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#1
10-08-2019, 04:55 AM
 Guest Join Date: Oct 2001 Posts: 1,352

## What practical methods did Kepler et al use to compute orbits?

How did they convert 2D maps into 3D theoretical models, eg when Herschel discovered Uranus?

For a given planet or comet, I'm guessing that the raw data was nightly recordings of the object's apparent position against the fixed stars behind it, so that they had a record of the object's co-ordinates through time. These were presumably some measure of ecliptic latitude and longitude. They could then plot the object's course onto paper, to show its angle compared with the ecliptic, its periods of retrograde motion etc.

My main question is: how did they convert this data into units for applying trigonometry, or whatever other maths were required?

Is there a unit of speed that is used, eg "it averaged 7 units a day eastwards for 8 days at an angle of 2.4 degrees and then slowed down to 5 for 3 days" or something like that? What would the next stage be? (Or the first stage, if my assumptions are wrong?). Did they have to draw it out on paper or could they work out the orbit straight from the co-ordinates?
#2
10-08-2019, 05:29 AM
 Guest Join Date: May 2016 Posts: 3,852
We can get into the numerical details of determining orbits pre-and post-Newton, as well as Gauss, but there are never any units of speed used at first. What you do at the observatory is note the time and apparent position of the object using some coordinates like azimuth and altitude above the horizon; the appropriate units are degrees.

Then you can correct for various effects and express the observation in terms of an appropriate celestial coordinate system with respect to the centre of the Earth or of the Sun, so, basically, more angles; you can see why trigonometry comes in.
#3
10-08-2019, 05:39 AM
 Guest Join Date: Oct 2001 Posts: 1,352
Quote:
 Originally Posted by DPRK We can get into the numerical details of determining orbits pre-and post-Newton, as well as Gauss, but there are never any units of speed used at first. What you do at the observatory is note the time and apparent position of the object using some coordinates like azimuth and altitude above the horizon; the appropriate units are degrees.
But if the time is important then that is effectively a measure of apparent speed, no? I'm trying to intuitively (ie I didn't take trigonometry past the age of 17) understand how you would tell the difference between a highly elliptical comet and an asteroid with a more circular orbit.
#4
10-08-2019, 05:41 AM
 Guest Join Date: Mar 2012 Posts: 1,793
Here is a timeline of Human "scientific" investigation of the solar system.

It starts with Greek Astronomy around 280 BC, then there is a downfall in western astronomy picked up by Indian and Arabian astronomy and then much later comes Kepler / Copernicus / Brahe / Galileo / Newton.

I am not aware of all the units, but the units for other planets, used in Indian Astronomy was the Earth/Sun orbit radius ratio :

From :

Quote:
 Aryabhata gives the radius of the planetary orbits in terms of the radius of the Earth/Sun orbit as essentially their periods of rotation around the Sun. He believes ...
#5
10-08-2019, 06:08 AM
 Charter Member Join Date: Apr 2004 Location: Europe Posts: 763
Here is the original article by Anders Johan Lexell where he establishes some orbital parameters of Uranus from observations:

It's in French and heavy on observations but light on math.

He doesn't detail his calculations, from my poor understanding of French he just says "from these observations we calculate that...." and "our calculations show..." Would the calculations have been obvious to his peers?

Maybe someone can tell from that article roughly what his calculations are?
#6
10-08-2019, 06:43 AM
 Guest Join Date: May 2016 Posts: 3,852
Quote:
 Originally Posted by The Stafford Cripps But if the time is important then that is effectively a measure of apparent speed, no? I'm trying to intuitively (ie I didn't take trigonometry past the age of 17) understand how you would tell the difference between a highly elliptical comet and an asteroid with a more circular orbit.
If you assume the orbit is an ellipse or conic section centered on the sun- indeed, Kepler's name is associated with this- then once you fit it to observations the relevant parameter is the eccentricity, which is a dimensionless number with no units. Circular orbits will have eccentricity close to 0, while the comet will have an eccentricity close to 1.

If you look at the last couple of pages in Frankenstein Monster's link, you can see for instance how he rules out a parabolic orbit for Uranus, demonstrating it must be a genuine planet.

