Measuring The Distance From Earth To Moon 284
lewiz writes: "Tom Murphy at UW is attempting to measure the distance between the Earth and The Moon to the nearest millimetre according to this BBC News article. 'His tape measure will be a giant telescope at Apache Point in New Mexico. Retroreflectors left on the surface of the Moon by various space missions, including the Apollo 11 lunar landing, will also come in handy.'"
That is... (Score:1, Troll)
"The moon is moving away from Earth." (Score:2, Funny)
Re:"The moon is moving away from Earth." (Score:2)
Re:"The moon is moving away from Earth." (Score:3, Funny)
Oddly Enough... (Score:4, Interesting)
Re:Oddly Enough... (Score:2, Funny)
Re:Oddly Enough... (Score:5, Informative)
Factor that in (Score:1)
Re:Factor that in (Score:2, Insightful)
We know that the moon recedes at 1 nanometer per second. Since they're trying to determine nearest millimeter they don't care about how much it's receding or even if it is. As far as they're concerned, it doesn't move farther away in a day or even a few months.
Also, it takes just over 2 seconds for light to get there and back so we're looking at a 2 nanometer change in the whole time. Also, this nanometer movement will change as the moon gets farther away.
Thus they don't have to worry about movement (tis less than the 5% error range they're "allowed")
hope that's kinda helpful
Re:Oddly Enough... (Score:1, Interesting)
Hmm, one significant digit isn't very precise. And significant digits are all that matters: any measurement in centimeter/year can be converted to nanometer/second (or to attometer/nanosecond, or to meter/century...) it's just a speed that is measured.
Re:Oddly Enough... (Score:2)
Re:Oddly Enough... (Score:2)
Re:Oddly Enough... (Score:2)
Re:Oddly Enough... (Score:2)
Re:Oddly Enough... (Score:2)
Am I reading this right? (Score:1)
Huh? since when did forces have weight
Re:Am I reading this right? (Score:3, Informative)
the formula for Force is: F=ma, replace a with the acceleration caused by gravity and you got weight.
Re:Am I reading this right? (Score:3, Insightful)
Weight is the measure of attraction between two masses, such as the gravitational force between the Earth and you. But force certainly does not equal weight.
Force is also a vector, which weight is not.
Re:Am I reading this right? (Score:1)
Re:Am I reading this right? (Score:2)
Re:Am I reading this right? (Score:2)
The gravitation force isn't a bi-directional force per se, but it can still be broken down into x, y, and z components (and thus a vector) in order to simulate gravitational forces between multiple particles. All you have to do to find Fx, Fy, Fz (and make a new vector F to be = {Fx, Fy, Fz}) is to multiply your scalar F by a trigometric function, and now you can treat gravitational forces between particles as vectors, and can do stuff like summation of forces of multiple particles on a specific particle to find out where it'll be pulled.
You are in effect saying that gravitational forces are not vectors only because the equations you presented only yield scalar answers. You neglected to consider that these forces in fact behave as vectors when you have the interaction of multiple particles, or when one particle is moving with respect to the other. Your equation is a scalar only because the two particles are on the same axis, thus requiring no axis component breakdown into a vector (if that makes any sense).
Re:Am I reading this right? (Score:2)
I think you're debating English, not the nature of gravity. Who says that the Earth doesn't weight 150 lbs on you? I think the only reason why you weigh 150 lbs on Earth is because you are smaller than the Earth, and therefore it's easier for the human mind to imagine that image.
If you have a really big rock, it still weighs something "on" the Earth. OK, what if you have an even bigger rock, say the size (and mass) of the moon. Now who weighs what on what? It's totally arbitrary, and stems only from a human's view of the universe from our extremely low-altitude vantage point.
Weight is most certainly a measure of the attraction between two objects, relative to *each other*. Which one of the objects we pick to be the "base" for that comparison is determined entirely arbitrarily.
Re:Am I reading this right? (Score:2)
The only thing keeping our moon from crashing into us, or Pluto and Charon from crashing into each other, is the orbital velocities. If you held Pluto and Charon at a standstill away from each other, they would both move towards each other and collide (with Charon accelerating towards Pluto slightly faster).
