This Joseph Kittinger article is pretty fascinating on the NASA skydives from the edge of space.
The wiki article is interesting as well, and has a video link.
But imagine being the unlucky dude who gets walloped by the pillow from space.
Yes, there is a real difference. Due to the gas laws, a more-compressed gas has a higher temperature than a less-compressed gas. This has nothing to do with friction. It is because the gas contains a fixed amount of heat, and when that heat has to fit into a smaller volume, it is measured as higher temperature. This is the principle that makes heat pumps, air conditioning, and vapocoolant sprays to treat injuries possible.
It is the motion itself, not the bumping of gas molecules, that is measured as temperature.
The misconception is that the heat of re-entry is caused by air molecules rubbing against the surface of the projectile, but it is actually the compression of the air on the front (“front” meaning relative to the direction of travel) of the projectile.
The article I read starts with Kittinger and then talks to people who want to push his record out to 80km. They suggest using John Carmack’s Armadillo Aerospace rocket to launch a naked (ie spacesuited but no enclosure or cockpit) skydiver to 80km. At the top, the diver just steps off the rocket, and the rocket and diver descend separately. :eek:
The ideal is an emergency reentry system from low earth orbit, but anything in orbit needs to handle re-entry heat, unlike a suborbital diver who has minimal heat to deal with.
Si
/trying to picture it…
how about a floating object on the path of the Earth’s orbit? is it like dropping something onto a propeller?
what causes the speed increase? is it the act of stepping out of the atmosphere? what determines the maximum height for the skydiver not to burn up coming down?
There ain’t no such animal. It’s either falling towards the Sun (or heading out from the Sun) on an elliptical orbit like a comet’s, or it’s orbiting the Sun at the same orbital speed as the Earth - 'cos any object with any mass whatever can only occupy a given orbit at a given speed, which at 1 AU is the figure quoted.
If you want to imagine something suddenly coming into existence just as the Earth arrives, and “stationary” in that frame of reference in which the Earth is orbiting the Sun at 66,000 mph, then yes, the Earth will smoosh into it at that very hearty speed, which will be much the same as re-entry from low orbit, only at an even higher speed - and with nearly 16 times the kinetic energy to lose.
It enters the atmosphere at the speed of the earths orbital velocity (110k km/h) , plus some additional gravitational acceleration, and burns up across the sky. Happens several times every year when we pass through comet tails, like the Persids and Leonids meteor showers - but see Malacandra’s post on the fact that there is nothing actually “floating” - but a comet fragment could have a velocity vector almost 90 degrees to the earth’s velocity vector, so the earth smacks into it at almost full speed.
Gravity 
Once you are in freefall your speed will accelerate at 9.8m/s every second. In a vacuum this will continue until you splat on the surface of the object you are falling towards. But the atmosphere supplies frictional resistance that slows you down. And the density of the atmosphere decreases as you get higher. So for the 80km diver, they will reach some terminal velocity without too much friction (as the friction is pretty low for a long time). As the density increases, the terminal velocity decreases - there is a balance between the frictional resistance and the gravitational acceleration that stops the diver speeding up too much, and the increasing air density slos the diver down slowly, so the heating effects can be coped with.
I think that the article suggested that 80-100km was pretty much the limit for a suborbital dive without major heatshielding.
Si
I was bored:
We can do a massively simplified thought experiment in which the rotation of the Earth is abstracted away - imagine the Earth is the only thing in the universe, is not rotating, and that we’re gently prodding a pillow towards Earth from an initial distance of infinity. Let’s make it a kingsize, soft filled, hypoallergenic pillow; there’s going to be a lot of napping to be done before we reach Earth, and with infinity to travel over, the probability of anything occurring during our journey approaches one - and that includes hyperallergenic Space Cats.
Anyway; that’s an 18oz pillow, or 0.5kg (for simplicity’s sake). At infinity, our pillow’s gravitational potential energy is zero. Let’s assume the atmosphere starts at 100km from the surface (this includes 99.99997% of the atmosphere, which seems sufficient). The gravitational potential energy at this point is:
U = -GMm/r = ( 6.67*10[sup]-11[/sup] * 5.9742 × 10[sup]24[/sup] * 0.5 ) / 6,378,260
where G is the gravitational constant, M is the mass of the Earth, m that of the pillow, and r the distance from the centre of the earth. Thus our pillow’s change in potential energy is:
delta_U = -31,237,292 joules
Assuming that all of this energy has been converted into kinetic energy, and not waylaid by Space Cats, this gives us
0.5mv[sup]2[/sup] = 31,237,292
=> v = 2 * sqrt( 31.2*10[sup]6[/sup] )
= 11.2 * 10[sup]3[/sup]
So our pillow is travelling at about 11km/s as it enters the lower thermosphere. More significantly, perhaps, it’s carrying some 31 megajoules of energy that must be dissipated in a non-conflagratory manner before it nestles to earth in a manner that supports crucial vertebrae while suppressing the causes of snoring. It only picks up another 500 or so kilojoules from here to the ground, so let’s ignore that.
