Built on Facts

All right, I’m gonna delay the next installment of the quantum bouncing ball for a brief diversion. I have a friend who’s also a physics grad student, and he suggested that we along with a few other fellow students form a yet-to-be-named unofficial club whose raison d’etre is to get together every few weeks and do interesting Mythbusters-style science for no good scientific reason. After all, we are fortunate enough to be surrounded by expensive cutting edge equipment and brilliant people, and we should have some scientifically interesting fun.

One suggestion for something we might try was inspired by a Big Bang Theory episode, which was itself inspired by a real experiment. The idea is to measure the distance to the moon with a laser. Shoot a pulse of light at the moon, measure the time it takes to bounce back, and using the known speed of light you can calculate the distance to the moon. Since the speed of light is about 1 foot per nanosecond, measuring with nanosecond accuracy (actually pretty easy) could mean measuring that rather vast distance to within the length of a ruler. In practice other effects like atmospheric dispersion increase the uncertainty, but these can be corrected for. While the lunar surface itself is reflective enough to do the experiment unassisted, the Apollo astronauts also left behind retroreflectors which make this experiment much easier and much, much more precise. (It turns out to increase the total return of photons by about a factor of 100, and eliminates the problem of varying elevations in the topography.)

i-77e4580d39c7836ec968c178aff930dd-280px-Full_Moon_Luc_Viatour.jpg

Fig 1: A harsh mistress.

So to do this you need a laser that spits out short (ie, nanosecond or shorter) pulses with as high an energy per pulse as you can muster. You need high energy because you ain’t gettin’ much energy back from the moon. We should try to put numbers on this, to see if anything we have around the lab might work.

First, we need to see how much of the laser light will make it to the retroreflector on the moon. Lasers travel in pretty straight lines, but as anyone with a laser pointer can verify, they do spread out some with distance. Typical values for the lasers we use might be somewhere in the 1 milliradian range. The distance to the moon is roughly 400,000 km, so the portion of the lunar surface illuminated by our laser will be about (0.001)*(400,000 km) = 400 kilometers in diameter. This is an area of about 125 billion square meters. If the retroreflector is one square meter, only about 1 part in 10^11 of our emitted light even makes it to the reflector. Now the reflected light has to make it back to the earth. If we’re extremely optimistic, we might say that the reflector introduces no extra angular spread, and its reflected light is spread over a 400 km diameter of the earth’s surface. To a first approximation this just means the total there-and-back efficiency is about 1 part in (10^11)^2, or one photon in 10^22. This, my friend, is Bad.

But hey, we can use a photomultiplier tube to count individual photons, and photons are pretty tiny units of energy. Our laser sends out a heck of a lot of them. The energy of one photon of wavelength λ is:

The laser we’d likely use has a wavelength of 532 nanometers, and plugging into the equation we find that each photon has an energy of about 3.7 x 10^-19 joules. Therefore we’d need about 1000 joules per pulse to get up to the neighborhood of 10^22 photons per pulse. And we need 10^22 photons just to get one photon back per pulse on average.

We have a few compact 15 mJ/pulse q-switched Nd:YLF lasers with 1 KHz rep rates, and we even have a few not-so-compact 2 J/pulse with a 10 Hz rep rate. With a 2 second round trip time, the repetition rate isn’t so relevant since we can effectively only use one pulse per round-trip time. Even 2 joules won’t cut it unless we’re willing to do many thousands of shots worth of statistics. And that’s without even thinking about noise, which will not be inconsequential even with good filtering.

So if it’s that hard for us, how does NASA do it? First and most importantly, they use a laser with a divergence of a microradian or so. This reduces the spot diameter by a factor of a thousand compared to ours, and thus increases the intensity at the lunar surface by a factor of a million. There’s an inverse relationship between beam divergence and beam waist size, so with some beam-expanding optics this is something we could do in theory – but it’s not trivial and at that point it’s not so much a fun project to spend a few hours on as it is a major research effort, especially since a smaller illuminated lunar area means aiming is that much trickier. Also, NASA’s lasers tend to have higher pulse energies.

Thus I regret to say that lunar ranging is probably not in our club’s future. It’s also probably not something the intrepid cast of the Big Bang Theory could have done on their roof with a few shots from a small laser, given that it wasn’t passing through collimating optics (ie, a large telescope). Still, it’s not impossible and I think a dedicated and well-equipped amateur group could probably do it. If you succeed, let me know!

Comments

#1 Lassi Hippeläinen
December 16, 2010

It’s easier with radio. Even ham operators do it.
http://en.wikipedia.org/wiki/EME_%28communications%29
#2 Ned Wright
December 16, 2010

The APOLLO (see http://en.wikipedia.org/wiki/Apache_Point_Observatory_Lunar_Laser-ranging_Operation) experiment gets average return rates of 0.25 detected photons/shot with 0.115 J/shot of 532 nm light.
See http://arxiv.org/abs/0710.0890

This is 1.2 billion billion photons transmitted for every one received.