Every single particle in our universe has a gravitational force with respect to every other object in the universe... but small masses or great distances make them irrelevant.
A human standing on Earth will exert their weight on the Earth, and also feel the reaction of that weight on their feet. Therefore, the weight acts in both directions (remember Newton's equal and opposite reaction). A human in mid air will accelerate towards the Earth pretty fast, and the Earth's acceleration to the human is so minute (on the order of 10^(-22) m/s^2) that it can be ignored.
Re:Am I reading this right? (Score:2)
That really was my only point -- I guess I got a little bit carried away
Re:Am I reading this right? (Score:3, Insightful)
force. Remember, mass is a number, weight is
a vector.
I think when they say weight of gravity they
mean that gravitons or grabity waves have
non-linear dynamics, i.e. they interact with
themselves and Einstein equations are needed to
deal with this self-consistently. But the simplest
way to correct Newtonian gravity is to analyze
corrections from gravity interacting with itself,
which could be worded as measuring weight of
gravity.
Re:Am I reading this right? (Score:3, Insightful)
Well, yes and no. I know the equation suggests it is (W = m.g, where bold represents vectors. But weight is not used as a vector, and it leads to silly conclusions if it is.
For example, I weigh myself here to be 100kg (or 980N) - what is the direction of this weight vector: towards the centre of the earth. Thought experiment time. Two 100kg men weight themselves, on at the north pole, one at the south pole. What is the sum of their weights?
Well, if weight is a vector then the sum is zero. If, however, you take weight as a scalar then it is 200kg. What people mean when adding weights only works if weight is a scalar. Basically, defining weight as a vector fails the common sense, similar to defining glass as a liquid as opposed to a solid.
All that is happening is that specialists are taking a word that is in common usage (weight in this case) and defining a new (and different) techincal meaning for the word, which is similar to but distinct from the everyday meaning. Then they complain when people use it in the everyday sense.
Re:Am I reading this right? (Score:2)
Re:Am I reading this right? (Score:2)
Here is a more formal statement of Newton's Second Law for you:
The acceleration of an object as produced by a net force is directly proportional to the magnitude of the net force, in the same direction as the net force, and inversely proportional to the mass of the object.
Force != Weight. As far as everyday life is concerned they might be pretty similar, but they are not one in the same thing. Acceleration does not have to have anything to do with gravity either. In many cases it does, but not always.
Weight IS a particular force in certain instances. When you're sitting here on earth, the force of gravity on you is equal to the product of your mass and ~~ 9.8 m/s^2.
Just don't go throwing around statements like "Force = Weight" when you mean that weight is a particular force in certain instances. You are going to mislead high school physics students reading /. and then we'll all be in for it.
How do you check the accuracy? (Score:2, Insightful)
I mean, it's all well and good to say you're going to measure this sort of thing, but how can we be sure he's correct? I could just as easily say that the exact distance from the center of the earth to the center of the moon is 385,137.473 KM.
In fact, what if these scientists are just using this as a cover to get a lot of funding money. Look for the research team to be on vacation in the Bahamas next month. When they come back, they've got a request for funding on a project about angels and pinheads...
Re:How do you check the accuracy? (Score:2, Informative)
Re:How do you check the accuracy? (Score:2, Insightful)
Re:How do you check the accuracy? (Score:2)
Yes there is. Newton's laws, modified by relativity, and a big of geometry allow you to calculate the distance using only the mass of the earth and the orbital period. In theory, if you had the mass & the orbital period known to enough accuracy, then you could calculate the distance.
The newtonian formulas are:
a=v^2/r, a=acceleration, v=orbital velocity, r=radius.
a = G m / r^2, G=Newton's constant, m = mass
v=circumfrence / orbital period
With known m & orbital period then you can solve for r.
Of course, in real life, it's going to be much easier to measure the distance to the moon than the mass of the Earth, and the theory ignores the influence of the rest of the universe, so it's not going to be 100% accurate, so even in Newton's time they measured it, except they used parallax methods.
Umm, the distance isn't constant (Score:4, Interesting)
Re:Umm, the distance isn't constant (Score:2, Funny)
So is he trying to get a precise estimate?
