Suppose I'm in a rotating space station (that is somewhere in free space) and there is no other force. Now how am I supposed to fall to the circumference of the station if nothing pulls me? I will certainly be at the same point at all the times since there is no friction, neither attractive nor repulsive forces. Then how would the artificial gravity work?


2 Answers 2


How am I supposed to fall to the circumference of the station if nothing pulls me.

You won't fall if you have no velocity relative to an inertial reference frame at rest with the centre of rotation of the space station, as for example A or B in the animation below, so your initial intuition is correct. A and B will remain where they are relative to the inertial reference frame, whether the station is rotating or not.

enter image description here

In the rotating reference frame (e.g C') object B appears to be orbiting internally, but accelerometers on B would confirm it is completely inertial as is not experiencing centripetal force. C on the hand is located on the rim of the station and has velocity relative to A and B due to friction from the rim and also experiences a centripetal force that keeps it moving in a circle, and it will feel as if it is pulled down to the inner surface due to centrifugal force.

But why should I fall back...if I'm just in the vacuum part of station after the jump?

If a person jumps 'upward' then they have upward velocity due to the vertical jump and an additional 'horizontal' velocity component, due to the momentum they had when standing on the rim. Once they leave the rim, they are completely inertial and travel in a diagonal straight line (e.g D) as seen by inertial observers A and B and their momentum carries them across the vacuum until they hit the rim again. From the point of view of the rotating reference frame, they follow an arc (e.g. D') that looks like they have gone up and returned back to the surface, a bit like the trajectory of a ball thrown in the air on the surface of the Earth.

Obviously, this artificial gravity has some differences to real gravity and in fact looks like an inverted gravity field. For example, object B' is like an externally orbiting satellite around a gravitational body like the Earth, which travels in a circle while feeling no proper acceleration, except in this case the orbit is internal. Another difference is that objects can only internally orbit in one direction.

Here is the link for the interactive animation app. The initial angle and speed of the particle leaving the rim can be changed by sliders at the bottom.

  • $\begingroup$ This was a great explaination.But y do some stuff do go to the rim as i hv seen in case of centrifuge?i know there are other forces too,but i like 2 be taught the same way u taught me above $\endgroup$ Commented Feb 14 at 6:30
  • $\begingroup$ If a particle is stationary in the the inertial frame (Like point B) and any motion at all is imparted to it it's trajectory inevitably leads it to the rim (like (D). In a centrifuge with for example test tubes, the sides of the test tubes impart 'horizontal' motion. If the centrifuge is full of liquid or gas, the particles at the rim that have to circulate due to friction with the rim, drag other particles further inside the centrifuge and eventually impart motion to any stationary particles inside the cavity. There is no force acting on isolated particles in the cavity. $\endgroup$
    – KDP
    Commented Feb 14 at 12:40
  • $\begingroup$ I have added the animation app link to the end of the answer so you can try it out for yourself. See if you make a love heart trajectory for Valentine's day! :-) $\endgroup$
    – KDP
    Commented Feb 14 at 14:12
  • $\begingroup$ If you can find a playground with a "merry-go-round" and a couple of friends, I recommend trying to toss and catch a ball while the merry-go-round is spinning. Trajectories to the people next to you are pretty normal, but trying to toss across the central axis is quite remarkable in the rotating frame. $\endgroup$
    – rob
    Commented Feb 14 at 15:15

How am I supposed to fall to the circumference of the station if nothing pulls me ?

If the space station is rotating then nothing needs to pull you. In an inertial (non-rotating) reference frame there are no forces acting on you so you travel at a constant speed in a straight line. In the rotating reference frame of the space station this straight line trajectory looks like a curve that takes you away from the centre of the space station and towards its circumference.

Of course, this begs the question of what distinguishes an inertial reference frame in outer space. If the space station is the only object for, say, thousands of light years how do we know that the space station is rotating ? This is when we apply Mach's principle.

  • 1
    $\begingroup$ You know because objects in contact with the wall experience a normal force as if there is radial gravity. $\endgroup$ Commented Feb 13 at 12:21
  • $\begingroup$ @ChetMiller Yes, but how does anything in the space station "know" that the space station is rotating, especially if the nearest outside objects are many, many light years away. And how is this different from supposing that the space station is the only thing in the universe ? That is the problem that Mach's principle (which comes in various forms) is intended to address. $\endgroup$
    – gandalf61
    Commented Feb 13 at 14:21
  • $\begingroup$ @gandalf61 if i jump,would i fall back to the edge of the space station similar to the situation of gravity? $\endgroup$ Commented Feb 13 at 14:23
  • $\begingroup$ @SuhailSarwar Yes, similar to gravity, but not identical as your path in the space station's reference frame would be curved by Coriolis force - see en.wikipedia.org/wiki/Coriolis_force. $\endgroup$
    – gandalf61
    Commented Feb 13 at 14:37
  • $\begingroup$ But why should i fall back...if i m jus in the vaccum part of station after the jump? $\endgroup$ Commented Feb 13 at 15:30

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