# Moon orbiting backwards

1. Oct 7, 2012

### DeepSpace9

If the Earth still rotated and stayed on its course around the sun, but the moon was orbiting in the opposite direction. What would be the consequences, and would you actually be able to see the Moon moving with the naked eye, would the effects of the Moon moving in the opposite direction of the spin of the Earth be enough to see?

2. Oct 7, 2012

### Nabeshin

Well the moon moves ~12 degrees/day due to its orbital motion, or ~ 0.5 degrees per hour. It moves ~360 degrees/day due to the earth's rotation, or ~15 degrees per hour. So you're looking at a 0.5 degree perturbation on a 15 deg/hr angular speed. I wouldn't be able to tell, for one.

3. Oct 8, 2012

### DeepSpace9

But the Moon wouldn't be in the sky for 12 hours everyday right?

4. Oct 8, 2012

### Staff: Mentor

Its apparent velocity in the sky would increase by ~1/15. Instead of ~12 hours + 30 minutes, it would need ~11 hours and 30 minutes from horizon to horizon (neglecting issues with the precise arrangement of the orbits and rotations).
The dominant contribution to its apparent motion comes from the rotation of earth, the moon's orbit is just a small modification.

5. Oct 8, 2012

### Bandersnatch

You'd have the phases go the other way around. The Moon would brighten/darken from the right side toward the left, instead of left to right.

6. Oct 8, 2012

### cepheid

Staff Emeritus
This is already a hemisphere-dependent statement.

7. Oct 8, 2012

### Bandersnatch

The northern hemisphere is the ONLY hemisphere. Or at least the only one that matters. If you're looking at the Moon standing upside down, then you're doing it wrong, and it's your own fault.
Besides, everybody knows that there are no actual people living down under.
...

Yeah, I didn't think about it.

8. Oct 8, 2012

### cepheid

Staff Emeritus
Well, the other problem with that is that the moon actually brightens from right-to-left in the northern hemisphere. So I thought *you* were from down under. :tongue:

9. Oct 8, 2012

### Bandersnatch

Is there someplace I can hide?

10. Oct 9, 2012

### EricNoot

Forget all of that, we'd have bigger problems if the Moon orbited backward. A retrograde orbit with a tidal drag would cause the Moon's orbit to decay and come crashing back into the Earth. Don't go there!

11. Oct 10, 2012

### Staff: Mentor

That is ~4cm/year, if the motion would be reversed today the moon would decrease its distance by ~40000km in 1 billion years. Which is about the timescale we have left for life on earth (at least without massive human intervention).

12. Oct 10, 2012

### EricNoot

Sorry mfb that only works if you're assuming that the tidal force is just works linearly. It doesn't.

The differential force on the Earth caused by the Moon is dF_Earth ≈ r^-3 dr. Since the mass of the tidal bulge raised by this force is proportional to it, then the m ≈ r^-3 dr.

Now the net force on the Moon caused by the two tidal bulges is also a differential force, that in turn depends on the mass raised by the Moon's pull, so:

dF_Moon ≈ m r^-3 dr ≈ (r^-3) r^-3 dr ≈ r^-6 dr.

When the Moon was 1/4 its distance now, the tidal force was 64x what it is today, and the net effect of the tidal forces would be 4096x what they are now! Of course, when "Theia" hit the young Earth, splating and ripping out a huge amount of material creating a ring which coalesced into the Moon, there were no oceans, they came later, and so no ocean tides. Now we do, and they're responsible for not only helping to accelerate the Moon foreward in its orbit, thereby increasing its distance, but also to help slow our rotation down. The Moon retrograding now would encounter a stronger tidal interaction, that would really slow down our planet's rotation and cause the Moon to experience an ever stronger pull inwards, cause it to come crashing inwards much faster than the billion year timeline you give.

13. Oct 10, 2012

### Staff: Mentor

Would it? Perhaps for some range of initial values of angular momentum, but I would rather expect it to settle on some tidally locked system with both Moon and Earth facing each other with the same side. I can easily be wrong, but I don't see why such a system MUST end with a crash.

14. Oct 10, 2012

### Staff: Mentor

@EricNoot: We have oceans for at least ~3.5 billion years now, and since then our moon increased its distance, based on those tides. Let the system evolve backwards (not in every detail, just the moon's distance), and you see that it would not crash within 3.5 billion years, even if you take into account that the tidal drag gets a bit larger as the relative angular velocity increases.

A more mathematical argument: 40000km are a change of 10%. Even if the rate of orbital change is proportional to the 6th power of the distance (which we would have to check), this gives a factor of ~1.77 between "rate of orbital decay now" and "rate of orbital decay when moon is 10% closer", with an average somewhere at ~1.35. In other words, my quick approximation has an error of about 35-50%. That is fine, I can live with that.

15. Oct 11, 2012

### EricNoot

Borek, retrograde orbits (basically the orbiting body doing so against the rotation of the orbited body) with tidal interactions are calculated showing that the orbiting bodies will crash, just as it is predicted to do so with Triton around Neptune. Triton does this, for it is thought to be a captured body and not something that initally formed together with Neptune. As that moon got closer, the force would grow quite quickly feedbacking into the force. When our Moon formed at a few Earth radii away, the Earth rotated about once every 6 modern hours and the Moon orbited the Earth at speeds that would seem ridiculously fast to what we're used to today.