Re:Umm, the distance isn't constant (Score:2)
I'm guessing a critical detail got lost in the translation to journalist-english. That is, that they are not looking for a simple distance estimate, but a refined estimate of the moon's orbital parameters (or Keplerian elements or ephemeris datum, whatever term you prefer). From that, given a time you can solve for the distance.
3.8 cm (Score:1, Interesting)
I remember reading something a while back about ancient Egyptians believeing there were two suns int he sky. I wonder if what they saw was a much closer moon back then?
All throughout the Earth there is evidence of dramatic gravitational changes int he past. For example, below all the ice in the northern reaches of Canada there are vast reserves of coal. Since coal is made through the decay of organic matter, it means there had to be a massive forest up there at some time. Then in the deserts of Africa there are mountains where they find seashells like 1000 feet above sea level.
Perhaps the moon pulling away from the Earth has caused many of the major changes on the Earth...
--Jon
Re:3.8 cm (Score:2, Insightful)
The moon is currently around 260,000 km at its nearest point in orbit. 3.8cm per year is not a significant distance over a couple years, or even 10,000 years(by which the moon has drifted 380m).
Re:3.8 cm (Score:1)
Re:3.8 cm (Score:2)
Re:3.8 cm (Score:2)
Re:3.8 cm (Score:2)
This paleontological evidence comes in the form of tidal rhythmites, also known as tidally laminated sediments. Rhythmites have been subjected to intense scrutiny over the last decade or so, and have returned strong results. Williams (1990) reports that 650 million years ago, the lunar rate of retreat was 1.95±0.29 cm/year, and that over the period from 2.5 billion to 650 million years ago, the mean recession rate was 1.27 cm/year. Williams reanalyzed the same data set later (Williams, 1997), showing a mean recession rate of 2.16 cm/year in the period between now and 650 million years ago. That these kinds of data are reliable is demonstrated by Archer (1996). There is also a very good review of the earlier paleontological evidence by Lambeck (1980, chapter 11, paleorotation)
Re:3.8 cm (Score:1)
Not far enough to differe
Re:3.8 cm (Score:2)
The gravitational effects of the moon on the Earth are fairly minor. Losing the moon would kill the tides, and creatures that depend on them, but that's about it.
The standard explanation for coal in cold areas and seashells way above sea level is continental drift and tectonic action. No moon gravity explanation necessary.
Re:3.8 cm (Score:2, Informative)
His book, The Deep Hot Biosphere is very insightful and most of the information and theories in it are also covered in depth on his website.
Re:3.8 cm (Score:2)
Perhaps the moon pulling away from the Earth has caused many of the major changes on the Earth..
*cough*Plate Tectonics!*cough*
Coal was generally formed from about 300 to 65 million years ago. That's a lot of time for them continents to wander. That 'massive forest up there' was probably some massive forest down here 300 million years ago.
Re:3.8 cm (Score:2, Informative)
Ever since then the moon has been slowly drifting away from the Earth. It doesn't drift away because of the original impact, that energy long since dissipated; the moon is sliding back as tidal forces between the Earth and the moon dissipate small amounts of angular momentum. Eventually the Earth and the moon will become face locked, so that not only do we see the same side of the moon all the time, but the Earth will have slowed so that the same side is always facing the moon. This is a long time away since the earth day will have to slow till it's as long as the moon's period or about 30 times longer than today's day. The moon will not escape however.
To answer your question though the moon has moved less than 1/10000 of a percent since the Egyptians, so no it wouldn't seem that much bigger. In the last 50 million years the moon would have moved around 1%. Because the lunar interactions slow the Earth's spin we know that the Earth probably once spun around twice as fast as it does now (12 hour days). Yes the changes would have had an impact on the Earth and on life, but the change is very very gradual. If you want to look for big effects on the Earth you probably ought to consider more drastic influences like volcanism, earthquakes and large meteors. And just to confuse it all, plate tectonics has no end of fun moving stuff around on the surface of the Earth.
Proof of Moon mission? (Score:1, Insightful)
Re:Proof of Moon mission? (Score:2, Funny)
Re:Proof of Moon mission? (Score:3, Funny)
Re:Proof of Moon mission? (Score:2)
Alternative Solution (Score:5, Funny)
Re:Alternative Solution (Score:1, Offtopic)
:s/Odometer/odometer.
