Re: Will Somebody PleaseTell bz What an Inertial Frame is.

"Arthur Dent" <jp006t2227@xxxxxxxxxxxxxxxx> wrote in
news:1120166750.294775.113260@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:

>
>
> bz wrote:
>> "Arthur Dent" <jp006t2227@xxxxxxxxxxxxxxxx> wrote in
>> news:1120153968.252223.60740@xxxxxxxxxxxxxxxxxxxxxxxxxxxx:
>>
>> >
>> >
>> > bz wrote:
>> >
>> >> That usually happens with good ideas.
>> >>
>> >> > Newsgroups: sci.physics.relativity
>> >> > From: "Androcles" <androc...@xxxxxxxx> - Find messages by this author
>> >> > Date: 1999/02/23
>> >> > [quote]
>> >> > Send a laser
>> >> > and a maser into orbit, bounce a signal to the moon and back at
>> >> > moon-rise
>> >> > and time it. Repeat at moon-noon and moon-set 45 minutes later.
>> >> > Get the difference. Then we'll know.
>> >> > [unquote]
>> >>
>> >> If Henri is right, photons, even from a surface laser, at moonrise and
>> >> moonset, should do. Compare to 'at zenith. Correct for the atmosphere.
>> >
>> > Phooey, what does Henri know...
>> >
>> > Tangential speed of Moon relative to Earth: 360 degrees/month.
>> > Radial speed of Moon relative to Earth: (Apogee - Perigee)/14 days =
>> > practically zilch.
>>
>> tangential speed of earth = circumference/24 hr = 1038 mph
>
> Ok... that's quite lot slower than an orbiter, Concorde and most
> military fighters could outpace it.

Of course.

>> > Speed of light in atmosphere c/n, n the refractive index, but n tends
>> > to 1 as the density goes down.
>> > Speed of light from Earth to Moon, c.
>>
>> > Tangential speed of ISS and shuttle orbiters relative to Earth, > 5
>> > km/sec
>> > Speed of light from orbiter to moon, c + cos(phi) * 5 km/sec, not
>> > zilch.
>> > Beats me when people are going to wake up to the concept of relative
>> > speed, even for light.
>>
>> You might only be able to count on the c +/- 1038 mph for the earth to
>> moon leg. Figure that the trip back will be made at c.
>
> Wasn't it you that said a mirror wouldn't change the VELOCITY of light,
> and I that said it would, v = -v, and then you agreed? So how come you
> are contradicting yourself now? Get a grip, old son.

Lunar corner reflectors are not mirrors.
http://physics.ucsd.edu/~tmurphy/apollo/lrrr.html

>> 384400 mean
>> distance, 1.28 seconds, one way. 1.548 ppm due to c+v = 2 micro seconds.
>> You should see 4 micros seconds difference between moon rise and moon set
>> due to c+/-v. You will need to compensate for the different path lengths.
>
> Already cranked it out on a spread*** in 1999, mate. I know what the
> numbers are, and yours are close enough. I'd prefer 17,000 mph to 1000
> mph, though, less ambiguity.

Ok.

>> >> In fact, the data may be available, on line, now.
>> >
>> > So? It would be totally useless if it were.
>> > The most you could get out of it is the tangential velocity of the
>> > atmosphere, mean value being the rotation of the Earth, and you'd need
>> > upper atmosphere wind speed as well.
>> >
>> >
>> >> Much of the earth station
>> >> information was available on line the last time I googled for earth
>> >> moon distance measurement laser ranging [or something similar].
>> >
>> >
>> > The laser ranging system is designed to do just that, and it will
>> > accurately
>> > measure apogee, perigee and all ranges between from the equation d =
>> > time/c, at moon-noon (overhead).
>> > When the moon is on the horizon it appears red, (as does the sun) the
>> > atmosphere is so damned thick you can't see the reflection from the
>> > laser.
>>
>> A red laser, running backwards through a telescope to keep it, collimated,
>> should get through. You might have to up the power a bit to make up for a
>> few extra dB path loss.
>
> You are very good with your "shoulds", aren't you?
> I wonder if you have any idea just how difficult it is to detect the
> reflected few photons we receive back at the ranging station...

I have a pretty good idea. I have looked at the problems involved before.

