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makis
Greece
259 Posts |
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tvanflandern
USA
2793 Posts |
 Posted - 23 Nov 2004 : 09:37:12
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quote:
Originally posted by makis
Has the orbital speed of the Earth been measured experimentally and if so, do the results agree with the theoretical numbers obtained from the solution of the kinematic equations of its elliptical motion around the sun?
Yes and yes. For example, we measure high-precision doppler shifts in quasars and pulsars that would not be constant if our calculations of Earth's orbital speed were not right on target.
At a more basic level, speed is a distance traversed in a time interval. So measuring location and time eventually turns into measures of speed, and these (naturally) are consistent with doppler shifts. We now get 8-9 digits of precision in such measures with modern high-precision techniques such as VLBI or radar and laser ranging or spacecraft transponders. -|Tom|- |
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Jim
1607 Posts |
Posted - 24 Nov 2004 : 16:58:09
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| If there are very accurate measurments of the orbital speed can the data be accessed? How much is the speed effected by the moon? |
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tvanflandern
USA
2793 Posts |
Posted - 24 Nov 2004 : 17:28:43
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quote:
Originally posted by Jim
If there are very accurate measurments of the orbital speed can the data be accessed?
Same answer as many times before: See the National Space Sciences Data Center (NSSDC). Modern high-precision data is not trivial to use. But it is all available to everyone, and the knowledge about how to use it is taught in graduate schools.
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How much is the speed effected by the moon?
+/- 12 meters/second. -|Tom|- |
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makis
Greece
259 Posts |
Posted - 24 Nov 2004 : 18:12:21
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quote:
Originally posted by tvanflandern
quote:
Originally posted by makis
Has the orbital speed of the Earth been measured experimentally and if so, do the results agree with the theoretical numbers obtained from the solution of the kinematic equations of its elliptical motion around the sun?
Yes and yes. For example, we measure high-precision doppler shifts in quasars and pulsars that would not be constant if our calculations of Earth's orbital speed were not right on target.
At a more basic level, speed is a distance traversed in a time interval. So measuring location and time eventually turns into measures of speed, and these (naturally) are consistent with doppler shifts. We now get 8-9 digits of precision in such measures with modern high-precision techniques such as VLBI or radar and laser ranging or spacecraft transponders. -|Tom|-
If I understand correctly how ranging instrumentation works, the only possible measurement is the relative radial speed between two or more bodies in motion.
For a planar orbit in polar coordinates, the velocity v is given by:
v = (dr/dt)ru + r(d8/dt)8u, where the number 8 used used for the angle theta and ru and 8u are the untit vectors.
Thus, knowledge of the instanteneous radial speed dr/dt does not suffice to determine the magnitude of v ( v dot v). Knowledge of the instanteneous angular speed d8/dt is also required.
How is the instantenuous angural speed is determined to plot |v| as a function of t? If an elliptical orbit is assumed in the first place then it is only natural that experimental and theoretical data will match to a high degree.
In the case of the sun-earth system how to you use ranging equipment for an empirical determination of the orbital speed changes?
I doubt there is any pure empirical evidence based on measurements that confirms the elliptical orbit of the earth around the sun and I suspect the available data is a 'model', that is extrapolations via the use of ad-hoc hypothesis about the shape of the orbits in the planetary system.
By the way I think that JIM is right to protest this for some time now. There is plenty of data available but nothing to prove the data can replicate the theory sssolely on empirical grounds.
Makis
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tvanflandern
USA
2793 Posts |
Posted - 24 Nov 2004 : 19:53:26
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quote:
Originally posted by makis
If I understand correctly how ranging instrumentation works, the only possible measurement is the relative radial speed between two or more bodies in motion.
Sorry, but your understanding is not correct. Ranging measures the round-trip time of flight for a radar or laser signal between the source and target bodies. Time of flight multiplied by speed c gives distance between source and target bodies, with a typical precision of 8-9 figures. So ranging measures distances, not velocities. The time rate of change of ranges gives velocities.
It goes without saying that one must then allow for where the transmitting station was relative to the center of the Earth at transmission, where the reflection point was relative to the target body's center, and where the receiver was at reception. Such ranges give the best measurements we have for planet positions relative to the Sun and to each other. But ranging can't do it all because they do not indicate orientation. So this data is combined with optical observations of positions relative to the star background, so that we have all three coordinates for the position of each body relative to each other body.