ETA Kepler's laws also imply that the speed changes depending on the distance from the sun, so a comet moves faster through space as it approaches the sun

Last edited by DPRK; 10-08-2019 at 06:47 AM.
#7
10-08-2019, 07:48 AM
 Charter Member Join Date: Apr 2004 Location: Europe Posts: 763
Quote:
 Originally Posted by DPRK ...you can see for instance how he rules out a parabolic orbit for Uranus, demonstrating it must be a genuine planet.
Again from a poor understanding of French, I think it goes like this:

He takes two observations of Uranus with the largest separations in time.

He fits them to a parabolic orbit (I think also in the plane of the ecliptic) with a perihelion distance of 10 times that of earth orbit.

He shows that the observations inbetween do not match what they would have been with such an orbit (differences of as much as 20 to 25 minutes of arc).

He also tries a perihelion with 8 times that of earth orbit and shows that it's not much better.

He doesn't spell out his formulas or calculations. I suppose the formulas are just plain obvious to astronomers of that era. I imagine it's really just high school trigonometry.
#8
10-08-2019, 11:37 AM
 Guest Join Date: Oct 2001 Posts: 1,352
Quote:
 Originally Posted by Frankenstein Monster He takes two observations of Uranus with the largest separations in time.
I think he might cover this ('...one could find an infinity of orbits...'), but would you not need at least 3 observations to plot an ellipse or other curve?
#9
10-08-2019, 12:21 PM
 Charter Member Join Date: Apr 2004 Location: Europe Posts: 763
Someone else should have a go at this, I speak French extremely poorly. But according to slightly corrected Google translate, the argument is:

Quote:
 From this it is easy to conclude, that if one tried to find a parabolic orbit, which at the same time that it would satisfy the observations of the 9th of April, 1781 and the 2nd of October, 1782, was also In agreement with a third intermediate observation, this parabolic orbit would be found in conflict with a large number of observations by up to 30 arcminutes: thus that it is incontestably proved, that a parabolic orbit can not absolutely satisfy the observations;
So yes he finds parabolic orbits from two endpoints and a varying point inbetween and notes that they all then conflict with the rest of the observations. Or something.

Last edited by Frankenstein Monster; 10-08-2019 at 12:23 PM.
#10
10-08-2019, 12:48 PM
 Guest Join Date: May 2016 Posts: 3,852
To say more, to a reasonable approximation, it's not really a 3D problem, it's a 2D problem: the planet (or other body) orbits the sun in a single plane. You are definitely going to make use of that, and the fact that the orbit is assumed to be a conic section focused on the Sun, as well as Kepler's law stating that the radius vector sweeps out equal areas in equal times.

In notes like the one concerning Uranus it is certainly assumed that any competent astronomer knows how to calculate things like the geocentric equatorial coordinates of the Sun at any given time, convert between various coordinate systems, and other feats of trigonometry as may be necessary. As for mystery planets/asteroids, just as an example you could do something like, start with three observations (time and geocentric coordinates, e.g., right ascension and declination). Then you solve a system of equations to figure out the distance of the body from the Earth (and from the Sun) at each of these observations, taking into account light travel time and other necessary effects. Once you do that, probably via an iterative method, you know the focus of the ellipse and three points on it, so you are able to extract the complete set of orbital elements. Gauss's method is a famous, relatively modern (1801), example of this.

One good question is, how did Kepler originally figure out his orbital laws? Of course, he had access to extensive observations of Mars, so plenty of time to figure out that circles and other non-elliptical curves simply did not fit.
#11
10-08-2019, 12:50 PM
 Guest Join Date: May 2016 Posts: 3,852
Quote:
 Originally Posted by The Stafford Cripps I think he might cover this ('...one could find an infinity of orbits...'), but would you not need at least 3 observations to plot an ellipse or other curve?
If it's a parabolic orbit, that is, eccentricity = 1, that takes care of one degree of freedom, so you only need two more points.
#12
10-08-2019, 01:45 PM
 Guest Join Date: May 2016 Posts: 3,852
Quote:
 Originally Posted by Frankenstein Monster Someone else should have a go at this, I speak French extremely poorly. But according to slightly corrected Google translate, the argument is: So yes he finds parabolic orbits from two endpoints and a varying point inbetween and notes that they all then conflict with the rest of the observations. Or something.
I believe your description is consistent with what he writes. First he calculated a parabolic orbit matching two observations and having a perihelion of 10 AU, and noted that it did not match the intermediate observations. Then he tried 8 AU, and it also did not work. "Whence it is easy to conclude that if one tried to find a parabolic orbit which at the same time satisfied the observations of 9th April 1781 and 21 October 1782 and also agreed with a third, intermediate, observation, then that parabolic orbit would be found to be in conflict with a great number of observations by at least 30 minutes: so that it is incontestably proved that a parabolic orbit can absolutely not satisfy the observations; and the errors even becoming so considerable that it is right to presume that ellipses whose eccentricities are slightly considerable find themselves completely excluded."

