How hot would the space jumper get?

A new video from the Red Bull Stratos Jump guys came out. Here it is:

This reminds me of an unanswered question about the Stratos jump that I didn't address on my last post on this topic. Commenter Long Drop asked about how much Felix would heat up as he falls from 120,000 feet. This is a great question. The first, off the bat answer is that he won't heat up too much. Why do I say this? Well, when Joe Kittinger jumped from over 100,000 feet and didn't melt. Still, this is a great thing to calculate.

How do you calculate something like this? I will look at this in terms of energy. For simplicity, I will consider the jump from 120,000 feet down to about 30,000 feet. After that, Felix will pretty much be a normal sky diver. Here is the plot of speed vs. height from my previous post.


Just a note, the green line is the speed of sound, the red line is his speed if he jumped from 100,000 feet and the blue is from 120,000 feet. Think about this fall. If there were no air resistance, he would be going much faster and would have much more kinetic energy. So, without air resistance I could use the work-energy principle. If the Earth and the jumper are in the system, then:


But with air resistance, the jumper will not actually be going that fast with that much kinetic energy. So the missing energy had to go into an increase in thermal energy. This increase in thermal goes both into heating up the air and the jumper. But, how much goes into the air and how much into the jumper? I am just going to make a basic assumption that half of the energy goes into the air and half into the jumper. Simple, right? Now I just need to re run my numerical calculation and get the difference between the no air kinetic energy and with air kinetic energy. Here is a plot of kinetic energy vs. height (both with and without air resistance).


From this, I get the values:


That seems like a lot, even if only half of that went to the jumper. Instead of calculating the change in temperature, let me think about this in terms of power. That can give me the change in thermal energy, but how long did it take? From the numerical calculation, falling to 30,000 feet takes about 150 seconds. This would give an average power (so I could compare to an electric heater) of:


Still not very good. I just can't imagine having a 70,000 watt heater hooked up to you for even 2 minutes. Maybe the time is so short, it doesn't matter. Here is an idea. What if I do the same thing for a normal sky diver? Let me assume a sky diver jumps from 10,000 feet to 3,000 feet falling at 120 mph (constant the whole way for simplicity). This is a little bit simpler. I can calculate the change in gravitational potential energy for the fall and compare it to the kinetic energy of a dude going 120 mph.


Assuming the fall is at 120 mph, this would take about 40 seconds. The power for this case would be about 19,000 watts. Ok. I guess the stratos jump isn't too bad. Yes, it is more - but not way out of this range. So, maybe the jumper will get a little hotter - but he does have a space suit on.

One - one more thing

I asked the Red Bull Space jump guys for acceleration data when Felix actually does the jump, but I never officially heard back from them. Red Bull, if you read this - please?

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Google tells me that a joule is 0.239 calories, and one calorie will heat one gram of liquid water by 1 K. So if half of that energy goes to the jumper--lets call it 10^7 J for a 100 kg man. That's 2.39*10^6 calories going into 10^5 g of jumper, raising his average temperature from 37 C to 61 C if he's not wearing a space suit. (In reality, the leading edge of his body could get quite a bit hotter.) So he'd better be wearing that space suit, or else he will be toast.

By Eric Lund (not verified) on 19 May 2010 #permalink

How about a "space dive" from an orbiting vehicle? Say from the ISS.

At a mere 40 km, is the air really thin enough for a human body to hit Mach One? I would expect the wave drag to put a pretty serious limit on velocity at that point.

Secondly, I suspect that your 50/50 distribution of energy is pretty far off. First, for the near-sonic portions of the fall the compression heating of the air would be significant. Secondly, although the energy of impact between air molecules and the body would be distributed more or less evenly, there is also heat transfer from the body to the air -- and that air is cold. What's the wind chill of -40 C air at 300 m/s?

By D. C. Sessions (not verified) on 19 May 2010 #permalink

@D.C. Sessions,

I agree that maybe the 50/50 thing is a bit off. Look at a normal sky diver, they don't really get hot. It was just my first guess.

Re: Orbital Jump(#2)

Jumping from orbit is a little more complicated. You are moving at least 15,000 mph parallel to the ground. You will need a small rocket to change your orbit enough to intersect the atmosphere. When you hit the atmosphere, you will burn up unless you have a heat shield or tiles(i.e. shuttle) to convert all that velocity to heat that doesn't kill you.

An alternative would be a bigger rocket to dump all your orbital velocity, discard the rocket, and then fall with zero velocity relative to the ground. Would a small 3rd stage solid rocket give enough delta-V to kill your orbital velocity? Add in the control system also, unless manual control would be part of the challenge :-)


It was just my first guess.

I would expect it to be a very good one from first principles. The problem is that there are other heat transfer processes which seem likely to dominate.

The good news is that it shouldn't be all that hard to calculate the loss rates. Fun assignment for students.

By D. C. Sessions (not verified) on 19 May 2010 #permalink

here is an article i just stumbled across today:

American Journal of Physics

"High-altitude free fall revised"

Jan Benackaa
Faculty of Natural Sciences, Constantine the Philosopher University, Tr. A Hlinku 1, SK-94974 Nitra,Slovakia
Received 24 September 2009; accepted 4 January 2010


Thanks for the article - I am reading it now. How do I miss these things?

May I ask what software you use to create the equation images?