I hate it when I do that.
Re:Alternative Solution (Score:1)
I hate it when I do that.
You must be German.
Re:Alternative Solution (Score:1)
Re:Alternative Solution (Score:2)
Yes - it's a pun(k).
Re:Alternative Solution (Score:1)
If he stopped the puck, why on earth would he pass it to someone on the opposing team?
Re:Alternative Solution (Score:2)
Wouldn't it change? (Score:1)
But my old encylopedia... (Score:4, Funny)
All that and the Kaiser Wilhelm still doesn't know what to do with his country.
Re:But my old encylopedia... (Score:2)
Oh, and asbestos is a miracle mineral that will render the House Of The Future completely fireproof.
Confirming the moon landing? (Score:1)
Cryptnotic
Re:Confirming the moon landing? (Score:2)
Re:Confirming the moon landing? (Score:2)
Old news (Score:1)
It was the theme of a science exam I took in 1994 to enter a college. Every aspect of it, including the ways the moon mirrors are designed so that they always send the light back to its source (vaguely mentionned in the article)
Wipe-cut to the next scene... (Score:4, Funny)
Useful? (Score:3, Funny)
Current estimates predict that all known gorgonzola deposits on Earth will be depleted by 2016. We need to think about cheese-mines on the moon now, before it's too late...
Re:Useful? (Score:1, Flamebait)
Mooting points. (Score:5, Informative)
- The Earth's rotation is not circular (it "sloshes").
- The Earth's shape is not constant.
- The Earth's mass is not constant, so the general relativistic field in which the moon orbits is not constant.
- Okay, so we know where that telescope is relative to the moon. Now where is it relative to my house? To Washington? To Wendy's?
- Isn't this just an attempt by the Bush White House to wag the dog to distract attention from the fistfight the President and Vice President had during the game Sunday?
--Blair
Re:ReMooting points. (Score:2)
'retroreflectors'? (Score:1)
Re:'retroreflectors'? (Score:2, Informative)
To see this in person, walk up to a surveyor at a construction site sometime... if you're lucky and they're using an optical total station, there'll be someone walking round with a pole w/ a corner cube reflector (a type of retroreflector) on top. This is used so that the pulses of light coming out of the total station get reflected back to the station no matter how the pole guy has the pole oriented.
Another good example is the material highway signs and license plates are coated with -- they show up in your headlight beams so well because much more of the light reflecting from their surface heads back toward the light source (the headlights, very close to being in line with your eyes) rather than being scattered or reflected off into a less useful direction.
Check out http://www.leica-geosystems.com/ims/product/tps50
Re:'retroreflectors'? (Score:2)
Reflectors on the moon? It's a lie !!! (Score:5, Funny)
To sum up: "we never went to the moon, hence there are no reflectors on the moon."
Moderators: Put down the crack pipe and the mouse. Step away from the keyboard. Take a DEEP BREATH and READ. This is not a troll. I repeat. This is not a troll.
No, I don't believe the "moon hoax" loonies. Anyone with an above-kindergarten education can easily refute the "moon hoax" loonies claims.
For your amusement, this is the gallery of the 'barking mad'...
Some anonymous kook [geocities.com]
Bill Kaysing [okneoac.com]
Ken Overstreet [8m.com]
"mpeeters" [geocities.com]
Re:Reflectors on the moon? It's a lie !!! (Score:3, Interesting)
another question: don't we have telescopes good enough that we could see things that were "left" on the moon, like the lunar rovers, their boots and gloves, etc.? It would seem that would put the "did we really land on the moon?" controversey to rest....
Re:Reflectors on the moon? It's a lie !!! (Score:2, Interesting)
I also suspect that orbiting telescopes can't be configured to focus so close. ( like trying to use binoculars to look at your hand)
It does leave me wounding just how powerful the laser must be to be scattered by the atmosphere on the way up and the way back and still be visible.
