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If you calculate the orbital period of an earth sized object around the sun at 1 au, it is 31554651 in seconds (1 year). (Why does Google say a year is 31536000 seconds?!)

If you calculate the orbital period of a moon sized object around the sun at 1 au, it in 31554698 is seconds, that is 47 seconds longer than that of the earth.

That means that the moons orbit is 1.48947932e-6 times longer than earth's. Which means in 1.5 million years the moon would fall behind the earth by an entire orbit, hence it would take a faction of that time for the moon to fall behind the earth enough to no longer orbit the earth.

Is my math wrong? (I did double check it.) Is my understanding of physics wrong? Or is there some strange effect that is causing the moon's orbit to match that of the earth's?

EDIT

So the moon does not fall behind, that is good, I am rather fond of our moon XD

You answered part of my question.

How is the moon captured by the earth? The fact that the moon is close to earth does not mean the sun no longer pulls on it. If that were true the moon would spiral straight and not follow the earth in its path around the sun. Gravity pulls the moon toward the earth (technically both, together), the centripetal force, constantly pulls the same exact amount on the moon (assuming no imperfections in the shape of the earth and moon); when the earth moves toward the sun, from the sun's pull on it, why does the moon get pulled along as well, and more so than its own pull from the sun?

What additional force is causing the moon to stay with the earth? Or is it just a remarkable quirk in how the math works out where the pull of the sun varies as the moon orbits earth, causing it to end up moving along faster?

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  • $\begingroup$ The moon orbits earth, not the sun directly, so it isn't going to fall behind. $\endgroup$
    – Matt Samuel
    Jul 10, 2020 at 17:06
  • $\begingroup$ $31\,536\,000$ seconds is just $365$ days times $86\,400$ seconds per day. Most of the discrepancy is covered by leap years. ¶ The reason for the discrepancy you calculated (if correct) might be that the center of gravity is then closer to the Sun's center (marginally), and therefore the period is longer. ¶ The reason this doesn't make any practical difference is that the Moon is in fact gravitationally bound to the Earth. It can't fall behind; the Earth's gravity makes it keep up. $\endgroup$
    – Brian Tung
    Jul 10, 2020 at 17:07
  • $\begingroup$ By the way, you've calculated that the Moon's orbit (if it were alone) would be $47$ seconds longer (not shorter) than the Earth's. You may want to fix this in your original post. $\endgroup$
    – Brian Tung
    Jul 10, 2020 at 17:09
  • $\begingroup$ Counter question: By how much do you have to adjust the Moon-Sun orbit radius to obtain same orbit length? And by how much does the Moon "wobble" about its wannabe orbit by the presence of the Earth? $\endgroup$
    – Hagen von Eitzen
    Jul 10, 2020 at 18:07
  • $\begingroup$ What effect are you using to calculate that the Moon takes $47$s longer to orbit the Sun? $\endgroup$
    – Chrystomath
    Jul 10, 2020 at 18:17

2 Answers 2

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Expanding my comment into an answer: At any given separation, the common period of revolution of two objects is indirectly proportional to the square root of the sum of their masses. See the Wikipedia plot summary on the gravitational two-body problem for more details.

The Sun has a mass about $300\,000$ times that of the Earth; the Earth has a mass about $81$ times that of the Moon. Therefore, the Sun-Earth system "outmasses" the Sun-Moon system by very nearly one part in $300\,000$ (close enough for our purposes). Thus, we would expect the Sun-Moon system (if isolated) to have a period longer than the Sun-Earth system (if isolated) by about one part in $600\,000$. That is approximately what you found.

The reason the Moon doesn't in fact fall further and further behind is that the systems are not isolated. The Moon is gravitationally bound to the Earth, and is in some hand-wavy sense dragged along with it in their common orbit around the Sun.

To answer your other question: Google's answer of $31\,536\,000$ seconds is obtained by multiplying a simple year of $365$ days by $86\,400$ seconds per year. The discrepancy is mostly covered by the leap year system.

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  • $\begingroup$ Thanks. What force causes the "capture"? See edit to the question. $\endgroup$
    – Lupus Ossorum
    Jul 11, 2020 at 4:06
  • $\begingroup$ @LupusOssorum: Gravity. It's just a matter of relative forces. $\endgroup$
    – Brian Tung
    Jul 11, 2020 at 4:52
  • $\begingroup$ Um, what do you mean? The gravitational force between two objects is calculated with $g\ =\ \frac{GM_{1\ }M_{2}}{r^{2}}$. The added force between the moon and earth, and moon and sun, will not keep the moon in orbit around the orbiting earth. $\endgroup$
    – Lupus Ossorum
    Jul 11, 2020 at 21:35
  • $\begingroup$ @LupusOssorum: You'll have to back up your claim. If there were no Earth-Moon gravitational interaction, that is exactly when the Moon would fall behind the Earth. That applies equally to me and you, as well as the Moon; if gravity were not keeping us fixed to the Earth, then what you've discovered is that we would fall behind the Earth. $\endgroup$
    – Brian Tung
    Jul 11, 2020 at 22:03
  • $\begingroup$ You mentioned the expression for gravitational force, and then did nothing with it, and somehow concluded that it's not the force that keeps the Moon with the Earth. It's like saying "$F = ma$, therefore the ball doesn't fall to the ground." The one doesn't follow from the other, at least as far as I can tell. $\endgroup$
    – Brian Tung
    Jul 11, 2020 at 22:05
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The fact that the moon's orbital period would be longer than earth's, is false. The mass of the orbiter is irrelevant to it's orbital period.

Although the equation of gravity takes into account the mass of both objects (F = G * (M1*M2/D^2)) the equation of orbital period does not (P = tau * sqrt(r^3/G*M)) (link). This is because the mass of the orbiter is canceled out when applied to the velocity (V=F/M). Hence the equation for "gravitational velocity" is V=G*(M1/D^2).

With a "semimajor axis" of 92966500 miles around 1 sun of mass, regardless of the mass of the orbiter (planet) the orbital period will be 1 year.

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