SPOILER WARNING: Star Trek Into Darkness is still recent enough that I need to warn you.
SPOILER WARNING WARNING: Eventually I’ll do to these warnings what J.J. Abrams did to the colon in the title.
Picture the scene: a spaceship falls helplessly toward Earth, only to regain power at the last minute. It falls through a cloud, then ascends slowly from below it with triumphant music in the background. Or maybe it hits the ground in a crash landing and the beleaguered characters step off it accompanied by the same triumphant music. Either way, the audience cheers and nerds reach for their calculators.
In Star Trek Into Darkness, there was one throwaway line saying how far from Earth the Enterprise was before they started falling. I’m pretty sure it ended in “hundred thousand kilometers,” but it could be any number of hundreds of thousands. If anyone knows, I’ll update the math; if not I’ll wait until the next time I see the movie. EDIT: Bad Astronomy says it was 237,000 (which doesn’t end in “hundred thousand” when spoken), and that this number is probably a mistake.
But it’s canonical now, so we’re stuck with it. On the up side, now that we have the information we need the math is solvable . The distance is stated (237,000 km), and gravitation is pretty well understood. F=GMm/r^2, and F=ma, so the acceleration of the falling ship is GM/r^2.
G=6.67×10^-11, M=5.97×10^24. Hopefully, anyone in charge of piloting a starship is at least marginally familiar with those numbers. We can plug those in and we get that, at any given time, the Enterprise is accelerating by (3.985×1014)/r^2 meters per second per second, where r is its distance from the center of the Earth.
Normally, the acceleration due to gravity is a constant; the famous 9.81 m/s^2. That’s what you get if you plug in the radius of Earth (about 6,371,000 m) for r, since most of the falling things we care about are at or near the surface. So their distance from the center of the earth before and after the fall is pretty much the same, and it’s OK to round them off to the same number and say that g=9.81.
In this case, the Enterprise is falling 237,000 km, which actually is a lot compared to the 6,370 km radius of Earth. So we don’t get to treat the acceleration due to gravity as a constant. It’ll start off slow, because the gravity pulls weaker on the ship when it’s farther away, and get gradually faster toward the 9.81 number that we all know and love. (Yes, it won’t ever actually be accelerating that fast because that only happens at the surface and by that point there’s air resistance, but really? Air resistance? We can ignore that.)
Unfortunately, I did not have any idea how to set up that equation. I think I used to, and there’s definitely some way to do it using integrals, but I couldn’t manage it. If anyone wants to, your solution will be more precise and elegant than mine.
What I did instead was try to estimate it. First approximation: pretend it was falling at the same speed as things fall at the Earth’s surface. Since we know it’s falling slower, it’ll take at least that long to reach the surface. With acceleration of 9.81 m/s^2 and h=237,000,000 m, it takes √((2.37×10^8)/9.81)=4915 seconds to hit the surface. That’s eighty-two minutes, or about an hour twenty: way longer than what it seemed to be in the movie, but potentially short enough to maintain dramatic tension. We don’t know how long that character was in that place doing that thing. But remember, this was an absolute minimum.
Second approximation: Think of it in two halves. For the first 118,000 km, use the acceleration due to gravity that the Enterprise will be experiencing when it reaches the halfway point. For the second 118,000 km, use the surface gravity. At 124,000,000 meters from the center of the Earth*, the acceleration due to gravity is 0.02592 meters per square second. At that much acceleration, it would take √(124,000,000/.02592)=69160 seconds, or nineteen hours twelve minutes. But that’s just to get halfway. The second half will be quicker, though. At Earth surface gravity, a=g=9.81 m/s^2, which is way bigger than the .0259 from earlier. And the ship is already moving in a downwardlyish direction at 0.02592/s^2 x 69160 = 1793 m/s.
With initial velocity of 1793 m/s and acceleration of 9.81 m/s^2, it would take 3378 seconds, or about 56 minutes, to travel the final 118,000,000 meters and reach the surface. When hitting the surface, it would be traveling at a pulverizing 34,930 m/s, which is over one hundred times the speed of sound.
The total time there is about twenty hours, mostly from the first half of the distance and a bit from the rest. This is plenty long enough that nobody was actually in a hurry for what they were portrayed doing. Now, the precise number is definitely going to be more than 20 hours. That number came from assuming that it’s falling faster than it actually is. For the first half, it’s too far away for gravity to give it the kind of acceleration we assumed. For the second half, as soon as it was close enough to get the accelerate we had already been crediting it with, we increased the amount we assumed. So we know it’s over six hours.
The same thing works for finding an upper bound. For that, we just assume an acceleration that we know is lower than the true one. One good candidate for that is the starting value of G*M/r^2 where r is the starting distance from Earth’s center, or 243,000,000 m. That gives an acceleration of only 0.00675 m/s^2, which would take it 132200 s = a day and a half to fall half the distance to Earth. If we then switch acceleration values to the actual acceleration at 118,000,000m from the surface, we get a=0.02592 m/s^2 with an initial velocity of 892 m/s. It takes them 52484= 14.5 hours. The total in this estimate is 50 hours, and it’s moving at 2245 m/s. The real answer will take less time and be moving faster.
