AFAIK all the celestial objects have a spin motion around its axis. What is the reason for this? If it must rotate by some theory, what decides it's direction and speed of rotation?

Is there any object that does not rotate about its axis?


In general yes, everything rotates. It is to do with something called angular moment. Gravity is the central force in the Universe, because it is the only one which has a significant pull over large distances. When things collapse under their own gravity in space (i.e. clouds of gas and dust), any small amount of asymmetry in the collapse will be enough start it spinning. Even if it spins by a tiny amount, as it collapses, angular momentum conservation will mean it spins more and more quickly - just like an spinning ice-skater pulling their arms into their body and spinning more quickly. This means that all coherent masses are spinning - e.g. asteroids, neutron stars, galaxies, quasars.

The Universe is a complex place so something may be slowing down (because the gravity of other objects is putting on the brakes) or some things may appear not to be rotating (e.g. the Moon rotates but at the same rate as it goes around the Earth).

Huge clouds of gas and dust tend not to be spinning as a whole because they are expanding to fill the available volume - like a bad smell in room! - and not necessarily gravitational bound together. However they might have little pockets which start are turbulent, collapse under their own gravity, spin and form stars.

  • $\begingroup$ I do not believe this answer: If the cloud is isolated and not spinning already, no amount of collapsing should make it spin: where would the momentum come from. $\endgroup$ – babou Nov 14 '14 at 21:55
  • $\begingroup$ @babou A perfectly symmetrical cloud collapsing on itself in an orderly manner likely won't spin. The chances of a perfectly symmetrical cloud existing in complete isolation from destabilizing forces is highly unlikely. Random space-rock passes by, gravity wave disturbs dust cloud, end result is a spinning body. Maybe not pulsar speed but it will rotate. $\endgroup$ – paul Nov 15 '14 at 0:00
  • $\begingroup$ @paul I disagree. Assymetry cannot create global spin. But it does create local spin. See my question. Of course, external perturbation can induce spin through angular momentum exchange. $\endgroup$ – babou Nov 15 '14 at 0:55
  • $\begingroup$ @paul - A perfectly symmetrical cloud is also highly unlikely. Look at the clouds photographed by Hubble. They are anything but perfectly symmetrical. $\endgroup$ – David Hammen Nov 15 '14 at 8:33
  • $\begingroup$ @babou - You are wrong. Asymmetry coupled with a nearby gravitating body (another gas cloud, a star, ...) can create rotation. It's called gravity gradient torque. Gas clouds are not spherical cows. They are asymmetric, lumpy things. $\endgroup$ – David Hammen Nov 15 '14 at 21:01

I'm not sure if you want an object that doesn't spin at all, or one that somehow doesn't spin on its axis. In the former, any structure large enough (e.g. superclusters of galaxies) that its dynamical time is longer than the age of the universe is effectively not rotating.

For chaotic spinning (often called tumbling), one example is Hyperion, a small moon of Saturn


All celestial objects are formed from larger, more diffuse collections of matter (such as a nebula which collapses to form a star). These larger objects typically have some very small net angular momentum (spin). That total angular momentum is conserved, and as the object collapses it causes the rate of spin to accelerate in order to maintain the same degree of angular momentum. It's the same phenomenon as the spinning ice skater pulling in their arms except in this case the amount of contraction is a factor of millions so even though a proto-stellar nebula may not be rotating much the sheer size difference between a star and a nebula will result in the star rotating a considerable amount.


Objects in orbit tend to lose their spin on their own axis.

However they do not completely lose their rotation and end up rotating with a period that is the same as the orbital period, so that they face always the same side towards the other body. The best know example is the Moon that shows always the same side to Earh.

The phenomenon is called tidal locking.

Tidal locking may happen to both the main body, and the smaller one orbiting it, but is much faster for the smaller one. Double locking is faster when the two bodies have similar mass. Double locking is observable for Pluto and Charon.

The angular momentum cannot be lost and is conserved as orbital momentum.

Of course, a lone (i.e. far from any other body) rotating body cannot transfer its momentum and will just keep rotating with the same period.

But can there be bodies that do not spin. I guess not to spin is with respect to the distant stars. I am no expert, but I do not believe that any physical law prevents that, even for smaller bodies, given that the spin of a body could be compensated by opposite spin of the larger structure it belongs to. It is probably rare, even considering approximations, i.e. extremely long periods, in particular because small structures tend to spin much faster than larger ones. But we have seen that they can be slowed down.

  • $\begingroup$ Nice. I was just on the point of adding something similar myself and then read your answer. However I was thinking of close binary stars. Anything less than a few days period is quickly locked and this has been observationally established. $\endgroup$ – Rob Jeffries Nov 14 '14 at 21:46
  • $\begingroup$ @RobJeffries Binaries are probably good candidate, as they are more fluid than rocky planets, and tidal bulges seem to play a major role in the phenomenon (but this is mostly guess). I guess their rotation is measured by Doppler shift differential (not an expert). How long is quickly? $\endgroup$ – babou Nov 14 '14 at 22:00
  • $\begingroup$ Within a million years. You can measure the orbital period, inclinations and stellar radii from eclipses and then get the projected rotation velocity from spectra. They imply the rotation periods are the same as the orbital periods. $\endgroup$ – Rob Jeffries Nov 14 '14 at 22:03

protected by Qmechanic Nov 14 '14 at 15:04

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.