9
$\begingroup$

Our moon and Saturn's moon, Titan, have this feature. Because of this we only observe one side (hemispere) of our moon. Why is this? What is the Newtonian or astrophysics that would explain this? It seems that it would be some kind of equilibrium that is arrived at over time, but I don't understand it.

$\endgroup$
  • $\begingroup$ Youtube links aren't recommended, but it's a nice simple explanation: youtube.com/watch?v=6jUpX7J7ySo Look up Tidal Locking for more info. $\endgroup$ – userLTK Jul 6 '17 at 14:41
  • 1
    $\begingroup$ I was just wondering this yesterday, weird $\endgroup$ – MCMastery Jul 6 '17 at 19:39
  • 1
    $\begingroup$ It's not just our moon and Titan. To the best of my knowledge, every large moon in the solar system exhibits this behavior. Tidal locking is a powerful phenomenon. $\endgroup$ – Mark Foskey Jul 7 '17 at 13:44
14
$\begingroup$

The answer to this is certainly tidal forces, but that doesn't explain the exact mechanism for how tidal forces result in tidal locking, i.e., an orbiting body showing the same face to the central body as it orbits due to the rotation rate and revolution rate being equal. I'll describe this mechanism using the Earth-Moon system so I can be specific, but it equally applies for any system.

To start, tidal forces are a result of differential gravitational forces across the distributed mass of a body. The Moon is not a point mass, it has an extended size. The force of gravity on the Moon by the Earth is dependent upon distance (as is the force of gravity for anything). What this means is that on the Earth-facing side of the Moon, the force of gravity is stronger and as you progress through the Moon to the Earth-opposing side, the force of gravity gets weaker. This means the Earth-facing side of the Moon gets pulled stronger and closer to the Earth while the Earth-opposing side, while still getting pulled towards the Earth, doesn't get pulled as strongly or as close. Ultimately, this results in the Moon being deformed such that it becomes slightly oblate and stretched out in the direction of the Earth. This flexing of the surface is known as tides.

Now let's assume the Moon is not currently tidally locked with the Earth, and in fact spins slightly faster than it orbits. The Earth is causing tides on the Moon and the Moon is spinning on its axis. The tides, caused by the tidal forces, want to remain aligned with the Earth-Moon line since that's the direction the tidal forces are applied. However, it takes time and a lot of energy to deform the Moon. Once the Moon is deformed, it's going to spin along and try to pull that tidal deformity along with it, effectively moving the tidal bulge ahead of the Earth-Moon line. The Earth is still applying the tidal force along that Earth-Moon line to try and pull the tidal bulge back in line. This constant force trying to pull the tidal bulge back (or ahead if the Moon is spinning too slow) allows for a transfer of momentum to slow down the Moon (or again, speed it up if it is too slow). The main point here is that tidal locking is a equilibrium state because if the Moon is spinning too slow or too fast, the Earth trying to pull the tidal bulge into the Earth-Moon line will change the rotation rate of the Moon until it becomes tidally locked. Once it is tidally locked, that tidal bulge will always be along the Earth-Moon line and this force will disappear.

That covers about half of the answer though. When trying to tidal lock the Moon to the Earth, you have to consider two time periods. The first, discussed in the preceding paragraph, is the rotation time of the Moon around its axis. The other is the revolution time of the Moon around the Earth. Both must match. The previous paragraph described how the rotation time of the Moon around its axis could be affected, but there's also a means of affecting the revolution time of the Moon around the Earth. Fortunately, this is by a nearly identical mechanism as above. In effect, the Moon also causes tidal bulges on the Earth and since the Earth is spinning these tidal bulges won't be directly aligned with the Earth-Moon line. This unaligned tidal bulge on Earth acts to transfer energy to the Moon's orbital speed, causing it to speed up or slow down. Incidentally, through conservation of angular momentum, this necessarily causes the Moon to drift away from us at a small but persistent rate.

In summation, tidal forces cause tidal locking, but it happens through intricate and slow forces over a long time period affecting both the Moon's orbital velocity and rotation rate until a equilibrium is found. That equilibrium is tidal locking.

$\endgroup$
  • $\begingroup$ If this goes for any system, does that mean Earth will eventually be tidal locked to the Sun? $\endgroup$ – Liren Jul 7 '17 at 5:43
  • 2
    $\begingroup$ @Liren Yes, although it will take a lot longer. In fact, the Earth will become tidally locked to the Moon first. $\endgroup$ – Mark H Jul 7 '17 at 5:52
  • $\begingroup$ I wonder which countries will get to see the moon... what a tourist attraction! :) $\endgroup$ – NikoNyrh Jul 7 '17 at 8:23
  • $\begingroup$ @Liren Technically it could happen if there was only the Earth/Sun. The problem is we have lots of other planets like Jupiter which have noticeable gravitational perturbations on the Earth that prevent us from ever reaching that equilibrium of tidal locking. But, we definitely do see planet's tidally locking to their stars, e.g. Mercury (albeit at a 2:3 resonance rather than a 1:1 like the Moon) or the planets in the recently announced TRAPPIST-1 system. $\endgroup$ – zephyr Jul 7 '17 at 17:35
  • $\begingroup$ @NikoNyrh Probably none of them. It's predicted that the Earth will take upwards of 50 billion years to tidally lock to the Moon (the Moon's affects on the Earth are just too small to do it sooner) and we know the Solar System will be long gone by then. The Sun will go Red Giant in about 5 or so billion years. $\endgroup$ – zephyr Jul 7 '17 at 17:36
3
$\begingroup$

The simple answer is: tidal forces, which are a secondary effect of gravity. In the same way that the Moon causes low and high tides of the oceans here on Earth, the Earth also has a similar effect on the Moon.

The force is of the same origin, however far stronger due to the mass of the Earth. These tidal forces cause a torque on the rotation on the Moon and this is why it only shows the same face to the Earth.

Fun fact, Pluto and it's only moon Charon show only one face during their orbit due to the tidal force. Also the tidal forces are slowing down the rotation of the Earth too.

$\endgroup$
  • $\begingroup$ Do tidal forces depend on asymmetry? IOW, if the moon were more perfectly symmetrical all around, would tidal forces still exist or have the same effect? $\endgroup$ – 0tyranny 0poverty Jul 6 '17 at 14:51
  • $\begingroup$ They depend upon mass and the non constant nature of the gravitational force. That is, the gravitational attraction between two bodies is more strong at the sides facing each other. Hope that helps. $\endgroup$ – Rumplestillskin Jul 6 '17 at 15:02
  • $\begingroup$ @0tyranny0poverty Sort of. If you could make the Moon's mass distribution perfectly symmetrical (the lunar mass distribution is notoriously uneven) it would reduce the rate of change of the Moon's rotational period, but tidal forces themselves create an asymmetry in the surface: the tidal bulge. To eliminate the effect of tidal locking you'd need to have a perfectly rigid body, and that's physically impossible. $\endgroup$ – PM 2Ring Jul 6 '17 at 15:57
  • 1
    $\begingroup$ @0tyranny0poverty I suspect it's because it's a complex system with lots of moons, as well as the rings. This results in a chaotic mixture of gravitational effects, and it takes some time and luck for a satellite to synchronize. $\endgroup$ – Barmar Jul 6 '17 at 22:48
  • 2
    $\begingroup$ @0tyranny0poverty Because tidal forces are a result of differential gravitational forces. Basically you need to differentiate Newton's law of gravity, resulting in the tidal force being proportional to $r^{-3}$. See the wiki page for more details. $\endgroup$ – zephyr Jul 7 '17 at 17:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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