So, it does not sound like he is deriving rigorous error bars on the value of the eccentricity, more like claiming it cannot be too great and still agree with all the observational data. [Actual eccentricity of Uranus's orbit: about 0.046]. Indeed, on page 74 he discusses fitting parabolic orbits to the data of 17 March 1781 and 23 January 1782, and notes there is no way to match them up with a couple of intermediate observations, but notes that ellipses will produce smaller errors, the more observational data the more shapes can be ruled out, and that it will take several years of observation to determine the precise value of the eccentricity.
#13
10-08-2019, 01:55 PM
 Guest Join Date: Oct 2001 Posts: 1,352
So would he use the length of time between the 2 observations and the angle between them to first get a ballpark figure for the object's distance from the sun?

And then use that distance to work out plausible parabolic (ie non-orbital) curves? Would any other variables be required before you start comparing the hypothetical parabolas with the actual 3rd observations?
#14
10-08-2019, 02:43 PM
 Guest Join Date: May 2016 Posts: 3,852
Quote:
 Originally Posted by The Stafford Cripps So would he use the length of time between the 2 observations and the angle between them to first get a ballpark figure for the object's distance from the sun? And then use that distance to work out plausible parabolic (ie non-orbital) curves? Would any other variables be required before you start comparing the hypothetical parabolas with the actual 3rd observations?
On Page 71, Lexell states that he got a ballpark figure of 18.92 AU by taking two observations and also assuming a circular orbit. He does this a bunch of times with other pairs of observations, getting values like 18.86, 18.89, 18.83, finally concluding that the orbit is pretty circular, but not exactly. Then he says he wants to rule out parabolas and highly eccentric ellipses, which is why he tries fitting various parabolic orbits.

Now, an orbit is determined by 6 degrees of freedom, so for example three observations will do it. I am not sure why he did not try taking three accurate, well-spaced observations, computing the exact orbital elements that fit those observations, and seeing how well those match up with the rest. Perhaps he felt he was no Gauss, or that the data was not precise enough to justify the effort (which, remember, involved a lot of scratch paper, ink, almanacs, trigonometric and logarithmic tables).

As for the parabolas, two observations yield four constraints, setting the eccentricity = 1 one more, so there is still one more variable missing before you can work out an actual parabola. Which is why he was able to find different parabolic orbits by assuming varying perihelion distances.
#15
10-08-2019, 03:00 PM
 Guest Join Date: May 2016 Posts: 3,852
Ah ha! I found this book excerpt purporting to give Kepler's original method: essentially, you can figure out the sidereal period of a planet and, given loads of observations, compute many points on the planet's orbit, enough to prove conclusively they are ellipses and not some other curve.
#16
10-08-2019, 03:47 PM
 Guest Join Date: Oct 2001 Posts: 1,352
Thanks, DPRK, I can just about grasp the bit about Kepler. But when it says

"...the orbital motion of a planet and the earth moving from an initial position with respect to the sun (opposition) to a position that repeats the initial alignment. This associated time interval is known as the synodic period of planet p with respect to the earth"

does it not contradict the previous line,

"The synodic period is simply the length of time required for the planet to return to the same place in the sky as seen from the earth",

since opposition does not mean being in the same place in the sky (with respect to background stars)?
#17
10-08-2019, 03:51 PM
 Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 85,111
Of course, Kepler had the advantage that all of the objects he was studying were periodic, and he had observational data spanning multiple periods for each of them. That makes the task much simpler, because you can find times when the planet you're looking at was in the same position, but the Earth was in different positions, and so triangulate its position at that time.
#18
10-08-2019, 03:52 PM
 Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 85,111
The synodic period is when the object is in the same position relative to the Earth and Sun, not relative to the background stars. To use the simplest example, the synodic period of the Moon is the time from one full moon to the next.
#19
10-08-2019, 03:56 PM
 Guest Join Date: Oct 2001 Posts: 1,352
That's what I thought. So the 2nd sentence I quoted (which is the first one in the actual text) is wrong?
#20
10-08-2019, 04:34 PM
 Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 85,111
It is "the same place in the sky as seen from the Earth", but the quote doesn't say same place with respect to what. You were interpreting it to mean "same place with respect to the background stars", while the writer meant (but could have expressed more clearly) "same place with respect to the Sun".
#21
10-08-2019, 06:32 PM
 Isaiah 1:15/Screw the NRA Charter Member Join Date: May 2003 Location: SoCal Posts: 51,649
James Burke did a very nice little explanation of Kepler's calculations in one of the The Day the Universe Changed episodes. #5 in fact. Go to 31:00 here. Not a lot of math, but it gets the idea across.

