Earth's Shadow (APOD 20 Aug 2008)

[NoelC]

Great post, neufer. I learned a thing or two, and you got me a thinkin'...

The moon probably has a liquid core of about 20% of the Moon's radius.

I can only guess that they must see it wobbling or something to infer this.

The universal force of gravity is very stable. The experiments have put an upper limit on the change in Newton's gravitational constant G of less than 1 part in 1011 since 1969.

I had no idea there was a concern over whether the force of gravity might be changing over time. I view the concept as ultimately possible, I just didn't realize anyone thought it might be something worth testing. A change of 1 part in 1011 isn't really small potatoes. Do they really think it might be changing near that much? I'd think we'd have no luck at all with missions to the outer solar system if we didn't have gravity nailed down to eleventy seven decimal places...

Oh and lastly...

The moon is spiralling away from Earth at a rate of 38 mm per year

Anyone got a good reference for an explanation of why this is? Seems to me if the Moon is dragging the tides around on the Earth, it ought to be losing energy, not gaining it. Is it because the Earth is spinning some 28 times faster than the moon's orbital period? Is the Earth slowing down?

-Noel

[henk21cm]

G'day Art,

You quoted the wiki on the reflectors on the moon. More specifically: the accuracy which was improved to near mm accuracy. That means that the time of flight must be accurate to 10 ps, (100 GHz being 3 mm). The time of flight is about 2.5 s, so a remarkable accuracy. That has rizen a few questions, which i boldly will pose.

• The space between moon and earth is not empty. Solar wind spoils the pure vacuum. Therefore the speed of light is somewhat slower than in vacuo. What is the magnitude of the correction for non-vacuum conditions? And how do the fluctuations in solar wind influence these measurements?
• Why is the laser light relefected so preceisely back to earth and not somewhere near the earth? Due to the non constant orbital velocity of the moon in comparison with the constant axial rotational period, the moon shakes her head, so it is virtually impossible to align a mirror that it reflects the light back to earth.
• How does the ever changing path (different heights above the horizon) of the laser through the earths atmosphere influence the measurements?
• From what you describe i guess that one needs rather elaborate equipment to send and receive pulses. Other than moonbounce, as done by radio amateurs on 2m and 70 cm (and 23 cm) is it possible for us, the common amateur astronomer, to send out light pulses to the moon and receive these pulses back?
• On radio frequencies the Faraday rotation is a real pain in the neck. To overcome this phenomenon, you need to switch to circular polarisation, since the resulting polarisation is all over the place. Are there any problems due to Faraday rotation with these laser pulses?

[Chris Peterson]

The moon probably has a liquid core of about 20% of the Moon's radius.

I can only guess that they must see it wobbling or something to infer this.

To be fair, this is not established. There are various well regarded theories about the Moon's core, including little iron or a small, solid iron region.

I had no idea there was a concern over whether the force of gravity might be changing over time.

There is a strong interest in measuring all physical constants to extreme precision in order to determine if they are, indeed, constant. The gravitational constant is one of the hardest to measure, and has only been established to about one part in 10^13. Many physical constants have been determined with much higher accuracy.

The moon is spiralling away from Earth at a rate of 38 mm per year

Anyone got a good reference for an explanation of why this is? Seems to me if the Moon is dragging the tides around on the Earth, it ought to be losing energy, not gaining it. Is it because the Earth is spinning some 28 times faster than the moon's orbital period? Is the Earth slowing down?

The tidal drag is slowing the Earth down. The rotational angular momentum of the Earth is being transferred to orbital angular momentum of the Moon. So the radius of the Moon's orbit increases.

[Chris Peterson]

Why is the laser light relefected so preceisely back to earth and not somewhere near the earth?

The reflectors are arrays of corner prisms, aka retroreflectors. These cannot help but to send reflected light back along the same path it entered on.

How does the ever changing path (different heights above the horizon) of the laser through the earths atmosphere influence the measurements?

This has to be compensated for. I believe the current method involves measuring scatter and fluorescence from the beam in the atmosphere to determine the path characteristics.

From what you describe i guess that one needs rather elaborate equipment to send and receive pulses. Other than moonbounce, as done by radio amateurs on 2m and 70 cm (and 23 cm) is it possible for us, the common amateur astronomer, to send out light pulses to the moon and receive these pulses back?

It would be difficult. Even using large mirrors (several meters) to collimate the beam, the laser pattern is kilometers across by the time it reaches the Moon. Only a tiny fraction is reflected by the retroreflector array. So you need a big telescope and a powerful laser (powerful enough that FAA clearance is required to shine it upwards).

[neufer]

The moon probably has a liquid core of about 20% of the Moon's radius.

I can only guess that they must see it wobbling or something to infer this.

