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So basically astronauts are weightless b/c they are free falling. But to get to that state they need to be moving real fast. So why doesn't that speed along with the centripetal force affect them?

I get that we don't really feel speed just acceleration but i'm thinking when you're in a car and go around a curve you feel that pull, why don't astronauts feel that if they are moving real fast and going around. Sort of how pilots feel all those G's when they're in a centrifuge.

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  • $\begingroup$ A man takes a box of 40 kg and jumps from 4th floor -ask him does he feel the load during his journey and he will say 'no'. A free fall does not give a feeling of any force acting on him. If somebody is enclosed in a lift he will also experience the same thing during uniform motion.Looking at the perspective of man in the satellite he is always in uniform motion and the landmarks like other heavenly bodies are moving around him, so no feeling of motion or change of velocity. $\endgroup$
    – drvrm
    Apr 10, 2016 at 7:56
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    $\begingroup$ If an astronaut were zipping around in a circle the size of a typical earth orbit at 18,000 miles per hour but there were no gravity present, then they would feel exactly one "G", just like in a centrifuge. Google did the calculation for me: $((18 000 (mi / hr))^2) / (4100 mi) = 9.81307317 m / s^2$ $\endgroup$
    – user55515
    Apr 10, 2016 at 12:23

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It is acceleration, change in direction in this case, which makes the difference as does the fact that gravitational attraction is a non-contact force.

In space away from any large mass going at 1000m/hr in a straight line does not require a force to be acting on you.

When you go around a corner it is a localised contact force that provides your centripetal acceleration. Your body "feels" the effect/position of that localised force.
For example you feel the effect of the seat and seat-belt in the car pulling on you and the forces involved are contact forces.

When orbiting the gravitation attraction of the Earth provides the force for both you and the spacecraft of just the right amount so that no contact forces are needed for you to go "around the corner". So you do not directly feel by contact the force which is causing your centripetal acceleration.

The idea of having artificial gravity in a space station by rotating the space station would work because you would need an extra force to rotate with the space station and that would be because there would have to be a contact force between you and the space station, a force from the space station which pushes you towards the centre of rotation and you would feel it as an identifiable localised force.

As for the g-forces. In a space station even though the speed is very large so is the radius and thus the accelerations involved are less than 1$\times g$ and vary hardly at all across the expanse of the space station.

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  • $\begingroup$ Why the -1 for my answer? $\endgroup$
    – Farcher
    Apr 30, 2023 at 22:58
  • $\begingroup$ You said "you do not directly feel by contact the force which is causing your centripetal acceleration", which agrees with the comment physics.stackexchange.com/questions/248581/… by user55515. But your final paragraph appears to contradict that, and implies that the acceleration in the space station is small. Perhaps you could clarify that. $\endgroup$
    – PM 2Ring
    May 1, 2023 at 11:50
  • $\begingroup$ @RM2Ring When in "free fall" there is no contact force between you and the space station and you do not "feel" the gravitational force of attraction. In a rotating space station there are contact forces between you and the "outer" wall of the space station which "push" you towards the centre of rotation. You feel the effect of that contact force just as you feel the effect of gravitational attraction when you are walking or seated on Earth. $\endgroup$
    – Farcher
    May 1, 2023 at 14:12
  • $\begingroup$ I agree. But as user55515 said, if you were in free space (far from a gravity source) executing a circular trajectory then you would feel the acceleration. $\endgroup$
    – PM 2Ring
    May 1, 2023 at 14:19

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