....few photons 'we'... ??? I didn't know you worked at a ranging station.

> Upping the power isn't needed when above atmosphere, and we increase v
> by 17.

That is correct.

> HST was built to overcome atmospheric effects, it's reflector is
> nowhere near as large as Earthbound scopes and it produces clearer
> images. Why struggle with Earthbound equipment when the main problem
> is the atmosphere?

Active optics will soon rival HST from earth's surface, with less cost.
In orbit, active optics will be be really awesome.

> An orbitting laser could be 2 * 17 = 34 times LESS powerful than the
> current ranging lasers, AND do the same job. It could also range planets
> as well if we put the corner reflectors in orbit around them.

The small increase in accuracy of location would probably not justify the
cost for the planets. An orbiting corner reflector presents significant
problems. One being that the orbiting device would need prisms on all
surfaces or would need to keep the reflector oriented within 45 degrees of
the eart.

However, the main problem is that the corner reflector returns the reflection
along the incoming beams line. We can lead the target with our outgoing beam
to compensate for propagation delay but the return beam will miss us unless
we purposely spread the beam. If we do that, we increase the path losesses.

We already have them pretty well located and the uncertainties are not
particularly significant. There were some good reasons for better earth/moon
distance information. It allows us to 'weigh' the other masses in our solar
system more accurately.

> Why settle
> for one rib when you can have the whole hog, chops, ham, bacon,
> sausages and trotters?

Cost/return ratio?

>> >
>> > We cannot measure c+v = d/t with the laser ranging data, you need the
>> > laser in orbit to do that, and you need a known 'd' and 'v' to make
>> > that measurement.
>> >
>> > Go through the laser ranging data all you want, you'll only get c for
>> > your answer, and ("as every child at school knows"- Einstein) that
>> > would be all the proof a relativist needs. I'm a little more rigorous
>> > than that.
>> > If I recall correctly, Henri wanted an orbiter around the moon. I do
>> > not object to his experiment, but mine is cheaper and more practical.
>> > I'd use the ISS to mount the laser and HST to see the reflection, GPS
>> > to know the exact location of each.
>> > AD.
>>
>> Just use the ISS to mount the scope that is used to collimate the laser
>> beam and you use the same scope to detect the reflection.
>>
>> The corner reflectors send the beam back to where it came from so the
>> chances of ISS and HST being in line are nil.
>
> Phooey. I see the light from my headlights corner-reflected from the
> car in front, and my eyes are not down at the headlight.

That is because automotive tail-light corner reflectors are designed to
spread the beam by a certain amount. So are the glass beads use on signs.

The lunar corner reflectors are designed to avoid spreading the beam.

>> Bum idea unless you are
>> trying to bounce off the surface of the moon rather than the corner
>> reflectors. Use the reflectors, their positions are well known and you
>> don't have anywhere nearly as much path loss.
>
> No, old son, it is YOUR idea that is bum. In reality a beam that has to
> travel 384400 * 2 kilometers isn't going to come back with exactly the
> same diameter it left with, it'll be a kilometer wide.

It will get to the moon a couple of km wide. The small amount reflected by
the corner reflectors will be spread by about the same amount on the way
back. You would have trouble hitting the HST, as I said before.

In fact, the spread helps make it possible to do lunar ranging with lasers.
Without the spread, the earth's rotation would have moved the ranging station
outside the return beam. As it is, it is possible to use the same optics to
transmit and receive.

> That's a spread
> angle of 2.27e-8 degrees, or 0.0000817 arc seconds, and your corner
> reflectors are not that precise and nor is your collimating telescope
> mirror. What you want to build would cost as much as the HST, and gain
> nothing.

What I suggest would be no more costly that a small celestron, a 5 watt laser
diode, a ccd, and some computer power. Less than 20k bux should do it fine.

>> GPS is fine for knowing where the ISS is, but its orbit is known
>> accurately.
> That's no problem then. :-)

Not quite. The location of the ranging telescope is going to need to be known
very accurately.





--
bz

please pardon my infinite ignorance, the set-of-things-I-do-not-know is an
infinite set.

bz+sp@xxxxxxxxxxxxxxxxxxxx remove ch100-5 to avoid spam trap
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