Another silly question might be: won't he "burn up on reentry"? Clearly, he's falling from a different altitude than a rocket, and starting from 0, but what are the actual numbers like?
> Clearly, he's falling from a different altitude than a rocket, and starting from 0, but what are the actual numbers like?
Without an atmosphere, the result would be deterministic and trivial to compute:
acceleration = f(t) = g
velocity = ∫ f(t) dt = g * t
position = ∫∫ f(t) dt = 1/2 g t^2
(The value of g changes slightly for this problem, in this altitude domain.)
But because of the atmosphere, the calculation of velocity as a function of altitude is much more difficult. And it turns out that the atmospheric pressure as a function of altitude is not trivially characterized. And at high altitudes, it's not even constant -- it depends on temperature, the position of the sun in the sky, even the time in the 11-year sunspot cycle.
The air resistance of a falling object is some constant k (based on the object's size and surface roughness) times the square of the velocity times the air pressure. But the air pressure is changing as the descent unfolds, so such a computation ultimately relies on a numerical solution to a differential equation.
This is why one doesn't see a trivial equation describing descent velocity for a skydiver. I have worked out skydiver velocity profiles for constant air pressure nearer the surface:
So in the end, the terminal velocity will depend on:
- The guy's weight (contrary to a free-fall's acceleration which is independent of mass)
- The "contact surface" between the body and the atmosphere.
- The atmosphere's density (which is lower than on Earth's surface).
The guy aims at reaching Mach 1. Note that due to the lower atmospheric pressure at this altitude, Mach 1 is a bit smaller than it is on the Earth's surface ( 301 m/s at 29 000 meters and -48 degrees C compared to 340 m/s at sea level ).
Finally, the term "free fall" is not appropriate as the definition of a free fall is "any motion of a body where its weight is the only force acting upon it."( http://en.wikipedia.org/wiki/Free_fall ) This does not take into account the drag which is not negligible here.
> The air pressure is proportional to either v or v² depending on the speed and atmosphere.
1. I think you mean "air resistance". Yes?
2. If so, then no, air resistance transitions from (linear) Stokes drag at low velocities to (aptly named) quadratic drag at higher velocities, as a function of velocity, but not a function of air pressure. So not "either v or v^2", but a combination of the two factors.
It made my day to have my question answered by an honest-to-goodness rocket scientist. Thank you!
I think the numbers that best sum up the situation are provided by yardie and blaze33 though. I knew that stuff coming from space was obviously going to be moving faster than the guy's 0 velocity, but had no idea about the order of magnitude.
In addition to jlgreco's mention of 'sideways' momentum (like a shuttle re-entry) which this skydiver won't have, keep in mind he's falling from within the atmosphere so he's got resistance immediately and no chance to speed up to velocities that cause enough friction to burn up meteors. A meteor gets caught by Earth's gravity long before it touches Earth's atmosphere and has plenty of time to accelerate to fabulous speeds; when it hits the atmosphere, the speed makes for enough friction to burn/melt/destroy the thing.
No. Rockets are descending from orbital velocity (30,000mph) to terminal velocity (700mph). They have a tremendous amount of energy they have to get rid of as they come down which they do as heat. Baumgartner is taking a balloon up and then freefalling down. He's no where near orbital velocity so his energy expenditure will just be some air friction.
A great deal of the energy that needs to be bleed off during re-entry of a formerly orbiting spacecraft is the kinetic energy it had from basically moving sideways at an incredible speed.
