*Nine dayes they fell; confounded Chaos roard,And felt tenfold confusion in thir fallThrough his wilde Anarchie, so huge a routIncumberd him with ruin: Hell at lastYawning receavd them whole, and on them clos’d,Hell thir fit habitation fraught with fireUnquenchable, the house of woe and paine.*

– Milton,

*Paradise Lost*

The biblical description of Satan’s fall is sparse. *Like lightning*, says Luke’s gospel. Milton’s poetry fleshes this out with his typically beautiful and dramatic imagery. A fall of nine days into Hell.

Maybe it’s just me, but the physicist in me is immediately curious at this kind of particularly physical literary license. If the fall took nine days, how far into space was he when he started falling?

A long way. Way too long to just assume he’s accelerating at 9.8 m/s^2 the whole way. After all, the earth isn’t all that big in the grand scheme of things, and you don’t have to go too far before the gravity and thus the acceleration gets much weaker. Satan is going to spend a good part of those nine days accelerating very slowly at the beginning of his fall before quickly picking up speed toward the end.

Falling bodies obey Newton’s laws, and so Satan’s mass m times his acceleration will be equal to the gravitational force at his distance x from earth which itself has mass M:

Now here’s where we hit trouble. The mass m cancels, but after that we’re faced with a second order nonlinear differential equation. It’s autonomous, which means we have a definite chain rule solution method that’s available. But the resulting integrals are ugly. *Really* ugly. My guess would be they’re not solvable in closed form, or at least not usefully so. Typical, for a problem involving the devil.

But it wasn’t so long after Milton’s time that Newton, Euler, and others developed numerical methods for handling these sorts of equations approximately. They were still a pain in the neck to do by hand, but today we automate those procedures by computer. I’ve done this for our particular equation, assuming an initial velocity of 0, and adjusting the starting position so that the falling object has dropped down to the radius of the earth after nine days – which is 777,600 seconds. Here’s a plot, with the upper curve representing the distance from the center of the earth in meters as a function of time, and the line just above the x-axis is the radius of the earth for reference. When they touch, the fall is complete.

How high up did he start? By my math (I encourage you to recheck it!), the initial distance was 5.801 x 10^{8} meters. That’s somewhat higher than the orbit of the moon. This makes sense: Apollo 11 took around 3 days to reach the moon. Since fall time increases very rapidly at larger distances, this seems fairly reasonable as a rough estimate. You can’t tell at this scale, but if you zoom in on the edge of the graph right as the falling object approaches the earth’s surface and measure the slope, you’ll see that it’s about 11,000 m/s. This is very nearly the escape velocity of earth. No matter how far away from earth you start your fall, you can’t hit the ground faster than that under the influence of gravity alone.

I doubt Milton had any of this in mind. Poetry is not built to be beaten over the head with a differential equation. If my opinion is worth anything though, I think classic literature is improved by a little physics.