Now, pretty much all of this energy has to be dissipated within a distance of 100km (assuming our pillow isn’t coming in obliquely), a distance which at its present velocity it will cover in a little less than 10 seconds (obviously the atmosphere will have something to say about this). Let’s assume that this energy is dissipated at a constant rate (an outlandish assumption, but generous to the pillow in that it’s the maximum energy dissipation rate that will determine its fate, so linear dissipation gives the best chance of survival).
Let’s further assume that the pillow lands at its terminal velocity for sea level; assuming a cross section of 20"x5" (from the pillow website), a C[sub]D[/sub] of 2.1 (for the sake of argument) and density 1.225kg/m[sup]3[/sup], I get a V[sub]t[/sub] of 7.6m/s, giving a kinetic energy of 14 joules, which I will completely ignore because it’s so tiny relative to our starting point.
Now, how long will it take to reach the ground with these assumptions? Well, our kinetic energy is decreasing linearly, so our velocity is decreasing as the square root of time.
v(t) = 11.2 * 10[sup]3[/sup] + sqrt( 2 * delta_E(t) / m )
where delta_E is the decrease in kinetic energy at time t. But we “know” (read: have assumed) that the rate of energy decrease is linear:
delta_E(t) = - ( 31.2 * 10[sup]6[/sup] / t[sub]L[/sub] ) * t
where t[sub]L[/sub] is the time of landing. So now we have:
v(t) = 11.2 * 10[sup]3[/sup] - sqrt( 2 * 31.2 * 10[sup]6[/sup] * t / ( m * t[sub]L[/sub] )
Integrating this with respect to time from zero to t[sub]L[/sub] will give us the distance travelled by time t[sub]L[/sub], which is 100km. Then simply solve for t[sub]L[/sub]. I get 27s (this post is already long enough without me showing all those workings).
So our approximately 30MJ have been dissipated in approximately 30s. That means our pillow is shedding kinetic energy at a rate of about 1MW (in practice, the dissipation will not have been linear, meaning the actual maximum figure will be higher). Obviously a lot of that will end up heading into the atmosphere rather than the pillow itself, but it still sounds like an awful lot for a humble pillow to absorb in the form of heat; even a pillow that has survived the lasers of intergalactic cats. If our pillow were made of water, 1MW would be sufficient to raise its temperature by nearly 500K every second. Here we also have to take into account the insulating properties of down - the heat won’t be able to escape from the pillow’s surface very quickly at all, leading to a very rapid increase in surface temperature. While our pillow promises to be flame-retardant, I’m not sure that the Allergy Buyers’ Club have subjected their pillows to heat flows in the megawatt region.
In summary, our pillow is doomed.
All rather silly, I realise, and in particular ignoring the angle of re-entry renders the whole exercise rather academic, but still, it saved me doing some ridiculously involved integrations. 
Would a pillow burn up on reentry? My eyes glazed over several posts ago, because algebraic formulae that explain things I don’t understand are…kinda…thingy. However, I feel sure that feather filled pillows would not burn up. They’d burn down. 
To be precise/anal, there are two things wrong with that premise:
- If the Shuttle is at its normal orbital height (about 300 km) but not moving, it would fall towards the ground like a rock.
- A basketball wouldn’t survive in vacuum. It’d burst before you’ve even finished depressurizing your airlock.
But if you had a 300-km high tower and dropped something from the top, it wouldn’t need much shielding to survive. The amount of heat generated upon reentry would be 1/10 that of an identical object reentering from low-earth orbit.
And “sounding rockets” do just that - go straight up, and come straight back down. I’ve seen instruments for sounding rocket flights and the heat shield is pretty simple, I think it’s just a metal plate on one end.
You really want to be on the SDSAB, don’t you?
No, I just write like that. Bored, y’see. 
