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I am preparing myself for an interview for a quantitative analyst position and one of the sample questions asked in previous examinations was:

"Suppose the moon were to disintegrate, and fall to earth over 5000 years. How does this influx of power compare to that of the Sun? Much more, about the same, or much less?"

My idea was to consider the moon as a star and compute its hypothetical flux of energy, but the "fall to earth over 5000 years" puzzles me a bit.

How would you approach the problem?

Thanks

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  • $\begingroup$ I like this question. It's very complicated though. $\endgroup$
    – will
    May 24, 2016 at 16:24

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Here's how i'd have at it;

* I happen to know these are okay guesses.

** Let's assume it's just the potential energy, and that as the point of the "5000 years" part of the question.

  1. The moon came from the earth - likely it's crust since the idea is that it was formed from an impact. The crust of the earth is less dense than the core, so the moon is less dense than the earth. Call it 60% density.
  2. The moon is about the same size as a finger nail at arms length.
  3. The moon is 240,000 miles away*, my arms are just under 3ft long, and a finger nail is ~1cm in diameter. There are ~5000ft in a mile. Therefore the moon's diameter is 240000 x 5000/3 = 4000km diameter.
  4. The earth's radius is 6000km, so it's radius is 3 that of the moon. $3^3$ is 27 => the earth weights 27/0.6 = 45 ~= 50 times more than the moon.
  5. The earth weights 10$^{25}$Kg*.
  6. potential energy is $mgh$. By disintegrating the moon and slowly(5000 years) bringing it down to the earth, we'll be gaining at least it's potential energy**. So this will be $10^{25}\mathrm{Kg} / 50 \cdot 9.8\mathrm{ms}^{-2} \cdot 240000\mathrm{miles} = 9.8 \cdot 240000 \cdot 10^{25} / 50 \mathrm{Kg\ m\ miles\ s}^{-2} = 10^{34} K m^2 s^{-2} = 10^{33} J$.
  7. There are $365\ 3600\ 24 =\ \sim30,000,000$ seconds in a year. Power = $ 10^{33}J / (30000000 \cdot 5000) = 10^{23}W$
  8. Now we go back to the earth bits for a bit, remembering that number. The radius of the earth is 6000Km. So the surface area is $4 \ \pi \ r^2 = 4 \ \pi \ 6000000^2 = \pi\ 1.32\mathrm{E}^{14} = 5\mathrm{E}^{15}\ \mathrm{m}^2$
  9. Energy per square meter = $2\cdot10^{7}\ \mathrm{W}$.
  10. It takes 1 calorie to raise 1cm$^3$ of water by 1 degree. 1 calorie is 4.2J. So per square meter of the earth, we have enough energy to raise $5\mathrm{E}^6 \mathrm{ml}$ of water by a degree every second. Or, enough to (raise to a) boil*** $50\ell$ of water per second per square meter.

So i'd be saying a lot more.

Now, that seems like a lot. It's obviously more that the sun puts out. I'm not sure if i've made any horrible msitakes in here, i'll check later - i'm about to go home. A quick sanity check though, yields that the Vredefort crater was created by a metoer thought ot have been ~15-20 km in diameter. And that you would have to carve 150 lumps this size from the moon per day to whittle it down to nothing in 5000 years.

So that is a lot of mass that's getting dumped onto the earth each day. Granted, it won't be going as fast, but there's still a lot of substance.

This also makes the assumption that all of the potential energy is converted to heat. I don't think this is likely - what will likely happen is that it will change the angular momentum of the earth such that the total angular momentum of the system stays the same. If you look at the sun and the moon from celestial north, the earth orbits the sun, the moon orbits the earth, and both the earth and the moon rotate - all anti clockwise. Pulling in the moon will make the earth rotate faster and decrease the length of the day.

*** ignoring that the energy required to heat water is not constant

You also need to ask why it's falling to earth?

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  • $\begingroup$ Not as good an answer as Enrico Fermi would have given ( mathforum.org/workshops/sum96/interdisc/sheila2.html ), but a good attempt $\endgroup$
    – nbbo2
    May 24, 2016 at 17:57
  • $\begingroup$ @noob2 i was trying to remember the name of the approach! i thought it was Feynman. $\endgroup$
    – will
    May 24, 2016 at 17:59
  • $\begingroup$ From what I remember the own gravitational energy of the moon is way greater than that of the eath towards the moon. And to get the answer rapidly, may we consider that the moon has been formed by meteorites falling toward earth ? $\endgroup$
    – lcrmorin
    May 27, 2016 at 12:26
  • $\begingroup$ @Were_cat yah of course the moon's gravity is stronger when you get closer to it - if it weren't then if you dropped something on the moon it would fall to the earth instead. And the moon would crumble apart and fall to the earth too - like in the question. How can that consideration give you the answer rapidly? $\endgroup$
    – will
    May 27, 2016 at 13:06
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    $\begingroup$ Consider a sphere of rayon r ($m=\frac{4}{3} \pi r^3 \rho$), then calculate the energy to bring enough mass ($dm= 4 \pi r^2 \rho dr$) from infinity to form a sphere of r + dr ($dU= -\frac{G m dm}{r}$). Integrate from 0 to infinity, that would give: $U=\frac{3*G*M^2}{5*R}$ the energy required to disintegrate. According to wikipedia it is aroud $2*10^(32)$ for the earth, less than what you calculate for potential energy of the moon... That lead me to think/remember the use of formula $E_p=mgh$ is only valid near earth. The Ep value should be calculated with $ Ep = - \frac{GmM}{r}$. $\endgroup$
    – lcrmorin
    May 27, 2016 at 13:55
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Based on the way the question is phrased, the simplest approach is:

1) Estimate the gravitational potential energy of the Earth-Moon system. 2) Use result from 1) to estimate an average power as Energy/time. 3) Estimate the solar power on the illuminated half of the Earth. 4) The comparison of the result in 2) and 3) allows you to judge relative magnitude.

Some of these steps were indicated in other responses, but there are a few errors.

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    $\begingroup$ Would you care to elaborate what these "few errors" are, and provide an alternative solution? $\endgroup$
    – AdB
    May 29, 2019 at 7:41

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