The Apocalyptic Physics of J.J. Trek

Suspension of disbelief is the willingness to disregard one’s critical faculties and believe the unbelievable; sacrifice of realism and logic for the sake of enjoyment. In Star Trek, this is most frequently required where space travel occurs. We accept that since the U.S.S. Enterprise represents the pinnacle of 23rd century human technology, it can do things that our current space-faring vessels can’t. The Enterprise can generate artificial gravity for its passengers; it can withstand extremes of temperature, energy, and gravity in interstellar space; it can warp space-time to traverse distances that would otherwise be prohibitively vast. And most importantly for this blog post, the Enterprise has propulsion systems that automatically stabilize the vessel against any gravitational forces in its immediate vicinity.

All of these fictional technologies are so commonplace in space-travel sci-fi that they represent the status quo of the genre. But as soon as you turn those technologies off, the Enterprise is just another (admittedly high-tech and oddly-shaped) piece of space-junk that must obey the laws of physics, some of which we’ve known about for hundreds of years; laws like the conservation of mass or momentum are taught in high school classrooms, and it can be awfully jarring if they’re suddenly and brazenly broken on the big screen. Cue the following scene from Star Trek Into Darkness:

When Admiral Marcus’s U.S.S. Vengeance pursues the Enterprise and knocks it out of warp, the two ships seem to end up in a more or less stationary position on the far side of Earth’s moon. A string of various action sequences later, Enterprise loses main power, deactivating those automatic propulsion systems above; Enterprise begins to succumb to the gravitational forces nearby.

Okay, that’s legit; many other sci-fi movies just ignore gravity altogether. Except then the Enterprise begins to fall not toward the nearest massive body (i.e., the Moon), but toward Earth, which is a minimum distance of 384,400 kilometers (238,900 miles) away. What’s worse, Enterprise “falls” to Earth’s atmosphere in what could not have been more than 4 minutes of screen-time (and that’s being generous), which would mean that Earth’s gravitational force can accelerate “914,442 metric tonnes” (over 2 million pounds) to a velocity of more than 5,766,000 kilometers per hour or 3,585,000 mph (assuming the ship reaches such a velocity almost instantly, with almost no acceleration time). This is of course completely absurd and highlights the rather sloppy continuity editing that plagues both of Abrams’ Trek movies, so let’s do some math to find out how long it should take an object to fall to Earth from that distance, because we’re smart people and that’s why we like Star Trek. All of the following calculations ignore all gravitational effects other than that of Earth, because it’s easier (I have no idea how to calculate multiple gravitational influences on the same object), and because that’s what J.J. did. So there.

Acceleration of gravity (g) in meters per second per second = GM / d², where

G = the universal gravitational constant (6.673 * 10-¹¹N-m² / kg²)

M = the mass in kilograms of the larger object, in this case, Earth (5.98 * 10^24)


d = the distance in meters from the center of the object to the center of the Earth, (in this case, roughly 384,400,000m).


G * M / d² = 0.27m/s²

Therefore, completely ignoring the fact that Earth’s gravitational pull would steadily increase to 9.8m/s² as we near the surface (and would therefore increase the object’s velocity and decrease fall time), after 1 second of direct free-fall toward Earth from a distance of 384,400,000 meters, the Enterprise would be traveling at 0.27m/s. At 10 seconds, 2.7m/s. The formula for distance traveled by a free-falling object is as follows:

d = 1/2gt², where

t = time (240 seconds, or 4 minutes, the maximum fall time from the scene in Into Darkness)


1/2 * 0.27m/s² * 240² = 7776m, or 7.78km.