Re:Reflectors on the moon? It's a lie !!! (Score:2, Informative)
Re:Reflectors on the moon? It's a lie !!! (Score:2)
The definitive list can be found on crank.net [crank.net]. If you ever want a good laugh, try reading through the sites listed as "illucid". ;-)
Why do cranks always pick on either cosmology or the moon? I want to see a crank theory of horizontal branch morphology or AGB stars! :-)=
[TMB]
Ex-squeeze me? (Score:1)
Much easier to hit that suitcase-size reflector with a beam that's more than a mile wide than with one that's pencil thin!
My tax dollars at work... (Score:1)
This is like trying to measure Bill Gates' wealth to the penny.
The moon's orbit varies by far more than 1 millimeter all the time. There are all sorts of influences, including the earth's gravity, the sun's gravity, Jupiter's gravity -- as well as factors that will introduce error into the measurements, such as refraction of the laser in the earth's atmosphere.
Besides, I've already measured the distance with my trusty golf rangefinder.
-- CP
Re:My tax dollars at work... (Score:3, Interesting)
Therefore, by measuring the distance to an accuracy of a millemeter, we might be able to gain some insight about: earth's gravity, the sun's gravity, and Jupiter's gravity. The point of this experiment is not to know how far it is to the moon. The point is to better understand the moon's orbit, and the various influences which affect it.
Parallel space.com article (Score:1, Funny)
Murphy's real motivation is to test Einstein's General Theory of Relativity, which is
based on an assumption that gravity affects a feather and a bowling ball in the same
manner. You cannot test this in your home, because air keeps the feather aloft.
I forget -- was Einstein the guy who dropped the balls off the tower or the one who was hit in the head with an apple?
Which part of the Earth? (Score:3, Insightful)
Unfortunately, he's not really measuring the distance between the centers of the two planetoids... he's measuring the distance between the top of some piece of equipment on the Moon and the lens of his telescope.
This reminds me of the human genome project. At some point, the scientists announced, "we've finished mapping the human genome! It's finished!" And as it turned out, it wasn't anywhere near completion. I believe it's still unfinished.
Is modern science really so desperate to inspire interest in people? Lies are not impressive.
Re:Which part of the Earth? (Score:2)
I believe these people know more about what they are doing than you do. At least give them the benefit of the doubt. But wait... you don't even have a reason to doubt.
Well, failing that, don't talk unless you know what the hell you're talking about.
Oh no! (Score:2, Insightful)
Fellow worker: So?
Tom: I've got 385,137.471, should warn someone?.... ah, um... what are you doing?
Fellow worker: CALLING THE WHITE HOUSE!
What should we expect from this 'experiment'? A warning that the moon is going to crash into the Earth in 2003? Come on! Put the money towards something useful, like... figuring out why we aren't on Mars when NASA admits we could have been there in '85 if we kept on schedule.
Why not... (Score:2, Funny)
PING moon (212.58.226.40): 56 octets data
64 octets from 212.58.226.40: icmp_seq=0 ttl=2000 time=1.283 s
1.283 x 300000000 = 385000000
So the moon is 385000 km from earth. Easy!
Another totally worthless experiments (Score:4, Interesting)
The moon distance measurement is obviously good. It seeks to do something no one else has ever achived. Even if the results aren't interesting, the new techniques used are. Anyone that doesn't see its intrinsic value is shortsighted.
I can see it now... (Score:2)
careful, in another 100 million years... (Score:2, Interesting)
These two phenomena are actually related -- the orbital angular momentum that the moon is gaining (moving farther away) is taken from the earth's rotational angular momentum. Gradually, but measurably, the earth *is* slowing down. You might have heard of leap seconds? These are to compensate (partially) for the slowing in fact!
Eventually (ok, in maybe 100 million - 1 billion years) the earth and moon will orbit/rotate at the same angular velocity, so that at that time, we will always see the same face of the moon, and the moon will always see the same face of the earth!
Better pick which side of the earth you'll want to live on, otherwise if you pick wrong, you'll *never* be able to see the moon...
Interesting fact... (Score:2, Funny)
If all of the articles published on this topic were stacked one on top of the other, they would reach the moon and back 3.42 times.
Or should that be 3.46?
significant digits / error / order of magnitude (Score:3, Insightful)
That's pretty stupid, considering the distance will obviously change more than a millimeter all the time. Hell even the astronaut's footprints will be several millimeters thick, assuming they are undisturbed still. So...within a millimeter of *what*, exactly?
