# Measuring the exact number of millimetres between planets

humy
Science 13 Jul '13 18:51
1. 13 Jul '13 18:515 edits
http://phys.org/news/2013-07-interplanetary-precision-laser-mars.html

This laser ranging system should be a practical means to measure the distance between planets to an accuracy greater than within 1mm!
The only real catch is that you need the equipment to be at both ends of the distance you want to measure for this to work.
2. menace71
Can't win a game of
14 Jul '13 20:02
Originally posted by humy
http://phys.org/news/2013-07-interplanetary-precision-laser-mars.html

This laser ranging system should be a practical means to measure the distance between planets to an accuracy greater than within 1mm!
The only real catch is that you need the equipment to be at both ends of the distance you want to measure for this to work.
They do this already with the moon. I believe there is a plack on the moon and there is a university that shoots a laser at this plack to ascertain distance.

Manny
3. menace71
Can't win a game of
14 Jul '13 20:03
http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment

Manny
4. 14 Jul '13 20:103 edits
Originally posted by menace71
They do this already with the moon. I believe there is a plack on the moon and there is a university that shoots a laser at this plack to ascertain distance.

Manny
yes BUT, as the link I gave explains, that old type of laser ranging system used for the moon would not be practical for measuring the much greater distances between the planets. The new type they propose would be practical for measuring distance between the planets albeit with the the catch that you would still need to place equipment at both ends of the distance you want to measure. If you want to measure that distance without that catch, you could use radar. But radar would not give an accuracy of anywhere near within a fraction of 1mm!
5. 14 Jul '13 20:54
Originally posted by menace71
http://en.wikipedia.org/wiki/Lunar_Laser_Ranging_experiment

Manny
Why the improvement from 1/R^4 to 1/R^2
Surely the lasers now travel half the distance so it should be 1/(R/2)^4 ?
6. DeepThought
14 Jul '13 21:08
Why the improvement from 1/R^4 to 1/R^2
Surely the lasers now travel half the distance so it should be 1/(R/2)^4 ?
The laser source at the reflection end, say the moon, sends a signal back which deteriorates as 1/r^2. If the reflecting source is passive, i.e. a mirror, then because total flux the initial (earth-bound) source emits is distributed over a segment of the surface of a sphere, then the reflector only reflects ~1/r^2 of the emitted radiation, so when it gets back there is a 1/r^4 rule for signal strength (the inverse square law multiplies). When the reflector is active (in other words a powered laser) it produces a response signal that is as strong as it's designed to be and the returning signal is only degraded by 1/r^2.
7. sonhouse
Fast and Curious
14 Jul '13 21:35
Originally posted by menace71
They do this already with the moon. I believe there is a plack on the moon and there is a university that shoots a laser at this plack to ascertain distance.

Manny
The problem is the corner reflector left on the moon is only about a foot wide at best. It does reflect photons from Earth but only a few photons reach the receiver telescope on the ground. That is good for a couple hundred thousand miles but would not return enough photons a thousand times further out. You would be lucky to get one photon per day.
8. 15 Jul '13 05:20
Originally posted by DeepThought
The laser source at the reflection end, say the moon, sends a signal back which deteriorates as 1/r^2. If the reflecting source is passive, i.e. a mirror, then because total flux the initial (earth-bound) source emits is distributed over a segment of the surface of a sphere, then the reflector only reflects ~1/r^2 of the emitted radiation, so when it ge ...[text shortened]... al that is as strong as it's designed to be and the returning signal is only degraded by 1/r^2.
So is it a quantum/wave effect? ie even light not travelling in a straight line contributes to the signal. Because if light travelled in straight lines that wouldn't add up.
9. DeepThought
15 Jul '13 06:03
So is it a quantum/wave effect? ie even light not travelling in a straight line contributes to the signal. Because if light travelled in straight lines that wouldn't add up.
It doesn't really matter whether you treat it as a wave or a collection of particles. It's essentially impossible to get a perfectly collimated beam of light. In a particle model the particle's headings will be distributed over a solid angle, so although the particles are contained within a fairly tight solid angle (think of a shotgun) the number of particles going through a unit area goes as 1/r^2. In a wave model you'll get diffraction from any finite size aperture and the intensity of the wave front goes as 1/r^2. The better the collimation, the larger the constant of proportionality.

The difference between the particle and wave model is that to get the collimation perfect for particles you need the aperture infinitely narrow to get all the particles moving the same way. For waves you need the aperture infinitely wide to avoid diffraction effects.

sonhouse is right, if the reflector were big enough then all the light would be reflected and you'd get 1/(2r)^2.
10. 15 Jul '13 07:101 edit
So if I placed a plane halfway between Earth and Mars with a 1m diameter hole, the intensity of light from Earth reaching a given spot on Mars would be 1/R^4 ?
11. DeepThought
15 Jul '13 14:47
So if I placed a plane halfway between Earth and Mars with a 1m diameter hole, the intensity of light from Earth reaching a given spot on Mars would be 1/R^4 ?
Yes, here R is the distance to the plane rather than Mars. An infinite plane, strikes me as a little impractical though. You could use a disc with a hole cut in the middle, provided the diameter of the disc is big enough that it would eclipse the laser if the hole was filled in for any observer on Mars.
12. 15 Jul '13 15:45
Originally posted by DeepThought
Yes, here R is the distance to the plane rather than Mars. An infinite plane, strikes me as a little impractical though. You could use a disc with a hole cut in the middle, provided the diameter of the disc is big enough that it would eclipse the laser if the hole was filled in for any observer on Mars.
Suppose the observer on mars is a 1m diameter telescope. If we put a 1m diameter ball halfway between Earth and Mars then is the light that reaches our Mars telescope now more with my 1m diameter hole scenario?
13. menace71
Can't win a game of
17 Jul '13 02:07
Originally posted by sonhouse
The problem is the corner reflector left on the moon is only about a foot wide at best. It does reflect photons from Earth but only a few photons reach the receiver telescope on the ground. That is good for a couple hundred thousand miles but would not return enough photons a thousand times further out. You would be lucky to get one photon per day.
Interesting and makes sense

Manny
14. 17 Jul '13 07:18
Originally posted by sonhouse
The problem is the corner reflector left on the moon is only about a foot wide at best. It does reflect photons from Earth but only a few photons reach the receiver telescope on the ground. That is good for a couple hundred thousand miles but would not return enough photons a thousand times further out. You would be lucky to get one photon per day.
I watched this talk on SETI:

He says it is trivial to make a laser that can outshine the sun at interstellar distances in nano-second pulses. Surely with a 1m mirror on mars that would be more than sufficient to see the reflection?
15. DeepThought