Probably the lunar "libration" wobble has different dynamics
http://antwrp.gsfc.nasa.gov/apod/ap070602.html
(just as an uncooked egg twisting at the end of a rope will have different dynamics from a hard boiled egg.)

The universal force of gravity is very stable. The experiments have put an upper limit on the change in Newton's gravitational constant G of less than 1 part in 1011 since 1969.

I had no idea there was a concern over whether the force of gravity might be changing over time. I view the concept as ultimately possible, I just didn't realize anyone thought it might be something worth testing. A change of 1 part in 1011 isn't really small potatoes. Do they really think it might be changing near that much? I'd think we'd have no luck at all with missions to the outer solar system if we didn't have gravity nailed down to eleventy seven decimal places...

Of course that's 1 part in 10^11 which totally demolishes
Dirac's suggestion of ~ 1 part in 10^10 per year.
http://en.wikipedia.org/wiki/Dirac_larg ... hypothesis

The moon is spiralling away from Earth at a rate of 38 mm per year

Anyone got a good reference for an explanation of why this is? Seems to me if the Moon is dragging the tides around on the Earth, it ought to be losing energy, not gaining it. Is it because the Earth is spinning some 28 times faster than the moon's orbital period? Is the Earth slowing down?

Because the Earth is spinning some 28 times faster than the moon's orbital period the earth's (moon generated) tides actual precede the moon such that these tides are dragging the moon around. The earth thereby looses angular momentum which the moon gains. Days are getting longer.... but then so are months since the moon is flung outward to longer orbits of higher angular momentum.

[neufer]

[*]On radio frequencies the Faraday rotation is a real pain in the neck. To overcome this phenomenon, you need to switch to circular polarisation, since the resulting polarisation is all over the place. Are there any problems due to Faraday rotation with these laser pulses? [/list]

The part of the project I worked on (in 1968) was the corner reflectors:

A linearly polarized laser beam into the reflector comes back out the way it came in but...it comes out elliptically polarized. (This does not pose much of a problem however.)

[neufer]

Why is the laser light relefected so preceisely back to earth and not somewhere near the earth?

The reflectors are arrays of corner prisms, aka retroreflectors. These cannot help but to send reflected light back along the same path it entered on.

Of course the moon is moving 1.022 km/s so there is the problem of light aberration (which is different on the way out that it was on the way in); hence, the peak of the return beam misses the earth telescope by about 2 miles. (One can thing of it as the moon giving the reflected photons some angular momentum forward.)

http://en.wikipedia.org/wiki/Aberration_of_light

[henk21cm]

The reflectors are arrays of corner prisms, aka retroreflectors. These cannot help but to send reflected light back along the same path it entered on.

In the wiki i can read that it differs from a plane mirror. It is a constellation of three perpendicular perfectly flat reflecting planes. The reason (proof or explanation) why for any incident angle the light is reflected in the same direction, is missing. I guess its a matter of repetative application of Snellius law.

Regarding the ever changing path (different heights above the horizon) of the laser through the earths atmosphere

This has to be compensated for. I believe the current method involves measuring scatter and fluorescence from the beam in the atmosphere to determine the path characteristics.

First of all there is refraction, which means that the atmospheric pressure influences the speed of light and so the apparent distance between moon and earth. (The actual deviation as caused by refraction is not the key issue). As far a i understand scatter is mainly caused by microscopic airborne particles, next to Rayleigh scatter. The reason for fluorescence i do not understand. If you could be so kind to elucidate that aspect, it would be appreciated.
Second is a geometrical aspect: the location of the observer on earth is varying. A rather precise description of the geoide (the shape of the surface of the earth) is needed, since a simple sphere is too inaccurate.

Regarding DIY experiments

It would be difficult. Even using large mirrors (several meters) to collimate the beam, the laser pattern is kilometers across by the time it reaches the Moon.

A mirror of several meters is far beyond amateur equipment: i must forget it.

The part of the project I worked on (in 1968) was the corner reflectors:
A linearly polarized laser beam into the reflector comes back out the way it came in but...it comes out elliptically polarized. (This does not pose much of a problem however.)

So the detector must be of the non-polarized type. And again one loses some sensitivity.

[Chris Peterson]

In the wiki i can read that it differs from a plane mirror. It is a constellation of three perpendicular perfectly flat reflecting planes. The reason (proof or explanation) why for any incident angle the light is reflected in the same direction, is missing. I guess its a matter of repetative application of Snellius law.

It's simple to see why it works if you first consider the 2D case. The only rule you need is that the angle of incidence equals the angle of reflection. Snell's law deals with refraction, which isn't involved here.

