Weightlessness and falling sideways In Season 2, ep 4 of Star talk Radio, host Neil Tyson is talking about the international space station. He mentions that the space station is in a low earth orbit and is falling along the curvature of the earth due to the gravitational pull of the earth. This falling along the curvature of the earth allows the astronauts to still experience weightlessness.(Around minute 12 of the episode). I’m wondering how this works from a physics stand point as I’m having a hard time wrapping my mind around this.
Couldn’t you say that a flight on the ISS is the same as a flight on an airplane just at a much greater speed? Since the earth’s gravitational pull is roughly equal on the plane as it is to the ISS, why don’t passengers on airplanes feel the same effect? 
 A: You are correct, if airplanes only flew a bit faster (about 17,000 mph) then they would be weightless.  At that speed the centripetal acceleration would match the gravitational acceleration. (Let's neglect the extreme effects of atmospheric turbulence at those speeds).
Centripetal acceleration goes as $v^2/R$, so for a plane at say 500 mph, the centripetal acceleration is about $(500/17,000)^2$ or 1/1000 G's, which is why passengers don't notice it.
A: Passengers on an airplane do feel the same effect if the airplane is free-falling: http://en.wikipedia.org/wiki/Vomit_Comet The speed is irrelevant as long as both the vehicle and its occupants are in free fall. Similarly, if you're in an elevator and the cable breaks, you will experience apparent weightlessness (until it hits the ground -- ouch!).
The free-falling ISS, the free-falling Vomit Comet, and the free-falling elevator are all in orbit around the earth. The only difference is that in the case of the Vomit Comet and the elevator, their orbits intersect the earth's surface.
