Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

When a shuttle is in orbit, it is essentially rotating around the "centre" of the Earth at a great speed. So why does there seem to be no centrifugal force sticking them to the 'ceiling' of the shuttle?

Is it because the shuttle is not actually being rotated around a point but rather continuously falling or something else?


share|cite|improve this question
up vote 14 down vote accepted

By definition an orbit occurs when gravity balances with the "centrifugal" force. It is essentially a free fall situation.

So the answer is the same reason why you don't get stuck to the ceiling of a free falling elevator. Both the spacecraft and the occupants are moving in-sync.

share|cite|improve this answer
Thanks for your answer. I wondered about the "in-sync" thing myself but if you have a cup of water suspended from some string, and you spin it around, the water will stay in the cup, but isn't the water and the cup in-sync as well? – zzz Jul 23 '11 at 0:17
@zzz no because the string acts on the cup only, and not on the water. In order to remain in sync the force from the string has to pass to the water via the cup. In a spacecraft, everything is attached to the string (gravity) and so no interaction is needed between the container and the occupant. – ja72 Jul 23 '11 at 0:56
That makes perfect sense, thank you. – zzz Jul 23 '11 at 9:55

Actually, the astronaut would only float completely free in the middle of the space station. Elsewhere, he will stick slightly to whatever side is closer to the Earth than is the middle, or farther from the Earth than is the middle. The reason is the tidal force from the Earth, which will be very small but probably detectable. If the acceleration from gravity is $g = GM/r^2$, then the tidal acceleration will be the derivative of this, namely $dg/dr = -2GM/r^3$. so the maximum tidal acceleration that the astronaut will feel, at distance $D/2$ from the center of the space station, where $D$ is the linear size of the space station, will be $GMD/r^3$. This is basically equal to $gD/r$. Putting in some typical numbers, a 100 kg by weight astronaut in a space station of size $D = 10 m$, and $r = 6500 km$, we get a maximum tidal force of about $150$ milligrams weight!

The situation would of course be completely different in orbit around a neutron star, where an astronaut would float free at the center of a (very strong) space station, but would get crushed beyond recognition were they to venture away from the middle, as has been pointed out in various science fiction stories. The reason is that $M/r^3$ would be about $10^{14}$ times greater than for an orbit around the Earth.

share|cite|improve this answer

The Shuttle (or the ISS for a better example since the Shuttle isn't flying any more) isn't rotating at a great's rotating about once every 90 minutes, once every orbit. This does have an effect, but a very small one.

Its motion around the Earth doesn't produce any such effect because, as you guessed, it and everything in it are in free fall around Earth and are affected nearly equally by Earth's gravity (the very small differences due to different distances from Earth do produce tidal forces).

share|cite|improve this answer

protected by Qmechanic Jan 3 '13 at 16:26

Thank you for your interest in this question. Because it has attracted low-quality or spam answers that had to be removed, posting an answer now requires 10 reputation on this site (the association bonus does not count).

Would you like to answer one of these unanswered questions instead?

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