Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

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?

share|improve this question
1  
Just thinking it through here, aren't they two different things? The airplane fighting gravity via an upward thrust on the wings and an orbiting object falling with gravity, not fighting it? –  Aeo Jul 27 '11 at 16:36
    
@Aeo: you're correct, and in fact that line of reasoning could make a good answer if you expand on it a bit. –  David Z Jul 27 '11 at 18:59

2 Answers 2

up vote 2 down vote accepted

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.

share|improve this answer
1  
I think you mean centrifugal. the centripetal (radially inward directed) acceleration is gravity –  luksen Jul 27 '11 at 16:43
2  
No I meant centripetal. The centripetal acceleration just happens to match gravity at 17000 mph. At any other speed the centripetal acceleration is the difference between gravity and lift. –  user1631 Jul 27 '11 at 16:49
    
Luksen is right. First there's the ambiguity of whether user1631 us reasoning in the rotating frame or the nonrotating frame. In the nonrotating frame, there is a centripetal acceleration, which is caused by gravity; it's not that it "just happens" to match gravity. In the rotating frame, we have both a centriptal force and a centrifugal force, and it's the centrifugal force that "just happens" to cancel the centripetal force of gravity. –  Ben Crowell Jul 27 '11 at 17:09
1  
Sigh. An object moving in a circle has a CENTRIPETAL acceleration of $v^2/R$. In the rotating frame, the object is not moving and has a centripetal acceleration of ZERO. –  user1631 Jul 27 '11 at 17:19
    
xkcd.com/123 –  user1631 Jul 27 '11 at 17:40

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.