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Suppose the moon underwent a single, massive, large-object bombardment event. About what number (or range) of about what mean mass of objects could shorten the synodic month by about 1.5633%?

(Assume the moon is not broken up.)

Corollary: Should we expect the moon also to orbit more closely to the earth in result?

By the way the question is historical. There is at least one ancient text that states that rocks “hurled to the earth” from the skies, whereupon the synodal month shortened from 30 days to the current value. That being “outside my field,” literary scholars don’t generally tackle it. So, is there a general way to approach the problem for antiquarians lacking a technical background in physics?

Hoping for an answer such as, say, ‘If 100 objects of a mean mass of 100 tons each struck the moon at a typical velocity of __, over a period of three weeks, then the synodal period could shorten by 1.56%.’ Or, ‘It would be impossible to shorten the synodal period of the moon 1.56% in an asteroid bombardment, because’…

An answer might also help in interpreting the ancient astronomical vocabulary, which so far is translated word for word as is.

Also, if relevant, I date the text’s putative asteroid bombardment to around 3044 BCE.

Thank you!

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The best we can do is estimate.

Using $\frac{1}{S}=\frac{1}{P_{moon}}+\frac{1}{P_{earth}}$ (sign is positive as Sun and Moon moves opposite in the geocentric frame) you can get the corresponding sidereal period for the ancient 30 day synodal month to be 27.72300 days. Then using Kepler's Third Law $P^2=\frac{4\pi^2}{G(M+m)}a^3$ you will get the ancient semi-major axis to be $3.8850\times 10^8\text{ m}$.

Let's assume direct impact against direction of velocity. We can use vis-viva to calculate the velocities. $v^2=GM(\frac{2}{r}-\frac{1}{a})$ and assume the final orbit (current) to be circular (so the ancient one is slightly elliptical, and $r$ is constant), to get $v_i=1.2886\text{ km/s}$ and $v_f=1.2748\text{ km/s}$.

For elastic collision, there are infinite solutions that can cause such a change using $\Delta v_{moon}=\frac{2m_{asteroid}}{m_{moon}+m_{asteroid}}u_{asteroid}$. Two are $m=7\times10^{18}\text{ kg}$, $u=70\text{ km s}^{-1}$ and $m=10^{19}\text{ kg}$, $u=50\text{ km s}^{-1}$ with respect to the Moon, but this is just a very rough estimation because we assumed the asteroid collided with the Moon like a billiard ball when in reality the Moon could be blasted into chunks. Also $10^{18}$ means 1 billion objects of 1 million tons.

Corollary: Should we expect the moon also to orbit more closely to the earth in result?

Yes, from Kepler's Third Law, a shorter period means a shorter orbit.

Edit:

For inelastic collision (which ProfRob pointed out to be closer to the actual scenario), we can use $\Delta v_{moon}=\frac{m_{asteroid}}{m_{moon}+m_{asteroid}}u_{asteroid}$ and get infinite solutions including $m=2 \times10^{19}\text{ kg}$, $u=50\text{ km s}^{-1}$. Your answer should be roughly bounded by these two cases.

Also answering the question in the title "How much energy to shorten a synodic month by about 1.56%?", we can use $\Delta E=-\frac{GM_{earth}m_{moon}}{2}(\frac{1}{a_f}-\frac{1}{a_i})$ and get $\Delta E=-8.0\times10^{26}\text{ J}$

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    $\begingroup$ what is $v_A$; what is $u_B$, what are $m_A$ and $m_B$? Asteroids don't move at 500 km/s, I think the maximum speed you could asume is around 70 km/s unless hypothesising an interstellar asteroid. I'm also troubled by the assumption of an elastic collision. I think assuming the collision is inelastic would be much more accurate. $\endgroup$
    – ProfRob
    Apr 24, 2022 at 14:30
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    $\begingroup$ I agree. What would really happen in the case of the velocities and masses involved here would be profoundly inelastic collision, not to mention mass transfer and/or loss. It's a great answer as far as it goes, but I think it can be further improved by avoiding some simplifications. "The best we can do is estimate." Yes, but this is not yet the best that we can do to estimate :-) $\endgroup$
    – uhoh
    Apr 25, 2022 at 3:14
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    $\begingroup$ Thanks, Mr. Qin. I conclude it is impossible. $\endgroup$ May 8, 2022 at 4:39

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