Regarding the ever changing path (different heights above the horizon) of the laser through the earths atmosphere

This has to be compensated for. I believe the current method involves measuring scatter and fluorescence from the beam in the atmosphere to determine the path characteristics.

First of all there is refraction, which means that the atmospheric pressure influences the speed of light and so the apparent distance between moon and earth. (The actual deviation as caused by refraction is not the key issue). As far a i understand scatter is mainly caused by microscopic airborne particles, next to Rayleigh scatter. The reason for fluorescence i do not understand. If you could be so kind to elucidate that aspect, it would be appreciated.

The laser stimulates certain atoms in the atmosphere (such as sodium) to fluoresce. By picking of part of the return signal and passing it through the appropriate filters, this can be detected. By measuring the timing precisely, some characteristics of the atmosphere can be determined. I don't know how critical this is, as the passage through the atmosphere is very short. A simple model of density may be all that's required to reach the point where something else is limiting accuracy.

[NoelC]

Regarding the Moon orbiting the Earth...

Not to detract from the APOD that started all this, but I did an animation of a number of lunar still images taken at intervals of several seconds... I find viewing it helps make the rotation of the moon around the Earth seem more real. Note that the crescent moon eclipses a star at one point, then some thin clouds blow through.

There's a little bit of wobble in the sequence; that's my fault for not getting the frames perfectly aligned. Sorry about that.

Note that the part of the moon you can see mostly in this image is illuminated by earthshine.



-Noel

[henk21cm]

G'day Chris,

It's simple to see why it works if you first consider the 2D case.
The only rule you need is that the angle of incidence equals the angle of reflection. Snell's law deals with refraction, which isn't involved here.

It is not my intention to be rude, but the phrase "It is simple to see" is synonymous with "the author did not bother to elaborate, after three weeks of intensive calculations the result can be found". Additionally "Typical results are shown in the graph" should be read as "These are the best results". "The sample was accidentally strained" should be read as "It dropped on the floor". Nevertheless i'll do my best. By the way, Snellius law is involved, since a retroreflector has a boundary between the material it is made of and the vacuum between earth and moon. As far as the reflections are concerned, indeed angle of reflection is equal to the incident angle.

The laser stimulates certain atoms in the atmosphere (such as sodium) to fluoresce. By picking of part of the return signal and passing it through the appropriate filters, this can be detected. By measuring the timing precisely, some characteristics of the atmosphere can be determined. I don't know how critical this is, as the passage through the atmosphere is very short. A simple model of density may be all that's required to reach the point where something else is limiting accuracy.

Using IR one might be able to excite the rotational bands of CO2. Lets send out a pulse of light. It travels through roughly 60 km of atmosphere. Reflection is likely proportional to the atmospheric pressure (the more atoms, the more reflection). So during the first 0.4 ms a sharply decreasing reflection pulse is recorded. Assuming that the probability per atom for reflection is constant, it is possible to derive some sort of distribution for the pressure. Assuming that the speed of light as function of pressure is known, from the reflection curve during the first 0.4 ms the extra time delay (with respect to vacuum) can be deduced. Is that what you mean?

[Chris Peterson]

It is not my intention to be rude, but the phrase "It is simple to see" is synonymous with "the author did not bother to elaborate, after three weeks of intensive calculations the result can be found".

I simply meant that if you made a simple drawing of the 2D case, it should become pretty clear how a retroreflector works. But I can do that:



I've drawn two rays, and you can see that each exits at the same angle it enters, because at each reflection the entry and exit angle is the same. In general, for a ray that strikes the first surface at an angle theta with respect to that surface, it reflects at angle theta from that surface. It strikes the next surface at angle 90 - theta, and therefore reflects at angle 90 - theta. Since the second surface is rotated 90° with respect to the first, the total reflection angle is 180 - theta; that is, a 180° reflection.

By the way, Snellius law is involved, since a retroreflector has a boundary between the material it is made of and the vacuum between earth and moon.

That's Snell's Law, not Snellius (in spite of the fact that it's named after Snellius). And you don't need to use it. Retroreflectors made as mirrors have no boundary between different indexes of refraction, so the law doesn't apply. Retroreflectors made as prisms do show refraction, but if you apply Snell's law you'll see that the entrance and exit refraction cancel out, so again you can just consider the system as reflective.

Using IR one might be able to excite the rotational bands of CO2. Lets send out a pulse of light. It travels through roughly 60 km of atmosphere. Reflection is likely proportional to the atmospheric pressure (the more atoms, the more reflection). So during the first 0.4 ms a sharply decreasing reflection pulse is recorded. Assuming that the probability per atom for reflection is constant, it is possible to derive some sort of distribution for the pressure. Assuming that the speed of light as function of pressure is known, from the reflection curve during the first 0.4 ms the extra time delay (with respect to vacuum) can be deduced. Is that what you mean?