Last edited by silenus; 10-08-2019 at 06:32 PM.
#22
10-09-2019, 10:29 AM
 Guest Join Date: Feb 2009 Posts: 15,079
Just a note here- if you are working with degrees and minutes, 90 degrees and 60 minutes, a 4-digit trig table should be adequate for calculations (90x60=5400). They had better than that by the time of Kepler.

Quote:
 From Muslim Spain, trigonometric tables spread to Latin Europe. Regiomontanus (1436–76), German astronomer and mathematician, composed the first tables with decimal values. Similarly, Georg Joachim Rheticus (1514–74), a student of Nicolaus Copernicus, prepared a magnificent set of tables of all six trigonometric functions at 10″ increments accurate to 10 decimal places. Rheticus also took the decisive steps of defining the trigonometric functions in terms of angles rather than arcs and as ratios rather than lengths
#23
10-09-2019, 11:37 AM
 Guest Join Date: Oct 2001 Posts: 1,352
Quote:
 Originally Posted by silenus James Burke did a very nice little explanation of Kepler's calculations in one of the The Day the Universe Changed episodes. #5 in fact. Go to 31:00 here. Not a lot of math, but it gets the idea across.
I probably saw this programme when it came out. It does answer my questions quite well along with the passage cited by DPRK. I think the French guy must have had a more advanced method than the one described in the passage for calculating the distance of Uranus in such a short period of time.
#24
10-09-2019, 07:08 PM
 Guest Join Date: May 2016 Posts: 3,852
By French guy, I assume you mean Laplace -- of course he had a more advanced method than Kepler's, because as a starting point he assumed all of Kepler's laws and Newton's laws of motion. Anyway, by the late 18th century they also had access to more and more powerful astronomical instruments and observations, so precision of seconds of arc was possible.

ETA as for calculating the orbit in a short amount of time, since you know the dynamics you can do it with a minimum of 3 celestial observations, and they already had many more than that for Uranus

Kepler, as we saw, used a geometric triangulation technique to get many points on the orbit of Mars, which he was able to fit on an ellipse. If you don't care about a few percent error, you can even do this while assuming the orbit of the Earth is perfectly circular and that the Earth and Mars orbit in the same plane.

Last edited by DPRK; 10-09-2019 at 07:12 PM.
#25
10-09-2019, 08:06 PM
 Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 85,111
In practice, you'd do a first approximation based on assumptions like that, then you'd use the results of that to improve your model of the Earth's orbit, then use the improved orbit of the Earth to make a new model for Mars, and so on. It converges fairly quickly.

There are also methods for solving for all of the parameters of all of the objects at once, but those are impractical without a computer.
#26
10-10-2019, 03:33 PM
 Guest Join Date: Jan 2003 Location: washington, dc Posts: 995
It is fortunate, I think, that Kepler focused on Mars, and that Mars has a fairly eccentric orbit (it varies from 209 m km to 249 m km from the sun). If Mars' orbit were more circular, like that of Venus (varies from 107 to 108 m km), he might never have discovered that the planets orbit in ellipses.

Last edited by zimaane; 10-10-2019 at 03:33 PM.
#27
10-10-2019, 04:08 PM
 Member Join Date: Feb 2001 Location: Not the PNW :-( Posts: 20,306
Kepler was also fortunate to have some really great data on Mars collected by Brahe.
#28
10-11-2019, 08:14 AM
 Charter Member Moderator Join Date: Jan 2000 Location: The Land of Cleves Posts: 85,111
I think that he focused on Mars precisely because it's more eccentric. His first attempt would of course have been based on the assumption of circles. When he was trying to improve it, he'd naturally focus his attention on the cases where the circle model worked least well.

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