What size and/or mass of impactor would need to impact the moon to destroy its tidal lock so that it's orbit is noticeable within its phases? What side effects would this impactor cause, and how would it affect the earth's tides, if at all?
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$\begingroup$ "destroy" can literally range from "give it enough angular momentum for it to start doing a little spinny in our frame" or "destroy the galaxy and the concept of tidal locks in a 100000 light year radius" $\endgroup$– Varad MahashabdeCommented Dec 21, 2019 at 18:06
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$\begingroup$ To reduce the sidereal rotation period of the Moon from 27.321661 days to 27.321659 days is one thing, to reduce to 14 days is quite another. A quantitative definition of "destroy[ing] [the] tidal lock" of the Moon is absolutely needed. $\endgroup$– AlexPCommented Dec 21, 2019 at 18:09
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$\begingroup$ I suggest that “orbit” be read “rotation”. Then the requirement that the moons rotation be noticeable from earth during (each) of its phases offers some basis for establishing the new rotational period. I’ll propose that to move sufficiently to be easily visible as changing during its crescent phases the moons rotational period would need to be not less than about 14 days (336 hours). Can someone work with that? $\endgroup$– JimCommented Dec 22, 2019 at 12:23
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$\begingroup$ @Jim, or simpler yet, suppose we want the Moon's entire surface to be visible from Earth over the period of a year. The question would then be "How much energy would be needed to completely stop the Moon's rotation with respect to the visible stars (excluding Sun), and what mass and velocity would an asteroid need to have in order for a collision to cause this?". $\endgroup$– Ray ButterworthCommented Dec 22, 2019 at 14:15
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2 Answers
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What size and/or mass of impactor would need to impact the moon to destroy its tidal lock
The "tidal lock" is the end result of a continuous braking process that would gradually slow the rotation of the Moon (or any body orbiting the Erth). Kick the Moon up to speed and it'll be dragged down again over time. The only way to "destroy" it would be to either a) destroy the Earth, removing the source of the tidal forces, or b) move the Moon outside of the Earth's sphere of influence. Impact with another body is not a suitable way to accomplish either of those goals, though a suitably close encounter with a massive body could eject the Moon into a heliocentric orbit, which would do the job without smashing it to gravel.
so that it's orbit is noticeable within its phases?
I'm not sure what this is supposed to mean. Presumably you meant so that it's rotation is noticeable? That doesn't need to be a big speedup, as anyone familiar with the Moon's normal appearance would eventually spot that it looked a Bit Odd. If you sped it up by a single day you'd notice weirdness in a few months.
What side effects would this impactor cause, and how would it affect the earth's tides, if at all?
It would be a colossal impact, and it will probably throw a lot of debris up into space. How fast that debris is travelling is anybody's guess... modelling such things is too hard for me to attempt. Given the speed of a likely impactor though, you'd expect some of the debris to fall back onto the moon and some to fall into Earth-centred orbits. I'm not sure if they'd fall to Earth, or threaten any satellites though.
The effect on the Earth's tides would be negligible. The mass of the moon isn't evenly distributed, but it is far enough away that the effects of that uneven mass wiggling about are unlikely to be noticeable.
Here's some working on the impactor, if you were interested.
Given the Moon's moment of inertia (from here), you can see that it has a moment of inertia about its polar axis of $8.736 \times 10^{34}kgm^2$ an hence angular momentum of about $2.15\times10^{29} kgm^{2}s^{-1}$. To change its rate of rotation, you obviously have to change its angular momentum... to speed it up by one day, for example, you need to add $7.54\times10^{27} kgm^{2}s^{-1}$. The required angluar momentum change scales linearly with the required day-length change.
If you could somehow perfectly deliver all of the momentum of an impactor to a point on the moon's equator perfectly at right-angles to its radius (which obviously you can't) you'd need $4.34\times 10^{21}kgms^-1$. An impactor travelling at 10km/s (which is a little above average) would need to weigh 4.34x1017kg... which by a happy coincidence is pretty close to the estimated mass of the Chixulub impactor. The required mass scales linearly with the required changed in day-length, so you need about five billion tonnes of impactor per second of day-length change.
Problem: an impact at right-angles to the radius will skip right off and do nothing. To steep an impact will produce a big bang, but not enough force in the direction we want. I've no idea what the optimal angle is, but there are tradeoffs with losing momentum in the form of ejecta travelling faster than Lunar escape velocity. This is too complex for me to try to answer here, though. I can at least say that the total momentum of the impactor is a good four orders of magnitude lower than the moon's orbital momentum, so the moon's orbit won't be changed noticeably.
You probably want your rock(s) to impact at about 15-30 degrees from horizontal, and you might need a few bigguns to deal with efficiency problems where you're losing energy to escaping ejecta and heating and other boring things. Maybe budget twice the aforementioned mass, and drop it in the form of a bunch of small rocks. If you get things up to speed with fewer rocks than you need, great. Just blow the others up or deflect them or something.
The time for the moon to become tidally locked (again) can be approximated by $$T_{lock} \approx {\omega a^6 I Q \over 3Gm_p^2 k_2 r^5}$$ where $\omega$ is the moon's orbital semimajor axis, $I$ is the moment of inertia (which is 0.394 x the mass of the Moon x the radius of the Moon squared), $Q$ is the dissipation function, $G$ is the gravitational constant, $m_p$ is the mass of the Earth, $k_2$ is the Love number of the satellite and $r$ is the satellite's radius. For the Moon, $k_2 / Q = 0.0011$.
Because it is only an approximation, it won't give you trivially useable answers... if you put in the Moon's current rotational angular velocity, you'll get a $T_{lock}$ of 177000 years. Crank up the angular velocity so that the moon rotates a whole day faster, and you get a lock time of 183000 years, suggesting that once you've crashed your impactor(s) you've probably got of the order of 10000-100000 years before everything completely grinds to a halt again. So you haven't destroyed the tidal lock by any means, but it'll do.
answered Dec 22, 2019 at 22:03
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Consider an extreme example.
A lunar rover starts driving to the west. Its wheels push the Moon causing it to rotate slightly to the east, preserving angular momentum. Tidal lock has been "destroyed".
When the rover stops, the braking wheels push the Moon to the west, and tidal lock is restored.
The question of "destroy" really is meaningless in any practical sense.
answered Dec 21, 2019 at 18:46