Let's say there is a fan spinning and stops in exactly 1 minute on earth. Would it stop faster or slower or exactly same time in a spacecraft without gravity but exact same density of air. Btw: let's assume friction on the axis is zero. Just trying to understand if gravity has any effect of stopping a symmetrically spinning object like a perfect wheel or fan.

EDIT: Because air is complicating things let's just assume we test this experiment in vacuum, on earth and in space. No air involved. Friction on axis is neglected.

EDIT2: Why would it stop without any air and axis friction? That's right. Let's put air back in. Without the air the question doesn't make any sense.

  • $\begingroup$ One affecting force might be gravity (tidal drag/locking) dependent on axis, rotational alignment, and frame of reference for 'stopping', which leads to 'it depends'. Another might be vertical pressure variation under gravity leading to differing aerodynamic effects at different points on the fan that affect overall drag, but that also depends (look at the Power Curve for aircraft for an example of how the curves for aerodynamics can be complex). $\endgroup$
    – Danikov
    Sep 18, 2017 at 13:59
  • $\begingroup$ Is the fan's centre of mass at the axis? $\endgroup$
    – JiK
    Sep 18, 2017 at 15:33
  • 4
    $\begingroup$ If there's no air friction, and no axis friction, why would it stop? (That is to say, the effect of gravity on stopping may depend on what's doing the stopping.) $\endgroup$
    – R.M.
    Sep 18, 2017 at 15:56
  • $\begingroup$ The effect of gravity on the bearings (i.e. axis friction in your words) would be likely to be much greater than any other effect it could have (most fans are unbalanced with a horizontal axle so the masds of the blades loads the bearings). $\endgroup$
    – Chris H
    Sep 18, 2017 at 16:02
  • $\begingroup$ To answer your second edit it would stop due to friction on its axis, so unless you can make it spin in a levitation state you will have frictions other than air $\endgroup$
    – Rolexel
    Sep 19, 2017 at 7:10

3 Answers 3


Let's make your fan really big - say, as big as the moon.

You probably know the rotation of the moon is tidally locked to the earth - that is why we always see the same "face" of the moon.

The tidal friction is a real effect - it depends on the size of the object and the distance to the source of gravitational attraction, but it will produce a small decelerating torque on an object in a diverging gravitational field.

So your fan, in a spacecraft away from all gravity, will decelerate more slowly. Of course the difference will be incredibly small - for an object that loses all angular momentum in one minute there is no chance you could measure the effect. But that doesn't mean there isn't any.

There will be many other effects that would dominate the repeatability of your experiment before this effect comes into play - the unrealistically zero friction bearings, air currents in your space craft, thermal expansion of the fan blades, the pressure gradient in the gravitational environment, differential pressure on the fan blades due to solar wind particles, ... But if you start with a fan that will spin down in one minute, none of these other effects will come into play.


It is worth noting that quadratic drag (the normal form of drag force in air for macroscopic objects with Reynolds numbers above 1000) will never lead to an object stopping: the equation of motion would show velocity changing as 1/t , never reaching zero. Even Stokes drag (linear with velocity) would imply the object will never stop (exponentially decreasing velocity), but at least you can calculate the total distance it will move (integral of velocity with time is finite). But to actually stop a rotating object (zero velocity after finite time t), you need a component of force that is not dependent on velocity (like the drag in bearings). I wrote an answer relative to that recently - you might find it useful.

  • $\begingroup$ Don't forget the barometric pressure gradient, which is absent in zero-g. It might be dependent on whether the fan is upright or flat, and in which direction it is blowing. $\endgroup$
    – M.Herzkamp
    Sep 18, 2017 at 13:11
  • $\begingroup$ The amount of energy lost to tidal friction in 60 seconds for a human scale fan on Earth's surface is so small that I wonder if it's less than the Planck energy. $\endgroup$
    – EL_DON
    Sep 18, 2017 at 13:47
  • $\begingroup$ @EL_DON are you saying "it cannot be measured" or are you implying "it will lose zero energy"? Because if it loses zero in one minute, it will lose zero in one day, one year, one eternity. And that's patently false. $\endgroup$
    – Floris
    Sep 18, 2017 at 13:53
  • $\begingroup$ Tidal forces will certainly act on the fan, but will they do any net work on its rotational energy if the fan is rigid? I thought deformation of objects like planets due to tidal force was essential to the drag on rotation. $\endgroup$
    – EL_DON
    Sep 18, 2017 at 13:53
  • $\begingroup$ @Floris I'm suggesting that the drop in energy might be so low that yes, you might have trouble exciting even 1 quanta of whatever wave you're dissipating energy into (probably some vibrational mode in the fan). So, you could lose 0 energy in a minute. This would not make your answer wrong from a classical perspective, but it's fun to think about. $\endgroup$
    – EL_DON
    Sep 18, 2017 at 13:59

In ideal conditions it'll be the same.

The time to stop will depend on the fan container too, due to convection currents. And in case the fan motor is hot, convection set up by buoyancy won't take place in microgravity, so, e.g., a hot fan blowing air upwards would spin for a little longer in the presence of gravity.

But a "perfect fan" is probably at room temperature, and the ideal conditions should also include the same container in both situations. Then, as the drag depends solely on the properties of the fluid and of the moving object (size, shape, etc.), all paths through which gravity might make a difference seem to be cut (except tidal friction $-$ check Floris answer.)


The density of air is actually dependent on gravity and to an isothermal approximation, given by the Boltzmann Law. Of course, without gravity, there would be nothing to hold the air down to earth. But, in case you are able to maintain the same pattern of the density of the air, there will be no effect on the fan.

  • 3
    $\begingroup$ A spacecraft can be pressurized without gravity. Gravity maintains Earth pressure, but a sealed vessel in space doesn't have to worry about that. $\endgroup$
    – JMac
    Sep 18, 2017 at 12:20
  • $\begingroup$ I just imagine a small table fan floating free in a spacecraft. If you'd turn it on it would start to propel itself backwards while also spinning violently... :D $\endgroup$
    – Adwaenyth
    Sep 18, 2017 at 12:51

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