A deeper understanding

There is a strong pattern in the behavior of the stage, which shows itself as soon as one simulates a month or two of its orbit. The orbit oscillates, becoming more circular for about 12 days, and then moving back towards being more eccentric...more egg shaped. What is driving this pattern? In this post we will dig in to understand it in more detail.

This is a plot I shared in an earlier post, simulating the first two months after the stage was placed in orbit, and the oscillation is visible as the waviness of the minimum and maximum altitudes. The peaks occur about once every 25 days.

A plot of stage altitude versus days in orbit shows a pattern of oscillating eccentricity. This pattern, repeating every 25 days or so, remains a fixture of the orbit even when simulated to the present day.

Here is another plot, simulating the stage orbit in the year leading up to the 50th anniversary of the mission. In this plot only the perilune points of each orbit are shown, but the same 25-ish day oscillation is still a major feature of the orbit.

The 25 day period is intriguing, since it is very close to the 27.32 days required for the moon to complete one rotation around its axis. But the difference between 25 and 27 is critical, since it means that the stage orbit oscillation is not directly tied of the moon's rotation. At least not in any obvious way.

When I first started running simulations of the stage, I was hoping to find an impact crater, and this required that the stage decay out of orbit quickly. Once I got going with the simulations, and the stage seemed to be long lived, there was one thing I noticed which kept me looking. I was interested in the lowest points of the orbit, the ones where the stage got below 20 km. Interestingly, these points always occurred at a similar location, around 30 degrees east longitude.

This plot shows low perilune points from a simulation of 3000 days. All the lowest points were centered around 30 degrees East longitude.

This result gave me some sliver of hope that I might yet find a crater...perhaps there was a lone mountain on the moon at 30 East, near the lunar equator, that acted as a giant catchers mitt, snagging the stage as it came whizzing by on a particularly low pass. So as I began to run longer simulations, I didn't just record the altitude of the perilune points. I also recorded their latitude and longitude.

One day, by accident, as I was plotting the results from a simulation run, I happened to plot both perilune altitude and longitude on the same graph, and something like this appeared:

The blue lines are longitude, while the orange line is altitude, from a simulation of a few months of 2019. What is important to notice is that there is a direct correlation between longitude and altitude. The peaks in perilune always occur near longitude 180 (or minus 180...same place) which is on the far side of the moon.

OK. Wow! That was a surprise to me. But something was not adding up. The plot is saying that the stage orbit perilune point moves all the way around the moon in 25 days. More correctly, what is happening is that the moon is rotating beneath the orbit, and it should take about 28 days to complete a full rotation. So how is the perilune point getting all the way back to the same longitude in less than 25 days? The answer is "precession".

Perhaps you have seen a gyroscope precessing as it balances on a pedestal. There is a nice video example here. Orbits also precess, and you can see a nice illustration (greatly exaggerated) here. In the case of the stage orbit, it is also slowly precessing, in a direction opposite to the rotation of the moon. The illustration below is not to scale, but shows how the major axis of the stage orbit changes over one month. The net effect is that the moon does not have to complete a full rotation for a given longitude line to come back under the low point of the stage orbit. It gets there a few days early. Just under 25 days.

So now the 25-ish day period of the perilune altitude oscillation makes more sense. It is a combination of the 27.32 day rotation of the moon under the stage orbit, combined with a slower precession of the orbit. This precession may also tie in to the slower cycles that are apparent in longer runs. The rate of the precession suggests that the stage orbit would precess all the way around the moon in about 10 months, which is about the period of the longer cycle.

I don't know what exactly is driving the change to the eccentricity of the stage orbit, but the fact that it is tied to longitude certainly points to a couple of possible explanations. It could be mascons along the lunar equator, tugging the stage to a greater or lesser extent. Or perhaps, since the same side of the moon is always facing Earth, and therefore the moon's rotation is closely related to its position along its own orbit, the relative positions of the Earth and Moon are somehow conspiring to tug the stage orbit in different directions. It remains a mystery to me. Nonetheless, I am happy to have a better understanding of the most prominent feature of the orbit of Snoopy's tail end.

Atmospheric Drag

Suppose it is true that Apollo 10 Lunar Module descent stage, aka "Snoopy", has not decayed out of orbit due to lumpy lunar gravity. What other factors might have brought the stage down in the intervening 50 years? That's a long time, so even subtle effects could come into play, including the nearly non-existent lunar atmosphere.

The space around the moon is nearly a perfect vacuum, by Earthly standards. But "nearly perfect" means that there are still thousands of molecules in each cubic centimeter of space around the moon. In fact scientists refer to the lunar atmosphere as an "exosphere" because there are so few molecules that they rarely collide with each other. They behave more like an army of tiny satellites, moving in response to gravity or electrostatic forces.

This instrument was placed on the moon's surface during Apollo 17 to study the moon's exosphere.

Source

According to this source, there are around 155,000 molecules in each cubic centimeter of the lunar exosphere. Yikes! That seems like a very big number. When you think that the descent stage is 12 feet on a side, and moving at almost a mile a second, that is a LOT of collisions with lunar molecules. Then take that out 50 years...now it's REALLY a lot. If we calculate the mass of all those molecules, and compare that to the mass of the stage, we can get an idea about whether the exospheric drag might have brought down the stage. If the mass of the gas is even close to the mass of the stage, that would be a real drag.

First of all lets estimate how many cubic centimeters the stage passes through every second. Just to get an upper bound on it, let's assume the stage is always moving with it's largest side facing forwards, sweeping through the largest area. We can approximate it as a square that is 12 feet on a side, which works out to an area of 140,000 square centimeters. (To get to this square I add the area of the foot pads into the missing corners of the stage, which is actually octagonal. Remember we are just estimating here.)