That might be one approach. The method I was referring to depends on the knowledge of the height that the fluorescent species is found, and looking at the time of flight. I'm sure there are many methods that can be used for characterizing the atmosphere, assuming it is actually necessary for lunar ranging.

[henk21cm]

I simply meant that if you made a simple drawing of the 2D case, it should become pretty clear how a retroreflector works. But I can do that:

Thanks Chris, now i see it. During the Apollo flights i heared about these reflectors but i never understood how they worked. In 1969 the internet and Google were not even dreams in the eyes of Darpa, let alone a school boy. That retroreflector riddle nagged me for almost 40 years!

That's Snell's Law, not Snellius (in spite of the fact that it's named after Snellius).

Sorry, that is a language related item, in my language it is called "De wet van Snellius".

And you don't need to use it. Retroreflectors made as mirrors have no boundary between different indexes of refraction, so the law doesn't apply.

Just another fact (hollow structure) i did not know. I was thinking of chunks of glass, with perpendicular faces polished on them. And even for the solid structure: the incident angle and the returning angle inside the solid are the same, and since refraction at an interface is invariant under reversion, it does not matter.

assuming it (correction for air) is actually necessary for lunar ranging.

Interesting remark. It made me wonder and trigger some activity. In the handbook of Chemistry and Physics (52nd edition) on page E204 the refractive index of air is given. At λ=600 nm the refractive index is 1.00028. Lets assume that after 60 km the refractive index is 1. On average (very crude 0th order approximation, the air pressure does not decrease linearly) the refractive index is 1 + 1.5E-4 (= 1+a). The speed of light in the atmosphere is then

v = c/(1+a).

The time needed for a pulse of light to cross the atmosphere is:

Δt = H/v = (H/c)(1+a) = Δt_{0} + Δt_{extra}

The extra time needed due to the atmosphere is

Δt_{extra} = Ha/c => 60 * 1.5E-4 / 3E5 ≈ 30 ns.

30 ns is about 9 m in length. So neglecting the atmosphere is not an option, if you want to find something like 38 mm.

I found an empirical relation by Edien for the refractive index on page 5 of http://www.feanor.com/laser_principles.pdf, but i have serious doubts about the size of the Δn correction in its equation (5). According to the sensitivity 0.0039 in stead of 0.057 in the exponent is more likely. I can't trace down the original article by Edien.

[Czerno]

Re: lunar ranging. Can someone provide a calculated or estimated value for how much higher the intensity of the light received would be (per unit surface) if the detector were near the center of the returning beam ?

Assuming it would be some order of magnitudes higher than the meager count of photons detected at the transmitter site, then why not offset a smaller, lighter telescope to a location at the right distance west of the emittor ?

Understanding the "right" distance has to be varied, depending primarily on the actual earth-moon distance at the time of the experiment, such a smaller detector could be mobile. Quick mental calculation, knowing that actual eart to moon distance varies by as much as ~ +-10%suggests an 100-200 meter long, east-west oriented track, with its middle point some 2.5km west from the main apparatus (transmitter).

Does this scheme calculate out, both scientifically and economically or am I grossly wrong ?

--
Czerno

[Chris Peterson]

Re: lunar ranging. Can someone provide a calculated or estimated value for how much higher the intensity of the light received would be (per unit surface) if the detector were near the center of the returning beam ?

Just estimating, but I'd say there's little difference. The beam has lost coherency by the time it gets to the Moon, and the light that reflects back isn't a tight beam at all. It illuminates a large area on the Earth- much larger than the distance the Earth moves during the time-of-flight.[/quote]

Assuming it would be some order of magnitudes higher than the meager count of photons detected at the transmitter site, then why not offset a smaller, lighter telescope to a location at the right distance west of the emittor ?

In any case, the difference is not orders of magnitude. You wouldn't want to make the telescope much smaller, because you still have a weak signal. And keeping track of the exact time difference between the sending an receiving would be more of a challenge using different sites.

AFAIK, the setup in Texas receives plenty of signal for the necessary measurements. Since it isn't signal limited, there would seem to be no advantage in adding another scope, even if the return signal could be stronger someplace else.

[Czerno]

Just estimating, but I'd say there's little difference. The beam has lost coherency by the time it gets to the Moon, and the light that reflects back isn't a tight beam at all. It illuminates a large area on the Earth- much larger than the distance the Earth moves during the time-of-flight.

Ah, I see, so twas a false good idea. One more
Be as it may, centimetre to millimetre precision on those absolute distance measurements is fantastic ! Have the results already served as checks against the "theories" that were used previously for the calculation of ephemerids - Brown's, Spenser Jones's - or is it too early ?

thanks, and regards