Falling from Heaven

i-0c6e7e00045a24f493aa8f58cbbf2fed-satan.jpg

Nine dayes they fell; confounded Chaos roard,
And felt tenfold confusion in thir fall
Through his wilde Anarchie, so huge a rout
Incumberd him with ruin: Hell at last
Yawning receavd them whole, and on them clos'd,
Hell thir fit habitation fraught with fire
Unquenchable, 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:

i-d4bb9fe09d6b78eb832985821d8a307f-1.png

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.

i-acac9b0772fef257e874ca93ddd40001-fall.png

How high up did he start? By my math (I encourage you to recheck it!), the initial distance was 5.801 x 108 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.

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Hint: multiply both sides by dx/dt, then take anti-derivatives.

Poetry is not built to be beaten over the head with a differential equation.

This is true. Diff eqs are beautiful; poetry is not. ;)

What if you take the wind resistance in account?

There's pretty much no atmosphere above 100km or so, and thus for all but the last tiny fraction of his fall there's no air resistance. Even if his cross-sectional density if low enough to cause him to slow to a terminal velocity typical of skydivers in the last few miles, it should only add a few minutes of fall time.

Interesting analysis. Assuming Satan weights 100kg (and why not?) he lands with a kinetic energy of 6 gigajoules, or about one and a half tons of TNT. Maybe that's what started hell burning...

The approximate answer can be obtained without any calculus at all by anyone familiar with some basic orbital mechanics and the works of Arthur C. Clarke :-)

(I got the same answer, 5.8 x 10^8 m; I'll post the workings later if anyone is curious. Clue: "Jupiter 5")

Earth-moon center-to-center distance of 384,403 km plus a moon radius of 3474 km/2 gives 3.861x10^8 m - so about half-again distance for the inner edge of heaven. It's a reasonable safety margin for an empirically boring boundary. Hell is more interesting for being condensed phase.

Were Dante's nine days any more exact than Yahweh's six? (There's that half-again factor, again.)

Quite an epic fall by the end of it, but the first few hours must have been awkward.

"Holy crap, guys I'm totally falling, I think! Can someone toss me a rope? Or at least a sandwich?"

Poor satan, the atmosphere will totally turn him to a crisp when he falls through the ionosphere. Not only that, his first few days of falling must be super slow. He must have been bored to death. :)

At that distance aren't there other bodies (the moon for instance) that could exert a substantial gravitational force, acting to spped up or slow down the corporeal evil one?

Michael Martin said "Yes but Milton's units of time are not uniform because precise timekeeping was still uninvented."

That is true, but since the duration is in days, there wasn't much precision required. Even cavemen could distinguish different durations measured in days. There's a convenient periodic occurrence that facilitates day-counting. It's called the rising of the sun.

By Tim Whitley (not verified) on 28 Aug 2008 #permalink

Timmy

Isn't the error in the answer too much implied precision? if the time of falling is 9 days is 9+/- 0.5 days - so if the answer is taken as 6x10^-8 m (not 5.801 x 10^-8 with allitsa nasty implied precision)the quandry is removed.

KAM

At that distance aren't there other bodies (the moon for instance) that could exert a substantial gravitational force, acting to spped up or slow down the corporeal evil one?

That would be very true, and at a distance that far beyond the moon, it would be possible to even be ejected from the Earth-Moon system completely. That would lead to a much longer "fall." ;)

Ahhhh, except satan didn't FALL from heaven, he was CAST OUT of heaven. V-initial wasn't zero, so apparently heaven is within the orbit of the moon...

By Max Fagin (not verified) on 28 Aug 2008 #permalink

So THAT's how the dinosaurs went extinct...

And assuming he fell straight down. He could have been cast out horizontally and done a couple of orbits before landing.

By Anonymous (not verified) on 28 Aug 2008 #permalink

Max, if V-initial was nonzero and toward the Earth that would require heaven to be farther away to still have a 9 day fall, rather than closer.

Since the physics has been amply addressed and the subject is philosophical, let me approach it from that side.

One: Jean-Paul Sartre said that hell is other people.

Two: While we know that Satan was cast out of Heaven, the term "fall" could well be moral rather than physical.

Consequently, it is easy to imagine that Satan was cast out of Heaven, away from Angels and God, reduced to a quasi-mortal state, and forced to live among men.

Let us then imagine Satan's fall as a nine day trek across the desert in search of a settlement.

Just so you know, I am an atheist and this solution was simply based on Sartre's quote flashing in my brain.

I find it interesting that everybody agrees with the assumption that hell is on the surface of the earth!

If there were a shaft allowing Satan to continue to fall to (and beyond) Earth's centre of gravity, would he start to oscillate back and forth before coming to rest? And if so, for how many additional days?

:-)

DOn't forget to take into account the force imparted by Gods foot when Satan was kicked out

By Damien@macmeis… (not verified) on 29 Aug 2008 #permalink

Interesting.

