The U.S. and Russia routinely send laser signals to retro-reflectors that have been installed on the Moon in order to measure distances to our satellite accurately. The Moon is slightly over one light-second away and, therefore, the round trip for a photon would take a little over two seconds. A particle emitted from Earth would have trouble retracing its identical path after the Earth moved some 70 km from origin, especially if space is also warped. So how can the laser light be detected if the Earth has moved away from the return beam?


1 Answer 1


I presume your 70km comes from the $30{\rm km\,s^{-1}}$ Earth orbital speed about the Sun multiplied by the 2.5 second return trip time to the Moon for a pulse of light.

However, recall that, although the Earth orbits the Sun at about $30{\rm km\,s^{-1}}$, so too does the whole Earth-Moon system! To a very good approximation, the Earth-Moon system can be thought to be in freefall around the Sun (the only error is the Sun's tidal effects over the width of the system), and therefore the only relevant motion in this picture is that of the Moon relative to the Earth. The Moon's orbital speed about the Earth is about $1{\rm km\,s^{-1}}$, so it's simply a matter of aiming a little ahead (about 1 200 meters) of the retroreflector site to account for the relative motion. In any case, the laser beam is roughly seven kilometers in diameter by the time it reaches the Moon, so the method would still work even if one didn't take account of the Moon's motion. Presumably, the signal to noise is improved by accounting for the motion though, and this offset will become more important for future, more accurate experiments when the Moon retroreflectors are at last replaced.

General relativistic effects weigh on this problem negligibly: lightlike geodesics are radial lines from the Earth anyway and when we add the Moon's gravity to the mix, well, the deflexion of a light ray grazing the Moon's surface overestimates any deviation and, given the Moon's Schwarzschild radius $r_s$ is of the order of 90 micrometers (the Moon's mass is two orders of magnitude below that of the Earth and the SR of the latter is 9mm), the deflexion of such a ray is $2\,r_s/r_M$, where $r_M$ is the Moon's radius. Thus any deflexion is less than $10^{-10}$ radians, or $2\times 10^{-5}$ seconds of arc. As we are about to see, this is utterly negligible compared to the special relativistic aberration effects owing to the Moon-Earth relative motion.

The cleanest way to quantify the relativistic aberration owing to the Moon's relative motion is to calculate the images of the rays incident on and reflected from the Moon retroreflector under the Lorentz transformation linking the Earth and Moo frames. Assuming the boost between the frames to be in the $x$ direction, the image of the wavevector one-form $K=\left(\frac{\omega}{c},\,-k_x,\,-k_y,\,-k_z\right)$ under an $x$-boost of rapidity $\eta$ is $\left(k\,\cosh\eta+k_x\,\sinh\eta,\,k\,\sinh\eta+k_x\,\cosh\eta,\,k_y,\,k_z\right)$. By calculating the images of a wavevector and one travelling in the opposite direction in the Moon frame, {\it i.e.} by calculating the images of $K_\pm=\left(\frac{\omega}{c},\,\mp k_x,\,\mp k_y,\,\mp k_z\right)$, the deviation from $180^\circ$ in the angle between them in the satellite frame is readily found to be, to first order in $\eta$:

$$\Delta\Theta = 2\,\eta\,\sin\theta\approx 2\,\eta\tag{1}$$

where $\theta$ is the angle between the light ray and the boost in the Earth frame. With the Moon moving always tangentially (almost), $\theta \approx 90^\circ$ at all times.

For the rapidity of the Moon's motion at $1{\rm km\,s^{-1}}$ relative to Earth, this aberration amounts to $7\times10^{-6}$ radians, or about $1.4$ seconds of arc. This amounts for a pointing error of about $2.8{\rm km}$ in the return trip back to Earth. This probably wouldn't affect the Apollo retroreflectors too much as the laser beam is about $20{\rm km}$ in diameter by the time it reaches the Earth, but some improvement in signal to noise could be expected by accounting for the pointing error. The error can be nulled by offsetting the faces of the retroreflecting corner cubes so that the total deviation of the reflexion from $180^\circ$ is $1.4$ seconds of arc in the appropriate direction; I am not sure whether or not this is done for the Moon retroreflectors for the reasons cited above. $1.4$ seconds of arc is about the limit of accuracy of standard corner cubes (hollow cornercubes, comprising three separate orthogonal mirrors which are positioned by hand before gluing to minimize error); see for example, Edmund Optics' hollow cornercube offering here. Presumably the Moon retroreflectors are better than this and the small error will be worth accounting for in future systems.

One very similar case where I do know that the optical aberration is systematically accounted for is in the LAGEOS retroreflectors. See:

David A. Arnold, "Optical and infrared transfer function of the lageos retroreflector array", M.J. Enos Grant number NGR 09-015-002, Smithsonian Institution Astro-physical Observatory, Cambridge, Massachusetts, USA, May 1978

The LAGEOS satellites orbit at an altitude of about $6 000{\rm km}$ (about two Earth radiusses from Earth's center), thus their relative motion to the Earth is about $1/\sqrt{2}$ times the LEO orbital speed of $7.8{\rm km\,s^{-1}}$, that is, the aberration from (1) is about 4 seconds of arc, implying a pointing error of about 110 meters by the time the reflected laser reaches Earth. It is well worth offsetting the cornercube faces for these devices as the laser diameters are much smaller and indeed this is precisely what is done, if you look at the above document.

  • $\begingroup$ I've just asked: Have LAGEOS' germanium corner cube reflectors ever been used? That's great to know about the offsetting of the cubes' planes; I'll have to read more about that. Curious where the figure of 20km comes from (size of Lunar reflector footprint back on Earth). If I assume diffraction of a 500nm wavelength by a 5cm aperture I get an angle of order 1E-05. for the useful intensity peak. At 300,000km that's 3km. Is diffraction not the source of the size, or my starting values wrong? $\endgroup$
    – uhoh
    Commented Sep 13, 2017 at 10:19
  • $\begingroup$ re: "This probably wouldn't affect the Apollo retroreflectors too much as the laser beam is about 20km in diameter by the time it reaches the Earth" (tiny) photo of some ~2cm Apollo retroreflectors and a green laser can be seen here. $\endgroup$
    – uhoh
    Commented Sep 13, 2017 at 10:31
  • 1
    $\begingroup$ @uhoh No, your calculation seems good and agrees with mine. I'll have to look at the source: there is probably considerable atmospheric spoiling of the beam on the way out / in. Also, there is some uncertainy in definition: my calculation gives 5km 1/e^2 diameter for a laser that size over 400000km. There will probably be detectable light levels outside this region. $\endgroup$ Commented Sep 13, 2017 at 10:42
  • 1
    $\begingroup$ @uhoh BTW: the corner cube offset is four seconds of arc, not minutes. I've just corrected it. $\endgroup$ Commented Sep 13, 2017 at 10:44
  • 1
    $\begingroup$ Ya, when the backs of two people's envelopes agree within a factor of 10, it's time to look at the fronts of the envelopes. $\endgroup$
    – uhoh
    Commented Sep 13, 2017 at 10:48

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.