How much will wind affect the Red Bull Stratos Jump? Here is a quick refresher on the space jump details (in case you haven't been paying attention).
- Felix Baumgartner will get in a capsule attached to a balloon (with life support and stuff).
- The balloon will carry him up to an altitude of 120,000 feet.
- He then jumps out.
I have previously modeled the motion of a skydiver from that extreme height. How do you do this? If you consider a jumper falling straight down in a no-wind situation, you would have this force diagram.
So, we are dealing with two forces during this fall. First, the gravitational force. Even at 120,000 feet, it isn't a terrible approximation to say that the gravitational force is:
Where g is the gravitational field with a magnitude of 9.8 N/kg and pointing towards the ground (it is just about 1% less than the universal model for gravity - you know, the 1/r2 version). So, I will just say this gravitational force is constant.
The air resistance force is a little bit more complicated. Here, I will use this model.
Although you may have seen this before, let me point out all the details.
- ρ is the density of the air. This will clearly change with altitude.
- A is the cross sectional area and C is the drag coefficient that depends on the shape of the jumper. I will estimate both of these values based on the terminal speed of a normal skydiver. Also, C could probably change with super high speeds, but I will ignore that aspect.
- v - this is the velocity of the jumper. But really, this is the velocity of the jumper with respect to the air. If the air is moving, we call this wind.
- If you are wondering about that last v with the pointy hat on it, we call that "v-hat", get it? It is just a unitless vector in the direction of the velocity. This will make the air force also a vector.
Now what about this "velocity with respect to the air?" Let me draw another diagram for the case of a falling person with a horizontal wind.
I know this looks confusing, so let me explain. There are three velocities that are important.
- The velocity of the jumper with respect to the ground (labeled jg). This is needed to find out how far horizontally (and vertically) the jumper moves.
- The velocity of the air with respect to the ground (labeled ag) - yes, the wind.
- The velocity of the jumper with respect to the air (labeled ja). This is the velocity that goes into the air resistance force.
When dealing with relative velocities, I can say these three vector velocities satisfy the following:
Ok. I think I am ready for a numerical model. One more reminder of the numerical model methods. First, break the problem into a whole bunch of small time steps. During each short time interval:
- Calculate the forces on the jumper. This will include determining the altitude to get the density of air and the velocity of the jumper with respect to the air - both of which are important for the air resistance force.
- Use the force from above to determine the change in momentum of the jumper and thus the momentum at the end of this time interval.
- Use the momentum from above to find the velocity and the new position of the jumper.
- Update the time and repeat.
Simple. So simple even a computer can do it.
Here is my first plot showing the horizontal position of the jumper as a function of time with a constant 5 mph horizontal wind.
Odd. I really thought there would be a greater displacement. I know that the Stratos jump practices have been put on hold before due to high winds, so I am not sure what went wrong. Maybe 5 mph wind isn't that fast. Maybe they suspend jumps not so much because of the falling part but rather because of the balloon part rising and getting out of the jump area. Maybe the winds at higher altitudes are much greater than at lower levels. Really, what do I know about wind speeds? Clearly, not much.
So, what do you do when your model doesn't give you the results you expect? Run the model for a wider range of wind speeds. Here is a plot of the displacement as a function of wind speed up to 10 m/s wind (about 20 mph).
Why is this so linear? Essentially, the jumper has enough falling time to reach the horizontal speed almost equal to the wind speed. So, faster wind means greater horizontal falling speed. Of course, with a high speed the jumper can be off from the starting position by as much as 2km - but that is the extreme case.
How about a comparison? What if the jumper started at rest with respect to the rotating Earth? How much would the displacement be in that case? I don't even really need to model this one. Let me just take the falling time of about 300 seconds. How far horizontally would the Earth's ground move in this time? Of course this depends on the location of the jump. The official launch site is in Roswell, New Mexico. This is located 33.39° above the equator. Here is a diagram of its position on the Earth.
The rotational speed of the Earth is about* once a day, this is 7.27 x 10-5 radians per day. (* don't forget the difference between sidereal and solar days - but the difference hardly matters here). To find the velocity of a point on the ground, I need to radius of the circle that point is moving in. From the diagram above, this will be:
Using the radius of the Earth (6.38 x 106 m) and the latitude of Roswell, this gives an distance of 5.33 x 106 meters. The velocity of the ground will then be:
Putting in values from above, I get a speed of 387 m/s. So, in 300 seconds the ground will move 116 km (72 miles). Crazy, right? but remember in a whole day, this point on the ground has to go ALL THE WAY around the Earth. At this latitude, this is a path length of 20,000 miles.
So, why won't the jumper (Felix) be displaced by 70 miles when he jumps? Simple. He starts his jump with a velocity of about zero m/s relative to the ground. Yes, since he is higher up, he will have a different linear velocity than the ground - but the difference is super small.
Homework
What about the centrifugal and Coriolis forces? How much will these change the motion of a jumper from 120,000 feet?