Song, if there were a shaft for Satan to fall, and assume that there were only earth to consider, then he would oscillate for a long time, until all his body could not be harmed by many iteration of burning while we was travalling back and forth through the atmosphere. Of course, when he would stop depend on the 'shape of him'. I mean, whether he would fall in a vertical fashion like a rocket, or he would open his wings while he falls make a big difference.

Apart from the same, there are also objects above. Moon is the obvious one, which exerts gravitational pull and affects the path. The huge sun is also a considerable factor. So the path Satan would fall would not be linear.

The path also depends on the initial velocity and direction of the way he was casted out. In some occasions, he might crash some satellites, hit other planet, say Saturn, or Jupiter, or go directly to the sun. With some very rare chances, he may leave our Solar system.

So...what's the method for calculating this without calculus?

In Clarke's short story "Jupiter 5", the Professor correctly observes that the time taken for an object to free-fall from rest to a distant large mass is equal to 0.177 times the orbital period of a circular orbit at the same distance.

So, how long is that? In this case ol' Mr. Newton has already done the calculus for us; the period of an elliptical (or circular) orbit is

T = 2pi * sqrt(a^3 / GM)

where a is the semi-major axis (or radius in the circular case).

So rearranging (since we know T*0.177 = 777600), we have

a^3 = (777600 / (0.177 * 2pi))^2 * GM

The quantity GM happens to be known far more precisely than either of G or M independently: it's 398600 km^3/s^2 for the Earth (actually 398600.4418 with uncertainty 0.0008, but we'll round it off).

So a^3 = 194869346510155320 km^3, therefore a = 580000 km (to 3sf),
which is 5.8 x 10^8 m, matching the original post.

[So why 0.177? Oddly enough, you can derive that too from the original equation as follows: assume that the mass is not exactly at rest, but instead has a tiny velocity perpendicular to the direction of the Earth. In this case the object must be at the apogee of a very eccentric elliptical orbit; in the limit as the velocity goes to 0, the semi-major axis of this orbit goes to half the distance to earth. Since the period of the orbit goes as sqrt(a^3), and we're interested in only half an orbit, this gives us a factor of 1/(2*sqrt(8)) = 0.177.]

Sorry to be snarky, but Satan fell into Hell, not earth. He had to cross the abyss of the void to reach Earth after he forced open the gates of Hell.

By Anonymous (not verified) on 29 Aug 2008 #permalink

My favorite poem of this sort is by Kipling, Tomlinson, which discusses various astrophysical things such as stellar evolution, for example:

"The Spirit gripped him by the hair, and sun by sun they fell
Till they came to the belt of Naughty Stars [Paris Hilton?] that rim the mouth of Hell:

The first are red with pride and wrath, the next are white with pain,
But the third are black with clinkered sin that cannot burn again:

They may hold their path, they may leave their path, with never a soul to mark,
They may burn or freeze, but they must not cease in the Scorn of the Outer Dark."

By Carl Brannen (not verified) on 29 Aug 2008 #permalink

"Like lightning"

Ah, a test run of the tripod pilot transport system. How Spielbergian.

On a similar note, I'd like to point out _Hesiod's Anvil: Falling and Spinning through Heaven and Earth_, by Simoson.

Interesting that your result is not that far off from Medieval ideas about where Paradise was located.

But be careful. When this becomes a mandatory part of the curriculum in physics after Creationism becomes a required subject in the science classroom, you will need to get the mass and location of Hell correct. Many Creationists have views that include a changing mass of the Earth related to the Flood and changing rates of the speed of light or time to fix up problems with radioactive dating. Arguments that Satan would not have had enough energy to burn through the atmosphere and earth to reach Hell will not be popular.

Also note that if your calculation of the location of Heaven is correct, trips to places beyond Heaven (such as Mars) must have been faked by heathen scientists.

Hi -- I'm a prof at Harvey Mudd in Claremont, CA. I work on Milton, other early modern poets, and the intersections btwn math and poetry in the period. I'm teaching a Milton course this spring and would really like to use this in my class -- the physicists will love it! Would you mind?
Thanks!

By jacqueline wernimont (not verified) on 10 Oct 2009 #permalink

The Mythbusters did once successfully test that a rope made from human hair was sufficiently strong to allow someone to escape from prison.

First of all, when Satan was CAST OUT of HEAVEN, the Earth was NOT yet created, therefore gravity cannot be calculated into the equation. Good try.

By Anonymous (not verified) on 24 Oct 2009 #permalink

Maybe "cast out" means evicted and he was given 9 days to pack his crap and get out.

Actually, the equation of motion does have a closed form solution of t as a function of x and x0:

t = sqrt(x0^3/2GM)*(acos(sqrt(x/x0))+sqrt(x/x0)*sqrt(1-x/x0))

There is no closed form inversion of this solving for x (or x0) as a function of t.

If we assume that Hell is on the surface of Earth (mean radius 6371.000 km), then the radius to Heaven is 580437.3 km. If it is at the center of Earth, 580247.8 km.