In the four minutes (max) that it took this scene to play out in the movie, Enterprise would not have traveled even 8km. Let’s solve for “t” instead, to get an idea of  how long it should take the Enterprise to hit reentry. And just to be generous to J.J. and friends (and to avoid an obnoxiously long calculation), let’s assume a maximum, constant acceleration of gravity regardless of distance from the Earth (9.8m/s²), rather than beginning with 0.27m/s² and increasing our “g” value as Enterprise gets closer.

t = √[2d /g]

d = 384,400,000m

g = 9.8m/s²


1/2 * 9.8m/s² = 4.9

384,400,000m / 4.9 = 78448979.59

78448979.59^0.5 = 8857.14s, or just under 2 1/2 hours to plummet at the maximum possible Earth-acceleration of gravity from the Moon to the surface of the Earth. (Again, this assumes no other gravitational forces are acting upon the starship, which would not be the case in reality.) For some perspective, it took Apollo astronauts between 2 and 3 days to return from the Moon. Here’s why they took the slow train on the way back.

If you’re accelerating at 9.8m/s² for 2.5 hours in a vacuum, you continually accelerate until you hit some kind of resistance (i.e., Earth’s atmosphere). Enterprise‘s velocity at reentry can be calculated quite easily:

Vƒ = gt, where

Vƒ = velocity at reentry, or “final” velocity

g = 9.8m/s²

t = 8857.14s


9.8m/s² * 8857.14s = 86799.97m/s, or over 194,000 miles per hour. This answer violates the maximum Earth impact velocity (72m/s) for an object orbiting the Sun, which the Enterprise would be if it was stationary relative to the Earth and the Moon. However, given J.J.’s claim that Enterprise could somehow plummet to Earth in the space of 4 minutes (requiring a speed in excess of 3 million miles per hour), and because it’s more fun, we’ll go forward using 86799.97m/s as our final velocity at impact with the Earth.

To put some perspective on this, NASA’s space shuttles reentered Earth’s atmosphere at about 7800m/s (17,500mph) and executed several wide S-shaped turns in order to decelerate in time to land safely and dispose of their massive amounts of kinetic energy. The kinetic energy of the Enterprise, a 914,442 metric-ton starship reentering the atmosphere at 86799.97m/s, can be calculated as follows:

KE = 1/2mv², where

KE = kinetic energy in Joules

m = mass of the object in kilograms (914,442,000kg)

v = velocity of the object in m/s (86799.97m/s)

1/2 * 914,442,000kg * 86799.97m/s ^2 = 3,444,810,365,833,443,300 Joules of kinetic energy. Rounded to the nearest ten-quadrillions, that’s 3.44 quintillion Joules.

One metric ton of TNT releases a little over 4 billion Joules when detonated. The above kinetic energy of the Enterprise upon slamming into the Earth’s atmosphere would be equivalent to the detonation of an 823-megaton bomb (823 million tons of TNT). The largest nuclear weapon ever detonated to date was Tsar Bomba, with an estimated yield of 57 megatons; it produced a fireball 5 miles in diameter, a blast radius of 22 miles, and more limited damage at a range reaching hundreds of miles. If we guess that such destruction would increase in more or less direct proportion to the yield of the explosion, an 823-megaton detonation would produce a fireball 65 miles in diameter (Earth’s atmosphere is about 62 miles high); Enterprise‘s shock wave would level everything within about 312 miles of ground zero, equivalent to over 300,000 square miles (an area larger than Texas); and the range of limited damage and fallout would span continents.


And none of these calculations assume, as we must, that when Enterprise‘s warp-core containment fails an additional anti-matter detonation of indeterminate size and intensity will occur.

What they do assume is that Enterprise would explode somewhere in the atmosphere, breaking apart without impacting the surface. But why should we assume that? According to Abrams himself in the opening sequence of Into Darkness, the Enterprise is perfectly capable of planetary reentry. And elsewhere in Trek canon we find further examples of starships entering a Class M atmosphere without disintegrating; see Star Trek III: The Search for Spock and Star Trek Generations, in which both ships suffer massive damage to their hull structures before beginning reentry. What if the Enterprise‘s hull, shields, and/or structural integrity allow it to pass through the atmosphere more or less in one piece?