Tidal locking occurs when the period of rotation of the orbited body matches that of the period of the orbiting body. Then there will no longer be a fluctuating difference tugging against the bodies. Whether or not that would occur during such an inward spiral, I have not checked.

mfb, yes, in our lifespans we would not notice the changes very much at all. But in consideration of the Moon's history, its inital existence would have undergone significant changes. After Theia hit, the Moon could have formed from the resulting debris ring in a few thousand years. By 4.4 Gya, the Moon would have gotten to around half its orbital distance of today, However, I would like to point out that I did not say that the "rate of orbital change is proportional to the 6th power of the distance". The tidal differential force on the Moon goes to the inverse 6th power of the distance, the Moon's inertia mass will then affect its rate of acceleration.

As for checking the math, the dF is found by taking the derivative of Newton's law, F = GMmr^-2, then using that dF to project the mass of the tidal bulge, as described earlier, to find the differential on the Moon.

16. Oct 11, 2012

### Staff: Mentor

I am not questioning it can happen, I wonder if it is an universal end, or whether it depends on the initial conditions. You seem to be suggesting every system with a satellite on retrograde orbit must end with a crash, and I find it unlikely.

17. Oct 12, 2012

### Staff: Mentor

I think there is a simple way to determine whether a retrograde satellite will crash or not:

Calculate the minimal radius for an orbit - for most objects, it will be the Roche limit, for very small objects it might be smaller. Calculate the orbita period of the satellite in this distance. Assume that both satellite and main body rotate with that period, and that the satellite is in this orbit (with the same direction as the initial rotation). Calculate the total angular momentum. If the initial angular momentum is smaller, the system does not have a stable configuration - the satellite will disintegrate and/or crash. If the initial angular momentum is larger, there is some stable configuration with a larger radius - which can be calculated, too.

Numbers :).
Moon has an angular momentum of 2.87*1034 Js.
Earth has an angular momentum of approxinately 0.71*1034 Js (with the approximation of a perfect sphere of constant density).
I neglect the angular momentum from the movement of earth (around the common center of mass) and the rotation of the moon, as they are of the order of the last digit.

If our moon suddenly moves in the opposite direction, the total angular momentum of the system is the difference between those numbers, 2.16*1034 Js.

In equilibrium with locked rotation:
Capital letters (V,M,R) refer to orbital parameters of earth (around the common center of mass), lower-case letters (v,m,r) are parameters of the moon. ω is the common angular velocity.
I'll solve the system for r.
$v = \sqrt{\frac{\mu^2MG}{m^2r}}$ where $\mu=\frac{Mm}{m+M}$ is the reduced mass.
$MR=mr$ (center of mass system)
$MV=mv$ (center of mass system)
$\omega=\frac{v}{r} = \sqrt{\frac{\mu^2MG}{m^2r^3}}$

We have 4 contributions to the total angular momentum:
$mrv=\sqrt{\mu^2 MG r}$ orbital momentum of moon
$MRV=\frac{m}{M}\sqrt{\mu^2 MG r}$ orbital momentum of earth
$\frac{2}{5}m r_{moon}^2 \sqrt{\frac{\mu^2MG}{m^2r^3}}$ rotational momentum of moon (rmoon is its physical radius)
$\frac{2}{5}M r_{earth}^2 \sqrt{\frac{\mu^2MG}{m^2r^3}}$ rotational momentum of earth

Now we can use the known parameters of the system:
$\frac{m}{M}=\frac{1}{81.3}$
$m=7.35*10^22kg$
$\sqrt{\mu^2MG}=1.449 \cdot 10^{30}\frac{kg\, m^{3/2}}{s}$
$\frac{2}{5}m r_{moon}^2 = 8.87 \cdot 10^{34}kg\, m^2$
$\frac{2}{5}M r_{earth}^2 = 9.72 \cdot 10^{37}kg\, m^2$

and solve the equation (sum of all 4 components)=2.16*1034 Js for r. I drop the units, as they are all compatible.

$1.449\cdot10^{30}\cdot\sqrt{r} + 1.516 \cdot 10^{45} \cdot r^{-3/2} = 2.16\cdot10^{34}$

This has two solutions:
r1=21900km (15% above the roche limit for moon)
r2=212000km
The second solution is unexpected... maybe it is a stable solution, but requires different initial conditions to be reached.
Edit: No, the second solution is the relevant one. The orbit of moon is the dominant source of angular momentum, so it does not take too much to stop and re-accelerate earth in the opposite direction.

With the current orbit (same direction as rotation of earth), I get
r1=13500km (not possible)
r2=607000km (that looks good as final distance)

Last edited: Oct 12, 2012
18. Oct 14, 2012

### Ophiolite

I believe Eric is saying that your intuition is trumped by physics. Eric can you point us to any site, or accessible textbook with the relevant equations?

Oop! I posted that without seeing the last post from mfb.