Physics 101: Invisible Woman, SHIELD and Terminal Velocity

Remember that time when the Invisible Woman was on a mission for SHIELD, and it had to do with terminal velocity?

If your only exposure to Marvel’s SHEILD is the television series on ABC, then probably not, but in 2014-2015, Marvel published a SHIELD comic series that brought Agent Coulson, May, Johnson, Fitz, Simmons into the comic book Marvel Universe. Storylines centered on superheroes that Coulson picked for specific missions, and in issue #4, he needed the Fantastic Four’s Invisible Woman, Sue Storm.

The short version of the mission – rescue. Athol Kussar, who knows a ton of secret Hydra intel is at the bottom of a South African diamond mine. Sue’s job – go down there and get him out. There’s lots of other complicating factors (of course), but the topic today is terminal velocity. The mission comes to a point where the Invisible Woman has to get out of the mine in a hurry, and tells Coulson in regards to how she got down there, “Terminal velocity got me down here in two (minutes).”

Prior to this, we saw Sue jump off of a catwalk running across the top of the mine and falling down to the bottom, before forming an invisible parachute, which slowed her final descent.

So let’s dig in to the science of the fall – and terminal velocity.

First up, what is terminal velocity? Despite the vaguely threatening nature of the term, terminal velocity does not mean the speed at which you die. Rather, terminal velocity means the speed at which you can’t free fall any faster.

Think of it like this – if you jump out of an airplane (with a parachute) you’ll fall faster and faster – your vertical velocity while in the plane (assuming the plane was flying perfectly level) was zero, after all. When parachutists are falling, they generally assume a flattened-out profile, as if they were lying flat on the ground. Sue is shown to do basically the same in the comic.

In that flattened out position, a human being’s terminal velocity is about 120 miles per hour (about 56 meters per second). That means that as they are falling, they fall faster and faster thanks to the acceleration due to gravity (a pull downwards – toward the center of the earth of 32 feet per second, each second, 32 ft/s/s, or 9.8 m/s/s) until they reach 120 miles per hour. At that speed, the air that they are falling through is pushing back against them with a force equal to their weight. They’re no longer accelerating.

The person is still falling – but they’re no longer falling faster and faster. They’re falling at a constant speed of 120 miles per hour. You’ve probably seen scenes in movies where parachutists can change their profile (a.k.a., “go Superman”) and increase their speed. That happens when you change the amount of surface area that’s being presented to the air – decrease the surface area, and you’ll fall faster. Increase the surface area, and you’ll fall slower.

So how long does it take to fall for five miles? Let’s run some numbers.

Basically, this is a three-part problem:

• How long it takes Sue to reach terminal velocity,
• How far she falls while she’s accelerating to terminal velocity, and
• How long it takes her to fall to the bottom of the shaft once she’s reached terminal velocity.

First things first. How long does it take Sue to go from a standstill (just before she lets go on the catwalk) to 56 m/s (we’re going ot do everything using standard units, because…science)?

For this part, we need to use the definitional acceleration formula, that is:

Acceleration = change in velocity/change in time, or

a = Δv/Δt (Δ means “change in,” or final condition – initial condition)

We already know our acceleration (9.8 m/s², technically -9.8 m/s² as it points down, but we’ll let the “-“ go for now as long as everyone promises to remember that Sue is falling down, not up), our v_inital (0 m/s) and our v_final (56 m/s – terminal velocity). We need to solve this for time.

So rearrange that formula up there, and we get:

t = v_f – v_i/a

Numbers!

t = 56 m/s – 0 m/s / 9.8 m/s/s = 5.7 seconds.

It takes Sue 5.7 seconds from the moment she starts falling to reach 56 m/s. Things fall fast. Thanks, gravity.

For the second part of our problem, we need to know how far Sue falls while she’s accelerating up to that 56 m/s. Since we figured out the time, we know everything we need, and it’s a simple kinematics formula that we can use. The word version tells us that the displacement of an object (vertical in this case, since Sue is falling down) is equal to the product of the initial velocity of the object and time in motion added to the product of .5, the acceleration and the time in motion, squared.

In symbols, it looks like this:

Δy = v_it + 1/2at²

Our initial velocity, as mentioned earlier, was 0 m/s, so that first term cancels out, leaving us with Δy = 1/2at². Plug the numbers in, and we get Δy = 159 meters.

So – Sue jumps off the catwalk, falls faster and faster until she reaches 56 m/s. While she was falling faster and faster, she fell for 159 meters.

The third part of the problem is relatively easy – we just need to figure out how long it takes Sue to fall the remainder of the way while moving at 56 m/s.

The mineshaft is 5 miles deep, which converts to roughly 8047 meters. She’s already fallen 159 meters, so that means that when Sue hits terminal velocity, she’s got 7888 more meters to go. How long does it take?

Because she’s moving at a constant velocity (56 m/s) the basic velocity formula will work out just fine.

Average velocity equals displacement divided by time, or as a formula: v = d/t. Rearrange it to solve for time, and you get: t = d/v.

Put numbers in, and we get 140 seconds. Divide by 60 to get minutes, and we get 2.3 minutes. Multiply .3 minutes by 60 to get seconds, and we find that Sue’s total fall time at terminal velocity was 2 minutes and 18 seconds.

But remember – that wasn’t the total time – Su spent 5.7 seconds reaching terminal velocity, so the total time it would take for her to fall the 5 miles would be 2 minutes and 23.7 seconds (2 minutes, 18 seconds plus 5.7 seconds from part one).

In the story, Sue told Coulson that it took her two minutes to get to the bottom of the mine, but if we’re being fair, and preventing Sue from going splat at the bottom of the mine, it probably took her a little closer to three or four. After all, once she created her invisible chute, her velocity would’ve dropped to something far less than 56 m/s. But we’ll give Sue the benefit of the doubt on calling it at two minutes – it was a high stress mission, and when she told Coulson that she was able to make it in two, she was looking down the barrel of performing unassisted, radio-guided open-heart surgery on Kussar. Read the comic for the full story – it’s available online via Comixology  or in the back-issue bins at your local comic shop. SHIELD #4 was written by Mark Waid, with art by Chris Sprouse and Karl Story, and colors by Dono Sanchez Almara.

Some other neat topics surrounding what happened in the story:

• Sue asked Coulson about the heat. At five miles down, the temperature, thanks to the earth’s core being really hot, it about 105° F (about 41°C). Coulson says the heat is shunted.

• Coulson mentions radiation in the hole – generally speaking, radiation’s not a huge problem in mines. It can be an issue, with radon leaks and if you’re specifically mining say, uranium, but deep mines always being heavily radioactive? Not always the case.
• In our world (not the Marvel Universe), the deepest functioning mine is/are the TauTona and the Mponeng mines, both in South Africa. Both are about 2.4 miles deep and jockey for the “deepest” banner. Mponeng was profiled in The Wall Street Journal in 2012 by Matthew Hart who’d been there, researching his book, Gold. The takeaway from Hart’s stories – deep mines are not a fun place to work or even visit, rescue mission or no.

• Oh, and one last thing – if you’re looking for another pop culture explanation of terminal velocity, check out the video below, originally posted by ElasticScience. It explains the parachute jump from the beginning of Moonraker in terms of terminal velocity and how Bond and Jaws changed theirs during their respective falls. You get bonus points for spotting when Roger Moore was replaced by the stuntman. Not many bonus points though – they weren’t even trying to hide it.

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