Using this website, we can estimate the crater dimensions, thermal radiation output, seismic effects, ejecta, and air blast intensity at varying distances from ground zero. Assuming the following parameters:

distance from impact = 500km (311 miles, on the edge of our estimated blast radius above)

projectile diameter = 370m (the longest dimension of Abrams’ reinvented Enterprise)

projectile density = 4328.76kg/m³ (average, according to this site again)

impact velocity = 868km/s

impact angle = 90° vertical

target type = sedimentary rock (with an average density of 2500kg/m³)

Here are the highlights (full results here):

The final crater left by the Enterprise would be roughly 55.4km (34.4 miles) across and 991m (3250 feet) deep, melting or vaporizing 377km³ (90.4 cubic miles) of the matter at ground zero.

The visible fireball at 500km away would be 100.2km (62.2 miles) in diameter, and even at this distance the thermal radiation exposure would exceed 52 million Joules per square meter for over 15 minutes. This equals over 49,000 BTUs per square meter. According to my calculations, this would raise the ambient air temperature to something over 2250 degrees Fahrenheit. For over 15 minutes. At this temperature and duration, all exposed organic material is incinerated.

The impact of the Enterprise under these parameters would register 9.3 on the Richter scale, but damage due to seismic shocks would be minimal at a distance of 500km. Several minutes after impact, the area would receive a fine dusting of dust and debris reaching up to 4.08in thick.

25 minutes later, the blast of displaced air would arrive at 196m/s (439mph), gutting or leveling nearly all man-made structures. Up to 90% of the trees hit by this air blast would be blown over, and the rest completely defoliated.

Some margins of error in this scenario:

First, we should remind ourselves that modern astrophysics tells us that the maximum Earth impact velocity for an object orbiting the Sun is 72m/s; I have yet to discover why this is true, and if any readers could enlighten me I would be most appreciative. If the Enterprise is assumed to be “orbiting the Sun,” then we must reduce our impact velocity drastically, which would also drastically reduce the ship’s kinetic energy upon impact. Enterprise is not an especially large object, at least in terms of giant hunks of menacing space-junk that might hit the Earth, and most of its destructive power would come from the excessive velocity we calculated above.

Second, the listed “diameter” and density of the Enterprise are not necessarily very accurate; they are merely really good estimates regarding a fictional spacecraft (from one of the most devoted fan bases in the world). Enterprise is not a “solid” shape, meaning that it contains far less mass than a spherical object of similar diameter; we would find more reliable results if we could calculate using total mass rather than diameter.

Barring any of these margins of error, though, we can expect total annihilation of an area larger than Texas and global fallout produced by superheated debris being flung into the upper atmosphere and raining down all around the world, creating a literal firestorm that would ignite vast wildfires anywhere the ejecta returns to ground. Depending on the location of ground zero, casualties could be in the billions (especially if population growth continues into the 23rd century).

Though most available studies on potential climate changes due to such a disaster use much larger objects in their calculations (>1km in diameter), we can assume catastrophic loss of plant and wildlife in the ecosystems immediately surrounding the blast radius, with the possibility of food-chain disruption and extinction of some species. Secondary loss of environment and agriculture would occur across the globe.

The Earth Impact Effects Program features a Google Earth plugin that allows you to pinpoint an impact anywhere on Earth using latitude and longitude. The same input parameters as above are applied; they have even included a drop-down menu to view the radii of various effects (crater, air blast, etc.). Here are the results if the Enterprise in our above calculations were to strike the Earth at Starfleet Headquarters in San Francisco, CA. The City by the Bay is incinerated along with basically the entire West Coast of the United States.

And here are the results if we use J.J.’s apparent velocity from the film, something around 3.5 million miles per hour (the speed that could allow the Enterprise to reach Earth from the Moon in under 4 minutes). In this scenario, note that

a) the air blast at first looks smaller than the previous scenario, until you zoom out and realize that it’s glitching the plugin and it actually covers the entire globe,

b) the final crater would be 353 miles wide and over a mile deep, with 1,060,000 cubic miles of matter being melted or vaporized on impact, and

c) the thermal radiation from such an impact would apparently incinerate all of North and most of Central America.