The nature of your questions suggests some confusion about how we can be so sure it is not all a house of cards. Perhaps the best of several checks is provided by pulsars, which are effectively radio beacons transmitting rapid pulses on a known schedule. They are as good as atomic clocks, perhaps better in some ways. So if you were blind and could see nothing else, you could still tell exactly where you were in all three coordinates by observing these "time signals" arriving from pulsars in many different directions around you. Because the signals travel at the speed of light, if you receive "pulse A" from "pulsar X" early, you must be closer to that pulsar than you thought by an amount you can calculate. And likewise for measures from all the other pulsars.
So you see there is no room for ambiguity. Between pulsars, ranges, and optical data, we know our precise position and velocity relative to the Sun and the other solar system bodies.
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If an elliptical orbit is assumed in the first place then it is only natural that experimental and theoretical data will match to a high degree.
Few people today have any use for assumptions about "ellipses" because, at this level of accuracy, no orbit is an ellipse. We deal with the actual, 3-dimensional motion. Ellipses are still used by astrologers for horoscopes, but are too crude for modern high-precision work. -|Tom|- |
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makis
Greece
259 Posts |
Posted - 27 Nov 2004 : 04:32:50
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quote:
Originally posted by tvanflandern
quote:
Originally posted by makis
If I understand correctly how ranging instrumentation works, the only possible measurement is the relative radial speed between two or more bodies in motion.
Sorry, but your understanding is not correct. Ranging measures the round-trip time of flight for a radar or laser signal between the source and target bodies. Time of flight multiplied by speed c gives distance between source and target bodies, with a typical precision of 8-9 figures. So ranging measures distances, not velocities. The time rate of change of ranges gives velocities.
I basically said the same. Differentiation of position data gives velocities. I have done enough experimental work with interferometers to know that.
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It goes without saying that one must then allow for where the transmitting station was relative to the center of the Earth at transmission, where the reflection point was relative to the target body's center, and where the receiver was at reception. Such ranges give the best measurements we have for planet positions relative to the Sun and to each other. But ranging can't do it all because they do not indicate orientation. So this data is combined with optical observations of positions relative to the star background, so that we have all three coordinates for the position of each body relative to each other body.
The nature of your questions suggests some confusion about how we can be so sure it is not all a house of cards. Perhaps the best of several checks is provided by pulsars, which are effectively radio beacons transmitting rapid pulses on a known schedule. They are as good as atomic clocks, perhaps better in some ways. So if you were blind and could see nothing else, you could still tell exactly where you were in all three coordinates by observing these "time signals" arriving from pulsars in many different directions around you. Because the signals travel at the speed of light, if you receive "pulse A" from "pulsar X" early, you must be closer to that pulsar than you thought by an amount you can calculate. And likewise for measures from all the other pulsars.
So you see there is no room for ambiguity. Between pulsars, ranges, and optical data, we know our precise position and velocity relative to the Sun and the other solar system bodies.
Is this a wish or something that referes to actual, undisputable data? IMO based solely on measurements and observations, this is not possible. Do you know where I can get that Earth velocity graph relative to the Sun that is based solely on observations?
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If an elliptical orbit is assumed in the first place then it is only natural that experimental and theoretical data will match to a high degree. Few people today have any use for assumptions about "ellipses" because, at this level of accuracy, no orbit is an ellipse. We deal with the actual, 3-dimensional motion. Ellipses are still used by astrologers for horoscopes, but are too crude for modern high-precision work. -|Tom|-
I thought astrologers use the geocentric system, something very similar if not identical to the Ptolemaic system, where the motions of planets are epicycles resulting in retrogate motion.
So what is the 'actual', 3-dimensional motion of the Earth around the Sun? If it's not an ellipse, what does it look like? Any references or links to graphs?
Makis
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tvanflandern
USA
2793 Posts |
Posted - 27 Nov 2004 : 11:40:32
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quote:
Originally posted by makis
Is this a wish or something that referes to actual, undisputable data?
The latter.
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IMO based solely on measurements and observations, this is not possible.
You are curiously silent about what "measurements and observations" lead you to such a strange conclusion. 300 years of "measurements and observations" by astronomers worldwide say otherwise.
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Do you know where I can get that Earth velocity graph relative to the Sun that is based solely on observations?
JPL and USNO publish high-precision ephemerides of all major solar system bodies based on all available data. If you want the raw data, go to NSSDC. For the results, look up the JPL Development Ephemeris series, or one of the many USNO publications such as the Astronomical Almanac. It has a nice companion volume, the Explanatory Supplement to the Astronomical Almanac (P.K. Seidelmann, ed., University Science Books, 1992). But you might find the 1961 edition of this companion volume far more readable for non-professionals because there were fewer observation types back then, and you are obviously not interested in velocities to an accuracy of parts per billion.