They want to get 1 mm accuracy in the range to the Moon. Very cool physics.
#3 Chuck
December 20, 2010

Is there any reason that you cannot use pseudonoise ranging techniques? These allow you to integrate measurements over many pulses. I think they are described in Horowitz and Hill and they were used to measure the range to Venus in the 1960s. See http://ieeeghn.com/wiki/index.php/Oral-History:Paul_Green#Lincoln_Lab_at_MIT.3B_Venus_radar_experiment.

Chuck
#4 Anonymous Coward
December 20, 2010

To get your there-and-back efficiency, you had to plug in some number for the area of your light-collecting optics on the earth. I didn’t see that number, but since your earth-to-moon and moon-to-earth efficiencies were the same, I’ll assume you have a 1 meter telescope?

Since you have a 1 meter telescope lying around, you can use it for both transmission and detection. This could give almost three orders of magnitude improvement over the the 1 milliradian number you plugged in, giving plenty of photons even for your 15 mJ laser. You could do a pretty good measurement (and verify the existence of the retroreflectors) with a couple seconds of data.

But I still agree with your conclusion: setting it up isn’t trivial.
#5 ppnl
December 21, 2010

Why are you limited to one pulse per round trip? You just watch for a photon at some time window after a pulse. There is no reason you can’t have multiple flashes and multiple time windows. If you do ten pulses per second you just have about twenty time windows to look for a returning photon.
#6 complex field
December 22, 2010

“…and verify the existence of the retroreflectors…”

Didn’t you hear? The LRO placed the retroreflectors as part of the ongoing conspiracy to fool the world, etc, etc,…
#7 plumbing supplies
January 10, 2011

Lunar eclipse occurs when the moon passes behind the earth so that the earth blocks the sun’s rays from striking the moon.It looks cool to see and amazing how everything happen above.
#8 Lyle
January 14, 2011

#4 catches the trick NASA Uses, all be it with bigger scopes. Initially they used the 120 inch at lick way back when. Today as noted a 40 inch scope could do the job with the better electronics. Now one of the questions is what f number for the scope, the higher the f number the narrower the field of view (or in this case the spot illuminated running the telescope in reverse). A higher F number would mean a smaller spot. 40 inch dobsonians do exist in private hands and if at a major science school there might be an old 40 inch scope laying around, so its possible. I am not sure but I heard that for a while at least the old 100 inch at Mt Wilson was mothballed, so it is a candidate.
#9 BoltonAudrey
December 2, 2011

I received my first credit loans when I was 25 and that aided my relatives very much. However, I need the auto loan over again.
#10 uncle quentin
March 20, 2012

Well I don’t know much about it but I did hear that after a short distance laser spreads according to the inverse square law – it must, as electromagnetic radiation. Therefore the signal is spread over a diameter of maybe 70,000 km (or miles, I forget I tried to approximate it) at moon distance. In other words no apprieciable signal reaches the moon. I worked out that the light on the retroreflector would be the equivalent of less than a hundred millionth of a watt, give or take. An “attenuation” of the initial billion watt signal. For the return “trip” there will be a similar reduction in intensity to the already miniscule signal. It is very clever how they do it, especially when you consider that atmospheric scintillation limits views of the moon to a resolution of one second of arc or approx one mile, Even allowing for the good seeing at the observatory these errors compound. Presumably a “moonphoton” could have come from one mile to the left or the right so the uncertainty doubles. Then there is the same difraction (which past a point is unpredictable) to deal with for the RETURNING laser light…Centimetre/milimetre precision? It may be so but I take it with a severe pinch of salt.
#11 uncle quentin
March 20, 2012

I meant to say atmospheric “refraction” not that other word that I can’t spell.
#12 uncle quentin
March 20, 2012

I’m not scientifically trained but I do have a layman’s interest in astronomy. Therefore I may be talking nonesense but here’s another take on this. I remember reading that the best telescopes on Earth were sensitive enough to detect a lighted candle on the moon (if one were up there). If we take a candle as the minimum light-source we can detect and assume 100% efficiency for sun-seared, 40 year-old reflectors it means that the equivalent of one candle-power is falling on that square metre of reflector. That’s enough light to read by. Bear in mind how diffuse this beam would be. I guess it would at minimum several moon diameters. And thinking as I type this (never a good idea)if the moon’s reflectivity (albedo.07) is 7% of a perfect retroreflector isn’t almost infinitely more likely that a returning photon would have been reflected from elsewhere on the moon’s surface? The Apache Point mirror is three and a half meters by the way (respect!)_