# Conservation of Angular Momentum -- Earth-Moon System -- Earth spin backwards?

I am developing an exhibit for a museum. We want to show how the Earth's spin rate changes as the Moon drifts farther and farther out from the Earth. The visitor has a slider they can move to set the distance of the Moon from the Earth.

Assuming for a moment a closed system, we have conservation of total angular momentum in the Earth-Moon system:

$$L_{tot}=L_{moon}+S_{earth}+S_{moon}$$

where $$L_{moon}$$ is angular momentum of the Moon in orbit about the Earth, and $$S_{earth}$$ and $$S_{moon}$$ are the angular momentum of the Earth and Moon respectively due to their spin about their own axes.

$$L_{moon}=m\sqrt{G(m+M)r}$$

where $$G$$ is the gravitational constant, $$m$$ mass of the Moon, $$M$$ mass of the Earth, and $$r$$ the orbital radius of the Moon about the Earth.

Angular momentum due to spin is calculated assuming uniform spheres spinning about their central axes:

$$S_{earth}=\frac{2}{5}MR_{earth}^{2}\omega_{earth}$$

$$S_{moon}=\frac{2}{5}mR_{moon}^{2}\omega_{moon}$$

To figure out the spin rate of the Earth at some new $${r}'$$, I calculate $${L}'_{moon}$$ and $${S}'_{moon}$$ at the new radius as follows:

$${L}'_{moon}=m\sqrt{G(m+M){r}'}$$

$${S}'_{moon}=\frac{2}{5}mR_{moon}^{2}{\omega}'_{moon}$$

which, as long as we are within the tidal lock limit around 992,000 Km, $${S}'$$ can be found from the orbital period of the Moon at that distance:

$${\tau}'_{moon}=2\pi\sqrt{\frac{r^{3}}{G(m+M)}}$$

and

$${\omega}'=\frac{2\pi}{{\tau}'}$$

Applying conservation of angular momentum, I then solve for $${S}'_{earth}$$:

$$L_{tot}={L}'_{tot}$$

$$L_{tot}={L}'_{moon}+{S}'_{earth}+{S}'_{moon}$$

$${S}'_{earth}=L_{tot}-{L}'_{moon}-{S}'_{moon}$$

Using this setup, if I run the Moon in closer to the Earth to the distance at which astronomer's estimate the Moon originally formed, I get a spin rate for the Earth that matches what they estimate. Setting the Moon at its normal orbit of $${r}'=60 R_{earth}$$, I get the expected 24 hour period for the Earth and 27.3 hour orbital period for the Moon. So, at this point everything looks fine and self consistent.

What happens though is that at a distance of roughly $${r}'=93.4 R_{earth}$$ (595,700 Km) the Earth stops spinning. Beyond $${r}'=93.4 R_{earth}$$ the equations predict that to conserve angular momentum the Earth has to start spinning backwards!

This violates my gut intuition, and that of others, as to what would happen.

My question is, would the Earth actually start spinning backwards?

Is there a flaw in the analysis? Specifically, am I violating the requirements of a "closed system" by allowing the visitor to play God and instantaneously move the Moon to a new orbital radius. If so, how then should I account for the user intervening in the system, and correctly calculate the spin rate of the Earth for different Moon orbital radii?

• Earth-Moon is not a closed system. We must also take into account the orbital motion of the system around the sun. Mar 5, 2019 at 23:46
• Hi Alex,If we make the assumption that the Barycenter of the Earth-Moon system does not change enough in this process to make any significant difference in the angular momentum of the Earth-Moon system around the Sun, can't we just ignore this? At $200R_{earth}$ (1,275,600 Km) the barycenter is only 15,504 Km from the center of the Earth -- i.e. about 2.3 times the radius of the Earth. That is tiny compared to the Earth orbital radius of 150,000,000 Km. Mar 6, 2019 at 0:39
• The orbital angular momentum $L$ of the Earth-Moon system is much larger than the total moment $L_{tot}$. Even a small exchange effect between $L$ and $L_{tot}$ may have a noticeable effect on $S_{earth}$. Mar 6, 2019 at 1:41
• Ok, I will work on factoring that in and see what the results are. Thank you. Mar 6, 2019 at 2:07
• Use data and model on the movement of the pole of the earth from IERS Earth Orientation Centre on iers.org/IERS/EN/Organization/ProductCentres/… Mar 6, 2019 at 12:20