> 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.
Comments
> 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:
http://arachnoid.com/sage/terminal_velocity.html
But this equation, only an approximation at lower altitudes, is of no use at all for a problem like the Baumgartner jump.
The air pressure is proportional to either v or v² depending on the speed and atmosphere.
For a friction force Fr, the terminal velocity is reached when the acceleration becomes null:
mg + Fr = 0
mg - k* Vterm² = 0
k, the friction coefficient depends on different parameters: http://en.wikipedia.org/wiki/Drag_equation
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.
http://en.wikipedia.org/wiki/Drag_(physics)
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.
Yes I did mean that, thanks for correcting me. I was never very good at fluid dynamics :)