Of course, ephemerides such as these give 3-D positions as a function of time. But as you say, "Differentiation of position data gives velocities. I have done enough experimental work with interferometers to know that." So you should have no trouble getting an Earth velocity graph from the Earth-Sun position data.
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I thought astrologers use the geocentric system, something very similar if not identical to the Ptolemaic system, where the motions of planets are epicycles resulting in retrogate motion.
That is another misimpression. Read the "Explanatory Supplement" and you will understand astronomical coordinate systems much better. Both geocentric and heliocentric systems are in current use, depending on the application. For example, if tracking artificial satellites, we would use a geocentric system. But we also use it for the whole solar system when preparing tables of where the various bodies will be seen by Earth-based telescopes. The paths of the planets relative to the stars, as viewed from Earth, really are "epicycles", which is why the Ptolemaic system survived longer than the heliocentric system has even existed.
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So what is the 'actual', 3-dimensional motion of the Earth around the Sun? If it's not an ellipse, what does it look like? Any references or links to graphs?
The orbit of the Earth is a perfect ellipse to the accuracy that the human eye can perceive. It is only modern, high-precision data that can see deviations from a perfect ellipse, caused primarily by perturbations induced by gravitational forces from other planets. -|Tom|- |
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Jim
1607 Posts |
Posted - 27 Nov 2004 : 16:48:10
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| The data is way too hard to access but I have hope it will become an easy matter when enough people get involved. Anyway, the moon speeds and slows the Earth by 12m/s according to the post above. I assume the Earth is slowed most at 1st quarter and speeded up 24m/s by the 3rd quarter. Then slows down 24m/s getting back to 1st quarter. Please correct if I have it wrong, thanks. |
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tvanflandern
USA
2793 Posts |
Posted - 27 Nov 2004 : 19:58:10
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quote:
Originally posted by Jim
I assume the Earth is slowed most at 1st quarter and speeded up 24m/s by the 3rd quarter. Then slows down 24m/s getting back to 1st quarter.
You have it right. Looking at the in-plane longitudinal effect only, Earth's average orbital speed is about 30,000 m/s, increasing by about 500 m/s in January and decreasing by 500 m/s in July because the orbit is an ellipse, not a circle. Overlaid on that is the +/- 12 m/s effect from the Moon, just as you described. The next largest effect, about +/- 1 m/s, is due to Jupiter. -|Tom|- |
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makis
Greece
259 Posts |
Posted - 28 Nov 2004 : 07:33:49
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quote:
Originally posted by tvanflandern
The orbit of the Earth is a perfect ellipse to the accuracy that the human eye can perceive. It is only modern, high-precision data that can see deviations from a perfect ellipse, caused primarily by perturbations induced by gravitational forces from other planets. -|Tom|-
Thank a lot Dr. Van Flander. All these plus more tends to support the hypothesis that there must exist a carrier of gravitational energy to affect deviations from purely inertial paths, such as circles and ellipses. In this respect I think GR is problematic unless one assumes that the 4-D space-time acts as a medium of gravitational energy propagation, since the metric is just a mathematical description of the variable curved properties of space-time and cannot serve that purpose.
I think this is highly troublesome for a theory such as GR. Although I have been questioning strongly your unobservable graviton particle theory, I come to realize that an energy/momentum transfer mechanism between real bodies in gravitational orbits is mandatory to avoid considering other hypotheses such as for instance that our physical reality is a computer simulation.
Still, I'm very skeptical about the graviton hypothesis as a momentum transfer agent and essentially the cause of gravitation but I recommened to people I have discussions with to take an unbiased look at your theory.
One thing is for sure, the observable orbits of cellestial bodies entail much more background activity than what GR assumes to take place and such activity is physical and real. More important is where that activity takes place which is not accounted for in the foundations of GR.
Makis
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Jim
1607 Posts |
Posted - 04 Dec 2004 : 16:17:58
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| Hoping to keep this a simple a possible I'll focus on the Earth/moon motion and skip whatever effect Jupiter and Venus generate. The force of the moon causes the Earth to speedup and slow down is calculated from F=ma and amounts to ~24m/s as posted above. The acceleration rate changes from zero at new and full moon to the maximiun rate at 3rd and first quarter. My question: What is the effect of this force on Earth? Or how is this force observed(data) to manifest itself? In the barycenter model shown at JPL/horizons web site this force is applied in reverse and that causes the moon to be nearer to the sun at full moon and farther from the sun at new moon. |
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tvanflandern
USA
2793 Posts |
Posted - 04 Dec 2004 : 17:23:47
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Originally posted by Jim
What is the effect of this force on Earth?