The alternative was marking a whooole lot of newbie programmers’ first attempts at logic programming…
So semi-flatten it before you toss it out - if it’s about four pounds below atmospheric in your space shuttle, that’s about the same as ten pounds over atmospheric once it’s in vacuum. Also, don’t gently toss it - put it in the mass driver and fire it out backwards at 17,000mph relative to the Shuttle. Then, at last, we get a basketball at low-orbit height and dropping more or less vertically.
Larry Niven, The Integral Trees: “East takes you out, out takes you west, west takes you in, in takes you east. North and south bring you back.”
This staff report I wrote might also be of interest.
Your report is very informative, but it doesn’t answer the obvious next question of what happens if you do manage to accelerate the baseball backwards by about 7600 meters per second. Does it necessarily burn up when falling essentially straight down, or does terminal velocity limit it so that doesn’t happen? One possibility is that terminal velocity of a baseball is still fast enough to generate a small amount of heat which, applied for long enough, will cause the ball to burn up.
I also don’t know how the issue of a changing terminal velocity based on altitude comes into play: up high, it can obviously get going faster than the lower-altitude V[sub]t[/sub], but since V[sub]t[/sub] changes continuously, the ball is also constantly decelerating towards the new lower V[sub]t[/sub] and never exceeds it by much. In fact, this makes the general concept of “terminal velocity” seem a little broken when considered on a macro scale: if V[sub]t[/sub] is continuously dropping with respect to altitude, a falling object will never actually reach it, unless I’m missing something.
Conservation of momentum. In a geostationary orbit, you’re moving really fast in order to stay in the same place above the rotating Earth. If you try to move to a lower point, but remain above the same place, you’re actually trying to move to an orbit with a smaller radius (and therefore circumference) - smaller distance to travel around the Earth in one 24-hour period means you have to slow down - so in practical terms, trying to move straight down from a geostationary orbit would make you veer off in the direction of rotation.
Two easy ways to grasp this:
-A ballet dancer performing a pirouette with arms outstretched spins faster when she draws her arms in.
-Find a children’s playground with a roundabout of this general design - spin the roundabout and get aboard - stand facing the central pole and try to kick it - your leg will veer off to the side.
From the ISS or Space Shuttle, the basketball will burn up - it will have enough velocity by the time atmospheric braking starts to kick in to fry. From 31km, it won’t. From 80km - maybe, maybe not. If a human can survive the drop from there in a pressure suit, then a basketball will probably be ok (if it does not burst from internal pressure). But the pressure suit may have some cooling, so the naked basketball may need to be dropped from somewhat lower to ensure survival
Terminal velocity is a concept relating to an equilibrium - the point where atmospheric drag is the same as the gravitational force so velocity is not changing. It varies with a large number of factors. Re-entry calculations don’t really involve terminal velocity, as the reentry does not reach equilibrium until late in the process and deep in the atmosphere. The elements of the math are the same - altitude, density, drag coefficients, gravity, but also has elements of initial velocity, angle etc. It is a moderately complex exercise in calculus - but one that has been successfully solved on a large number of occasions. I can’t think of many re-entry failures involving objects burning up (that don’t involve additional issues contributing to the failure). And some of those re-entries involved atmospheres that we don’t have direct experience of.
Si
Ok, then: same question, but for a pillow. Will the higher drag relative to its mass make much of a difference, as suggested in post 5? If not, would a more extreme high-drag / low-mass example work? Enormous super-light wiffle ball?
I guess one important question is just how much speed an object in freefall from ISS altitude gains before atmospheric braking starts. Presumably, if there’s any atmospheric braking at all, an object with a large enough amount of drag can prevent itself from accelerating any more and even slow itself e.g. with a large enough parachute (maybe the parachute required would be impractically large, though).
I’m skeptical of an astronaut being able to survive re-entry in a spacesuit.
Anyone ever hear of SuitSat I?
It was a Russian spacesuit that was released from the ISS on February 3, 2006, carrying an amateur radio transmitter. The transmitter was logged from Earth stations for about 2 weeks.
The suit finally entered the atmosphere September 7, 2006, and burned up upon re-entry.
Suitsat II in now in development! When it’s launched, you will be able to listen and hear the transmission using a 2-meter amateur radio or a police scanner that can receive the 144-148 MHz band.
Ah, yes. James T. Kirk did this in the deleted opening scene of Star Trek Generations, and Torres did it on the holodeck in Voyager. Sounds like fun!