UPDATE: I fixed all the formulas with the WP LaTeX plugin, so you can read the sub- and superscripts now.

OK, here is your spoiler warning. If you want to avoid spoilers for Star Trek Into Darkness, stop reading…. now!

A lot has been written already about other problems with STID, including plenty on the infamous fan-service underwear scene and the whitewashing of a famous ethnic character. I’m not going to write about these problems. Nor will I cut into the overall problems with the plausibility of a Star Fleet ship undertaking a state-sanctioned assassination.

Nope, I want to harp on a single point that bothers me: the physics of the USS Enterprise falling into Earth’s atmosphere.

So, after apparently escaping from the Killerprise and evil Admiral RoboCop, the Enterprise is travelling merrily along at faster-than-light speeds towards earth. Then, surprise! The Killerprise shows up and blasts our heros out of the ‘warp stram’ and into regular space. Oh, no, says Mr. Sulu! We’ve come out of warp 237,000 km from our destination (Earth, remember?).

I was surprised at this number, when I saw the movie. I can tell why they chose it: the moon’s orbit is about 380,000 km from Earth, so they were able to show some sexy shots of the moon in the background while the Killerprise took the Enterprise apart. However, light travels at ~300,000 km/s, so they were less than one second away from their destination when they “crashed” into regular space. They were, in fact, close enough to the Earth that their battle should have been clearly visible to any on-planet observers. I could argue about the relative timing of deceleration from faster-than-light speed, or the scramble time of Earth’s defense forces, but all of those points could be waived away by invoking the fiction part of SciFi.

Instead, I’ll whine about one particular science fact that should play into the storytelling at this point: the acceleration caused by gravity. Remember from your physics class that the force from gravity is proportional to the product of the masses of the two interacting bodies divided by the square of their distances:

$F = \frac {G(m_1 m_2)} {r^2}$

Also remember that the acceleration of a body is equal to the force acting upon it divided by mass:

$a = \frac {F} {m_1}$

So the acceleration from gravity will be:

$a = \frac {G(m_1 m_2)} {r^2 m_1}$

And the mass of the object in question (say, the Enterprise) will cancel out, leaving:

$a = \frac {G m_2}{r^2}$

OK, so the acceleration of the Enterprise will be a function only of the mass of earth and the distance between their centers of mass. That should fit with your elemetary physics knowledge: objects in a vacuum all fall at the same rate, right? When astronaut David Scott dropped a feather and a hammer simultaneously on the moon, they hit the ground at the same time. Also, as the Enterprise gets closer to the Earth, the gravitational acceleration will increase. If we sub in the mass of the earth and the gravitational constant, G, we get:

$a = \frac {(6.67 \times 10^{-11} m^3/kg/s^2) * (5.97 \times 10^{24} kg)}{r^2}$ $a = \frac {3.98 \times 10^{14} m^3/s^2}{r^2}$

This works out, because the units of a are m/s/s, as they should be. At the earth’s surface, a=9.8 m/s/s, so we can check the math. In this case, the radius of the Earth works out to be:

$r = \sqrt { \frac {3.98 \times 10^{14} m^3/s^2}{9.8 m/s^2}} = 6,372 km$

And the actual radius of the Earth is similarly about 6,371 km, so we’re within rounding error (I switched from m to km there, if you weren’t paying attention!).

Teriffic! Now, in the movie, the Enterprise and the Killerprise go through a lot of shenanigans, but eventually they both fall into the Earth’s atmosphere and hilarity ensues. How long, exactly, would it take the Enterprise (or any object) to fall into the atmosphere from 237,000 km?

The position of an object is the second integral of its acceleration, right? (I’m going to skip the Wikipedia link for that one: you’ll have to Google it yourself.)

So, in this case the position of the Enterprise would be:

$r = \frac {G(m_2)t^2}{2r^2}$

This is a bit of a problematic equation: it’s nonlinear, because position (r) is a function of both time and itself. As the position changes, the force from gravity increases, so the initial acceleration will be lower than our usual 9.8 m/s/s acceleration near Earth’s surface. We can approximate the process by calculating the relative acceleration from gravity at the initial height of 237,000 km. At this distance, the acceleration from Earth’s gravity would be only…

$5.5 mm/s^2$

That’s right, each second the Enterprise would accelerate an additional 5.5 millimeters per second! After an hour, it would have the lofty speed of 20 m/s. If this acceleration were constant, it would take about 80 hours to fall into the atmosphere. While the rate of acceleration would increase over time, in the first hour they would only have moved a few tens of kilometers.

OK, OK, so the Enterprise might have still carried some momentum from its fall out of warp, or it might have been accelerated by all of the phaser blasts from the Killerprise. Sure, but contextually, the movie suggests that 1) the ship was ‘dead in the water’ and 2) all of a sudden the Earth’s atmosphere snuck up on them and put them in mortal danger. It would only take one line from Mr. Sulu to indicate that they were still coasting toward Earth and would reach it in a matter of minutes.

Maybe I should just lighten up, but gravity isn’t just a good idea: it’s the law.