Now, the stage is moving about one mile every second. It moves faster when it drops down closer to the moon, and slower when it reaches it's "apolune" high point, but we can average that out and just assume a fixed velocity, as if it were in a circular orbit. One mile is 1.6 kilometers or 160,000 centimeters. Every second. Day after day. Month after month. Perhaps even decade after decade?

Well, now it seems like we might have a problem. If the stage is sweeping through an area of 140,000 square centimeters, and covering 160,000 centimeters per second, that works out to 22 billion cubic centimeters every second. And remember, if every cubic centimeter has 155,000 molecules, it means the stage is colliding with around 3.5 quadrillion molecules. Every second. Day after day. OK, then, perhaps all these collisions slowed down the stage, and it hit the moon after all?

Now we have to figure out how much all those gajillions of molecules weigh. Fortunately, remembering high school chemistry, this is easy. With a periodic table of elements, we take the atomic weight of each molecule, and then add them up. So for instance each Helium4 molecule has an atomic weight of 4 (or close enough for this estimate) so 40,000 of them total to 160,000 atomic units. Total up all the other molecules in the same way and it comes out to about 2.5 million atomic mass units.

OK, so now how much is that in units I can comprehend? It seems like a lot! Well, it turns out to be very little, because each molecule weighs very, very, very little. For instance one Helium atom, in grams, weighs about 0.00000000000000000000000166 grams. We better start using scientific notation, so that would be written as 1.66E-24. Now multiply that by the 2.5 million atomic mass units, and it come out to 4.13E-18 grams per cubic centimeter. Which is not much. (Yes, I am assuming the lunar exosphere is uniformly dense, which I know is not true. At this point I just want to get a feel for the problem.)

So now we have a war between big numbers and small numbers, that is, between the very large number of cubic centimeters (or cc's) that the stage sweeps through every second, 22 billion, versus the very small weight of the gas in each cc. So 22 billion times 4.13e-18 comes out to 9.2E-8 grams. Let's write that out as 0.000000092 grams. Every second. So far the small numbers are winning. That is the weight of the stuff the stage is running into every second, and it isn't much. And if we imagine each molecule as stationary when the stage strikes it, and moving away at the velocity of the stage after the collision, each collision will rob the stage of that tiny amount of momentum. (Think of a billiard ball striking another, stopping the first ball and sending the second one off with the momentum of the first.) So again, 9.2E-8 grams of collisions every second.

So now we just multiply by the number of seconds in 50 years. One year is 31.5 million seconds, and 50 years works out to 1.58E+9 seconds. Now the seconds help the big numbers to win the war, but not by much. The 9.2E-8 grams times 1.58E+9 seconds comes out to 146 grams. About one-third of a pound. Compare that to the weight of the stage. It's dry weight was 4700 pounds. So colliding with one-third pound of lunar exosphere over the course of 50 years could have slowed down the stage by 0.007%...i.e. nada!

So far, it seems that lumpy gravity didn't bring the stage down, and neither did atmospheric (or exospheric) drag. Perhaps Snoopy is still out there!

Simulating Uncertainty

Previously I posted about a simulation of the Apollo 10 LM descent stage which shows that the stage remains in lunar orbit to the present day. How robust is this result? The data for the initial stage orbit comes from the Mission Report...no doubt it was the best information they had. But fifty years in lunar orbit is a very long time, and lunar gravity is notoriously "lumpy". What if a slight change in the initial stage orbit state meant the difference between stability and decay?

To answer this question, I ran a set of 50 simulations, each with the initial conditions randomly varied to cover any possible miss in the initial state of the stage. I tried to keep the variation wide, to insure I covered the real conditions, but I also stuck to reasonable limits. In fact the variations I applied were so wide that many of the orbits were not viable. To cover this, I ran each parameter set through one orbit, recording the apolune and perilune...the low and high points of the orbit. I cut any set that was lower or higher by more than 20% from the values NASA reported. Only about one third of the random sets passed this test. To get 50 sets for the final test I passed more than 150 sets through the initial 1-orbit screen.

Here is a plot of the perilune points for all 50 random parameter sets, showing their minimum orbit altitude after 10 years in orbit. It's a bit messy, as these orbits show quite a bit of variation.

But the important thing to note is that in all 50 cases, the stage was still in orbit after 10 years. Each one of these plots is very similar to the "nominal" orbit I simulated initially. What if we just find the one of these, out of the 50, that got lower than any of the others, and plot it out by itself? Here it is.

You can see that this one did indeed make a rather low pass, in December of 1979, to about 12 km above the mean radius. (Still well above any lunar mountains.) And if you saw my earlier post about the stage orbit behavior, you see the same patterns here. The oscillation over a period of 25 days, and a longer oscillation with a period of around 5 months. Why did this one get lower than the others? It was one of the lowest initially, so it is hardly surprising. The real question is whether this one is any less stable over decades  than the "nominal" orbit that I showed before. What happens if we simulate this orbit out to the present? Here is the answer:

It is every bit as stable in it's orbit as the "nominal" case. There is no long term decay in evidence, and the simulated stage remains in orbit to the present day.

To me, this represents rather convincing proof. The result I got the first time I ran a simulation out to 50 years was no fluke. The nominal stage orbit is just one of a family of similar orbits that all exhibit long term stability. If something knocked the stage out of orbit during those 50 years, it wasn't the moon's lumpy gravity.