Q:

I am currently teaching 8th graders dropped object problems in Algebra 1. We were discussing Felix Baumgartner's step into space 128,100 feet above the earth last October. We could not figure out how he could freefall for four minutes before deploying the parachute. If we apply the formula we had been using for dropped objects h=-16t^2 + s, he would have crashed in 89 seconds. What are we ignoring? The students thought the thinner atmosphere would make him go faster, not slower. Gravitational pull less the higher you go? I'd appreciate your help with this one.

- Susan Huffer (age 54)

Indianapolis, IN 46260

- Susan Huffer (age 54)

Indianapolis, IN 46260

A:

Funny you should ask- some news organization had asked me about this event at the time, and I'd done some calculations on it- all forgotten now, of course. ()

Nevertheless, here's the basics. The height wasn't very large compared to the radius of the Earth, so the gravitational acceleration was almost as large as the sea-level value of 9.8 m/s^{2}, equivalent to the value you used. It's also true that the thin atmosphere at great heights doesn't make as much drag as the thick atmosphere near the ground. So the terminal velocity (the velocity at which the drag just cancels the gravity, so acceleration ceases) is larger higher up. Still, he fell into atmosphere that was dense enough for the drag to be very significant. Therefore his acceleration wasn't given by the simple gravitational value. In fact, as he fell into denser atmosphere the drag got bigger than the gravity force, because he'd built up speed in the thinner atmosphere. So his speed actually started to go back down before he opened the parachute.

Mike W.

Nevertheless, here's the basics. The height wasn't very large compared to the radius of the Earth, so the gravitational acceleration was almost as large as the sea-level value of 9.8 m/s

Mike W.

*(published on 04/17/2013)*