It changes the Earth's velocity, as already described.
In general, the only effect of gravity on a rigid body is to change the speed of the body. For a slightly non-rigid nody such as Earth, the fact that the Moon's pull is stronger on the nearside than on the farside causes Earth to bulge slightly toward the Moon. This is the simplest kind of "tide", a bulge toward the source of gravity. There is also a solar tide, a small bulge toward the Sun.
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In the barycenter model shown at JPL/horizons web site this force is applied in reverse and that causes the moon to be nearer to the sun at full moon and farther from the sun at new moon.
The Moon's distance from the Sun is dominated by the ellipticity of Earth's orbit. The ellipticity can change the Moon's distance from the Sun by +/- 2.5 million km over the course of a year. The diameter of the Moon's orbit is about 800,000 km.
However, because the distance change from ellipticity cannot exceed about 600,000 km in just the two weeks between Full and New Moon, the condition you described is not physically possible for any consecutive Full and New Moons. You misunderstood something. -|Tom|-
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Jim
1607 Posts |
Posted - 04 Dec 2004 : 17:51:36
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| I need to edit my last post to read the Earth is nearer the sun at full moon-not the moon is nearer. |
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Jim
1607 Posts |
Posted - 04 Dec 2004 : 18:16:53
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| Are the magnetudes of the tidal bulges caused by the moon and sun proportional to the distance from Earth to them? As you explained in another post the diameter of Earth is the same in either calculation so is the moon making a bigger tidal bulge-how do you calculate that proportion? The diameter of Earth is 1.3x10E7 meters and the sun is 1.5x10E11m from Earth. How does that make a bulge almost as big as 1.3X10E7m and 4x10E8m? You did say the diameter of Earth was a factor in this-right? |
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tvanflandern
USA
2793 Posts |
Posted - 04 Dec 2004 : 19:16:32
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quote:
Originally posted by Jim
Are the magnetudes of the tidal bulges caused by the moon and sun proportional to the distance from Earth to them?
They are inversely proportional to the cube of distance.
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As you explained in another post the diameter of Earth is the same in either calculation so is the moon making a bigger tidal bulge-how do you calculate that proportion?
The gravitational force on Earth with mass m caused by another mass M at a distance R is GmM/R^2. If Earth's radius is r, then the gravitational force on the front side of Earth is GmM/[R-r]^2, and on the back side of Earth is GmM/[R+r]^2. So the excess force on the front side over the back side is (using some algebra and some reasonable approximations such as R >> r): 4GmMr/R^3.
So that is the size of the small force leading to a tidal bulge. Note that the albebra shows why it must be inversely proportional to R^3. When we calculate this force acting in one direction over several hours (because the solid Earth spins), and allow for Earth's resistance to deformation, we arrive at the result that the tidal bulge from the Moon is typically three feet (relative to the Earth's center); whereas that from the Sun (farther away but much more massive) is 2.2 times smaller, or a bit over one foot.
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The diameter of Earth is 1.3x10E7 meters and the sun is 1.5x10E11m from Earth. How does that make a bulge almost as big as 1.3X10E7m and 4x10E8m?
You lost me there. Solid-body tidal bulges are tiny -- a meter or less. -|Tom|- |
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Jim
1607 Posts |
Posted - 04 Dec 2004 : 19:30:49
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| I see you are using a lot of math to get from point A to point B that is not transparent. The meters in r and the meters in R are not at all as interchangable as the math allows don't you think? |
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tvanflandern
USA
2793 Posts |
Posted - 04 Dec 2004 : 22:57:47
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quote:
Originally posted by Jim
I see you are using a lot of math to get from point A to point B that is not transparent. The meters in r and the meters in R are not at all as interchangable as the math allows don't you think?
You can't ask "How do you calculate that proportion" if you haven't yet mastered even algebra, which is the math of symbols and equations.
If you just need the intermediate steps spelled out for you, that is one thing. If you understand F = ma (Newton's second law) but not F = GmM/R^2 (Newton's universal law of gravitation), then I would be at a loss for how to answer your "how do you calculate" question.
Would someone else like to have a go at it? (Show the steps in the formula manipulation and/or explain Newton's universal law and/or explain something else, as Jim wishes.) -|Tom|- |
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Jim
1607 Posts |
Posted - 05 Dec 2004 : 15:53:05
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| Well, as it usually is the way things go this is getting way off the point. Can we restart from a few laps back? Or lets get bogged down in calling me stupid and lacking in book learning. |
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Orbital speed of Earth