Which part of the moon? (Score:2)
OK, they say they are trying to measure the center to center distance, but they don't get that directly. The real measurement is from a telescope mount on top of a mountain on Earth, to a retro-reflector on the Moon. Do you actually know the height of that mountain at the observatory to 1 mm? And can you correct that height to the day the measurement was taken? (Some sorts of subsoil will shrink and swell depending on water contact, sometimes resulting in the ground rising and falling a few feet annually. I'd think that deep down in a mountain would be rock so it wouldn't do that, but in most cases the whole mountain is either rising or falling by at least millimeters a year, and if there is any soil cover weather changes might change the height by a few millimeters.)
And on the moon, you are measuring to a reflector which is basically laying where the astronauts dropped it 30 years ago. How would the distance from the reflector to the center of the moon be measured? Laser beams & navigational gear in satellites orbiting the moon? What satellites?
Wider beam makes hitting target harder??? (Score:2)
This is made even trickier by the fact that the beam will be about 2 kilometres (1.2 miles) wide by the time it reaches the Moon."/I>
Oh-kay.... so it would be less tricky if your laser beam was, say, 2 millimeters wide by the time it reaches the moon?
Re:So? (Score:1)
from somebody who set about to actually make
those changes occur. Innovation and
discovery result from generally fucking around with stuff.
Re:So? (Score:1)
Perhaps some scientist somewhere will do important research like where all the damn socks go that dissapear in clothes dryers.
In either case it's money well spent, it is.
-
Re:moon cheese (Score:2, Funny)
The Moon. For several years she has fascinated Mankind
And the Monty Burns classic:
For centuries, man has wanted to destroy the Sun
Re:Research? (Score:2, Informative)
(1) Before we can even discuss basic science research, we need to agree that such fundamentals are important and thus deserve tax dollars. I find it irrelevant that the money comes from NASA for this discussion.
(2) As mentioned in the article, this experiment hopes to measure to unprecedented accuracy the rate of change of the distance between the Earth and the moon. Why is this useful? If it can be done accurately (the conditions of which I will discuss in a moment) it would allow a determination of the self-interaction of gravity, e.g. graviton-graviton interaction. This is fundamentally different than Newtonian gravity, and, as mentioned elsewhere, the simplest way to explain in our nascent theory of quantum gravity the Einstein field equations without solving the actual math. On a much larger scale, the determination of the Hubble constant and how it changes with time also measures this. Finding the argument that it is preferable to do such measurements in one's backyard when possible I leave as an exercise to the reader.
With respect to the potential accuracy (vs. precision) of such measurements I will note the following. First, current gravity meters based upon atomic fountains are accurate enough to find Cave complexes in Afghanistan and see people moving around in them. (c.f. Steve Chu's recent work at Stanford with atom interferometers); we have a very detailed picture of our local gravitational field available to us. Second, considerations such as chaos theory and effects of the other planets are relatively straightforward to deal with. Back at the beginning of the 20th century they had already done it for Mercury and still had a discrepancy, at 43 arcseconds per century in its orbit(c.f. this explanation [mathpages.com]). That's over 10 times smaller all the other planets' influence, and that was calculated before computers. It seems to me the greatest unknown is the tectonic structure of the moon and the associated vibrations in the mirror. I suppose that radar rangefinding, given the scale of these variations, would be sufficient for most purposes.
Finally, some of the past results of this experiment, from the Nasa site [nasa.gov]
Re:Parallel space.com article (Score:2)
Re:Disprove this picture (Score:2)
"Oh look no stars..aaaaaaiiiyyyyy"
dude, there is too much light on the moon to see stars. when your on the side that happens to be where the sun light, or earthlight happens to be shining. I've been through all the "evidence" hat gets presentsed as "proof" the moon landing was faked, and its all explainable in clear easy consice maner, using small words.
Re:Hrmmm..... (Score:2)
Cut taxes, Gimme stuff!
Persoanlly I think all school should get a perfered tax treatment status, and everytime there is a budget hit, take it out of road maintenance first. Then will see how long people want to lower taxes.
Re:Already done with accuracy of 3mm (Score:2)
Yeah right. They have not.