It took me a several hours over the course of two days to relearn some basic physics, do some arithmetic, and type this out. Apparently J.J. was too busy with blowing shit up and telling Alice Eve to strip down to her skivvies and couldn’t be bothered to stop and consider how ridiculous this scene is.

Ironically, we are now forced to wonder why Admiral Marcus would wake up a 300-year old genetically modified superman to build top-secret weapons so he could blow up the Klingons when really, he could have just set his big jet-black spaceship on auto-pilot and programmed it to ram into the Klingon homeworld at warp speed. I guess you don’t get to run the show at Starfleet by being good at physics.

*          *          *

But wait! Bonus round!

In the scene from Into DarknessEnterprise is traveling at 3.5 million miles per hour when it reaches Earth. If we assume that this velocity is still slow enough that the crew actually have the time to process what is happening and react accordingly before impact, then we would need 3.44 quintillion Joules of energy to bring the ship to a full stop before it hits the atmosphere, at which point it’s already “Kobayashi Maru” for the Enterprise and her fearless crew. If we also assume that, based on my recollection of the film, we only have 5 seconds in which to complete this maneuver after engine power is restored, then:

3.44 * 10^18 Joules / 5 seconds = 6.88 * 10^17 Joules per second, or 688,000 terawatts. Another post from a previously quoted nerd-site argues that at a comfortable cruising speed, the warp core of the U.S.S. Enterprise-D (four generations of technology later) produces about 7.1TW. In the year 2006 the entire human race consumed only 16TW of power.

Bonus round 2:

In order to accelerate the Enterprise to 3,585,000 miles per hour over a distance of 238,900 miles under gravitational force alone, the Earth would have to be orders of magnitude more massive. To solve for the correct mass of a stellar body that could accelerate an object to these speeds in such a small period of time, we’ll need to return to our equation to find the acceleration due to gravity above. Our goal is to find “g” in the calculations below, which we can then use to find “M” in the very first equation we solved above (acceleration due to gravity at varying distances from the larger object).

d = 1/2gt², or

g = 2d / t²

t = 240 seconds (4 minutes)

d = 384,400,000m


2 * 384,400,000 / 57600 = 13,347.22m/s², the acceleration due to gravity required to accelerate an object fast enough to travel 384,400km in under 4 minutes.


g = GM / d², or

M = gd² / G

We know all of these values except “M”, so:

13,347.22 * 384,400,000² / G = 2.96 * 10³¹ kilograms, or, rounded up, a 3 with 31 zeros behind it. For some perspective, the mass of the Sun is roughly 1.99 * 10³ºkg.

The Schwarzschild radius is the radius of a sphere such that, if all of the mass of an object is compressed within that sphere, the escape velocity from the surface of the sphere equals the speed of light. In other words, any object that occupies a volume smaller than its own Schwarzschild radius is a black hole. The Schwarzschild radius is calculated as follows:

r(s) = Schwarzschild radius

c = the speed of light in a vacuum (299,792,458m/s)

r(s) = 2GM / c²


2 * G * 2.96 * 10³¹ / 299,792,458² = 43,970m, or 43.97km (over 27 miles).

The Earth’s radius is 6371km (3959 miles), so according to J.J.’s new film it falls short (or rather long) of the Schwarzschild radius. However, with a mass of 2.96 * 10³¹kg (and assuming the real-life volume), the Earth would have a density of 2.69 * 10¹ºkg/m³, which approaches the density of some portions of a neutron star.

An Earth of this density would have an escape velocity of about 25,000km/s, or nearly 56 million miles per hour. This is roughly 1/6 of the speed of light, which maybe seems a bit prohibitively high for the residents to invent space travel. Or do anything else, for that matter.

This has been fun. I’ve relearned a lot of physics and I got to pick on J.J. Abrams and his fucking hipster glasses at the same time. In the future he should either do a better job than Scotty at keeping the engines on the Enterprise running, or do enough math that any English major with Google and a calculator can’t make an ass out of him.

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