got in Physics 101 when the Earth was not curved and the gravitational field was constant, so that solution is valid for speeds much less than the orbital velocity. Cool. Also, the leading correction is positive; the range you get for a given velocity is always greater than the leading term because centrifugal force supports the projectile against gravity. cue hate mail about how centrifugal force is not really a thing.

## Burnout Speed

We are now equipped to determine what the missile’s burnout speed was based on its trajectory. Recall that it reached an altitude of 3700km (R = 1.58 in our units) and had a range of 1000km (D=0.156 in our units). We eliminate $\theta$ from the equations $d\left(v_b,\theta\right) =D$ and $a\left(1+e\right)=R$ to obtain:

$v_b = \frac{\sqrt{(1-2 R) \cos \left(\frac{D}{2}\right)+2 (R-1) R+1}}{\sqrt{R \left(R-\cos \left(\frac{D}{2}\right)\right)}}=0.86.$

Multiplying by the unit velocity 7.9km/s, we find that the missile can do 6.7km/s. This is after aerodynamic losses in the boost phase.

## Optimum Angle and Maximum Range

We now ask the following: equipped with a burnout velocity of 6.7km/s, how far can the missile go on an optimal trajectory? It turns out that launching at 45 degrees doesn’t give the greatest range in general, so we have to work out the optimum angle. Solving $\frac{\partial}{\partial \theta} d\left(v_b,\theta\right)=0$, we obtain:

$\theta_{max} = \frac{1}{2} \cos ^{-1}\left(\frac{v_b^2}{v_b^2-2}\right) = \frac{\pi }{4}-\frac{v_b^2}{4}+\mathcal{O}\left(v_b^4\right),$

so $\pi/4$ is optimal for small velocities, but in general the optimum angle is less than 45 degrees. As $v_b\rightarrow 1$, the optimum angle approaches 0, as the optimal trajectory connecting two antipodal points is just half a circular orbit.

In fact, looking at Figure 2, we see that the maximum range and the optimum angle are related in a straightforward manner, specifically:

$d_{max}\left(v_b\right) + 4 \theta_{max} \left(v_b\right) = \pi.$

Neat! Plugging the optimum angle back into the range equation gives the maximum possible range for a given speed:

$d_{max}\left(v_b\right)= 2 \cos ^{-1}\left(\frac{2 \sqrt{1-v_{\text{b}}^2}}{2-v_{\text{b}}^2}\right).$

## Results

Now, when I plug in the speed of the Korean missile for an optimal trajectory, I obtain a range of 8,000km. This is well within ICBM territory, but it is much less than the number that the Union of Concerned Scientists gave, 10,800km. I suspect they might be including a safety factor.  HAHA I’M DUMB, SEE UPDATE BELOW. UCS KNOWS THEIR SHIT. As an astrophysicist, I would normally wave away discrepancies within a factor of 2, but this difference matters because the range needed to hit LA (and me!) is 9,500km.

But what about Earth’s rotation? That complicates the problem beyond the scope I want to do here, but we can estimate the effect. Firing the missile eastward gives the missile a little speed boost along direction that the earth rotates. An upper bound on this speed boost is Earth’s rotation speed at the 38th parallel, which is 362m/s, upgrading the missile from 6.7km/s capability to 7.06km/s. When we plug that one in, we get a  range of 9,800km.

So, could Mr. Kim rain fire and fury on my beloved Institute? Probably, if the recent test flight actually represents the performance with a full payload. That’s a big maybe: the North Koreans want above all to scare others into giving them free stuff and leaving them alone, so it would make sense for them to demonstrate their missile’s performance under rather generous conditions.

Could they drop a warhead on the roof of the White House? Probably not yet, but have a look at panel 2 of Figure 2: the range rises very steeply as a function of speed as we approach orbital velocity. So it may be only a matter of time before they figure out how to pull off a few extra $100\mathrm{m\,s^{-1}}$ and achieve sufficient range.

UPDATE: David Wright of the UCS got back to me about the discrepancy, and of course it’s because his calculation is much fancier than this and models the dynamics of the boost phase. Implicit in our calculation was that the rocket’s burnout happened relatively quickly compared to the time of flight. This is OK for a first approximation, but actually not great. In particular, on the high trajectory, gravity losses were greater than they would have been on an optimal trajectory. Thus, we underestimated its capability. Thanks David!