(This was basically a Riemann sum with n=2. You see why I wish I had figured out how to set up the integral, and I’m still feeling stupid for not managing it. This could have all been done as one problem and ended up with a better answer.)
So, the time it takes is somewhere between twenty hours and fifty, and at the end it’s moving somewhere between 6.5 and 100 times the speed of sound. Bad news on the way-too-long-for-dramatic-tension front and worse news on the not-killing-everybody front.
Wikipedia tells me that the Enterprise (NCC 1701-E where this one was A, but it should be close) has a mass of 3,250,000,000 kg. This would of course result in the Enterprise being annihilated on reentry, so it was handwaved by Scotty getting the shields running in the nick of time, or something. Plugging in that mass for an asteroid, I’m told** that the impact would blast a crater 4.5 miles deep and twelve across. This would also kill the main cast.
Now, you might be thinking that I assumed the spaceship was at rest relative to the Earth when it started falling. I did. But if it didn’t, then it would either be moving toward the Earth and fall even faster and more fatally, or it would be moving away from the Earth and still end up crashing even harder. It’d just take longer for Earth’s gravity to turn it around, and then it’d be dropping from higher. Assuming that the ship isn’t moving relative to Earth is actually the nicest thing I could do. As for the true answer, well, it had come out of warp without getting a chance to move, so whatever the velocity is when it drops into normal space, that’s how fast it was moving.
Now let’s look at what would have to happen in order to save the Enterprise and the day.
Obviously if they’re going to turn it around it has to be at the last minute, for the same reason that all timers must be showing 0:01 when their bomb is disarmed. It has to be in Earth’s atmosphere, because otherwise it isn’t recognizably a Close Call.
So that means the turnaround has to happen within, say, 100 km from the surface. Any more and it happens in Space, which isn’t as excitingly last-minute. It’d probably be lower, but this gives us a good maximum altitude. The problem is that it would cover the distance from space to surface in less than ten seconds. (Probably a lot less, but that Purdue site says that 11 km/s is the minimum for objects freefalling from outside Earth, so let’s use that.) It’s got to decelerate from 11,000m/s to 0 in ten seconds
That kind of sudden stop, at 1100 m/s per second, is extremely unsurvivable. Everyone on the ship would be instantly transmogrified into a bloody pulp smeared across what used to be the floor, assuming the floor doesn’t collapse when the weight of literally every part of the ship increases by a factor of 112.1. Fortunately, it’s established that Star Trek ships have “inertial compensators,” which, well, compensate for inertia. We know this because of the lack of frequent transmogrifications into bloody pulp. So we can assume that machinery was working and the people in the ship were in no danger from a deceleration several times worse than ramming your car into a wall.
No, I’m more concerned with the effects on the Earth. The ship has a minimum of 3,250,000,000kg*1100m/s=3.575*10^12 kg*m/s of momentum. That’s three and a half trillion. Trillion, with a T. And the ship isn’t stopping by some magic futuristic technology; they specifically say that they’re using thrusters. In other words, rockets. Sure, they use deuterium fusion reactors instead of combustion to power those rockets, but they’re ultimately working by good old opposite reaction. So they have to blast out a lot of mass really fast to take care of that momentum. If the exhaust is moving at 3×10^8 m/s, which you may recognize as THE SPEED OF LIGHT, then they’d still need to be blasting down more than ten (metric) tons of helium per second at relativistic speeds.***
Since this is taking place in Earth’s atmosphere, it would be a ten-second explosion bigger than anything you want to be around. Helium is usually not very scary, but in this case it’s tons of plasma blasting down at you at a significant fraction of the speed of light, followed shortly by the world’s biggest sonic boom for good measure. It won’t be anywhere near planet-destroying, but it’d still be incredibly destructive. And the Enterprise is definitely not going to gently fall through a cloud and rise back up. Maybe it could avoid being obliterated by the shock wave from the helium thing, but I don’t like the survival chances of anyone standing underneath. And this is what happens if they don’t crash.
Good thing there isn’t an even bigger ship crash-landing at the same time and aiming to cause as much damage as possible. That would suck.
*”distance from Earth” usually means distance from the surface. So I added 6000km in some of the calculations to approximate the Earth’s radius because sometimes the number is measured from the center.
**OK, so this is kind of an extreme case and I don’t know if the Purdue people set it up to be able to handle this. Especially since, to simulate the ship not breaking up in the atmosphere, I told the asteroid calculator a diameter of 1m and an incredible density. 6207042780 kg/m^3, to be exact, since a sphere of that size and density ends up with the mass of the Enterprise. Anyway, accuracy of this simulation is not guaranteed but the point is that ouch.
***Not really. This is using Newtonian math because I don’t feel like doing relativity calculations right now, but the total momentum of all that helium will still be 3.575 trillion N*s. The speed and mass there are definitely going to be big numbers. It would be a very good idea to be somewhere else.