Selected Questions -Answers From All Experts Astronomy Forum (Tidal locking, resonance)

Question:

I don't quite understand how even if the moon is spinning, why is it not
subject to tidal braking? And what is different about Mercury that causes

its spin rate to be locked differently from the moon's?


Answer -

The basis of your question resides in the principle of conservation of
angular momentum. To put it in simple terms, two bodies in a joint system
cannot both 'spin up' at the same time, when tidal forces act. One will
slow (in this case, the Earth), the other (Moon) will acquire its 'lost'
angular momentum. Ignoring external effects, you can summarize the
transaction as a "zero sum" game.

Let's now fix ideas within the Earth-Moon system: The Earth rotates faster
than the Moon moves in its orbit. Because the tides are linked to the more
slowly moving Moon, they act by friction as a brake on Earth's rotation,
gradually slowing it down.

The angular momentum lost by the rotating Earth in this process is
transferred to the Moon's angular momentum. Thus, the Moon is accelerated
in its orbit, causing it to slowly spiral outwards, away from Earth. The
day and month are thus lengthening at different rates.

Calculations have actually been retro-worked to show how the length of
month differed when the Moon was much closer to Earth in the past. For
example, when the Moon was only 16,000 km away (10,000 miles) the month
was approximately seven mean solar days long.

Similar calculations based on the conservation of angular momentum also
allow us to project into the future. Thus, about three billion years
hence, the day and month will be equal - about 47 of our present days
long -  and the Earth will always turn the same face towards the Moon.

In the case of Mercury a number of reasons explain why its spin rate
is locked differently.

One is that its resonance period is simply different. Thus, 59 days
rotation period for Mercury equals roughly two-thirds of its period of
revolution (88 days). This suggests a resonance effect with the tidal
forces of the Sun. (Resonance effects always mean the rotation, for
example, is a whole number or integer multiple of the revolution(s). In
this case: 3 rotation periods » 2 revolution periods).

Clearly, since Mercury is much closer to the Sun than the Moon is,
the role of the Sun would figure much more prominently in such 'resonance
periods' for Mercury than for the Moon. This would surely lead to the locking

In of a different spin rate .

Another factor that may play a role is the particular shape of the Moon,
which is a tri-axial ellipsoid, not a perfect sphere. In this case, its
longer axis is always pointing more or less towards the Earth's center.
Thus, the Moon's shape may well contribute to some degree to its differing
'lockage' of spin rate.

Why Some Planets' moons recede and others don't

Of interest over the years in planetary astronomy is why some planetary moons (such as Earth's) recede or increase their orbital distance from the central planet over time, and why others (e.g. Mars' Phobos) do the opposite.  This recently came up in a question posed to me on an astronomy forum, and also whether or not recession or approach of planetary moons would not occur if such moon formed exactly at the "geosynchronous" orbit position, i.e. that for which the orbital velocity of the moon matches the planet's rate of revolution.

The issue of the Moon's tidal gravitational effects actually came up some months earlier in connection to the 'leap second' added to compensate for a slowing Earth's rotation, e.g. http://brane-space.blogspot.com/2012/07/where-did-that-extra-second-come-from.html

Still, the Earth's period for one rotation, 86,400s (or 86, 401s on the day the leap second was added) is much much shorter than the Moon's synodic period (29.5 days) or sidereal period (27.3 days). This leads to a situation such as depicted in the attached diagram. Here, I denote two tidal bulges on the Earth's surface by x and x' at essentially opposite ends of the line through the center of Earth. Many blog readers probably already know that two tidal bulges occur as a result of a differential gravitational force. (The general origin for a given tidal force on Earth due to the Moon is from the variable values for the Moon's gravitational attraction at differing locations inside Earth.)


In the sketch I've shown, note that bulge x is closer to the Moon than bulge x' on the other side of Earth. This difference results in a net torque exerted on Earth (torque is the product of the moment of inertia, I, and the angular acceleration, a). Specifically for our diagram, bulge x leads the Moon, and because it's also nearer the Moon than bulge x' ,  the force exerted by the Moon on x is greater yielding a net torque. This net torque slows the Earth's rotation, in effect, slows its angular momentum. Now, at the same time, bulge x is pulling the Moon forward in its orbital path, effectively speeding it up and causing it to move farther out. This is just a consequence of the conservation of angular momentum.


To fix this concept, consider two bodies m and M connected as shown with c the center of mass:


Mass M O---r1------c ---------r2----------o m

Both M and m rotate about c at distances r1 and r2 respectively. Since the angular momentum, e.g. L = Iw must be constant (where w = v/r)then if m somehow speeds up (higher v) then to compensate, r2 must increase. (w ~ 1/r and w ~ v) . Thus, the Moon constantly moves further out, and by about 3-4 cm a year.
Now, consider Phobos - which orbital period is shorter than a Martian day,since  tidal deceleration is decreasing its orbital radius at the rate of about 20 metres (66 ft). This is the converse of the case for our Moon. Thus, Phobos (orbiting faster then Mars is rotating) experiences the opposite of what our Moon does, i.e. a slowing down of its orbital speed. Again, by the law of conservation of angular momentum (and using the simple model shown above), if v is decreased then r must decrease. Mars' other moon Deimos, meanwhile, moves in its orbit faster than Mars is rotating and therefore speeds up like our own Moon, and hence increases in its orbital distance.

Therefore, the conjecture that if our own Moon "just happened to form"  at the geosynchronous distance the orbit would remain fixed and stable, is technically correct provided position only is considered, irrespective of dynamical interactions.. Since its "positionally assigned" orbital velocity is in synch with Earth's rotational velocity there is no net torque acting on Earth and hence, the Moon could still be orbiting. The downside  is that such a fortuitous happenstance has about as much probability as happening as a blind man playing eight -ball pool and getting all of his balls in before his opponent can!

Hence, a definitive answer to the question would hinge on: 1) the exact mass of the object captured, 2) the Lagrange point solutions based on this mass, and most importantly, 3) the assumptions used in considering the nature of the problem, i.e. two-body (ignoring Mars' small moons Deimos and Phobos entirely), or restricted 3-body, i.e. considering only the more massive, say Phobos.  

Re: "the Lagrangian points",  in general there are 5 such points in all which refer to the five positions for a hypothetical orbit for which a small object (i.e. tiny moon) can be incorporated without altering the overall Lagrangian pattern (which defines 5 different points at high and low gravitational potential, i.e. V = -GM/r).

Readers who wish to learn more can google "Lagrangian points".