Orbits of the Outer Planets

I notice that figures on various web pages and in books for
the mean radius of the orbits of Saturn, Uranus, Neptune, and
Pluto frequently disagree in the third or even the second
significant digit. Such a large disagreement between sources
surprises me. Is there any any explanation for it?

Here are the figures I'm currently using:

Mercury 57,910,000 km = 0.3871 AU
Venus 108,200,000 km = 0.7233 AU
Earth 149,597,870 km = 1 AU
Mars 227,940,000 km = 1.5237 AU
Jupiter 778,330,000 km = 5.203 AU
Saturn 1,429,400,000 km = 9.5388 AU
Uranus 2,870,990,000 km = 19.1914 AU
Neptune 4,504,300,000 km = 30.0611 AU
Pluto 5,913,520,000 km = 39.5294 AU

It is always possible that I've made gross errors, as well,
so please point out to me whatever you see. :-)

-- Jeff, in Minneapolis

Subtract 1 from my e-mail address above for my real address.
..

Dear Jeff Root:

"Jeff Root" wrote in message
om...
I notice that figures on various web pages and in books for
the mean radius of the orbits of Saturn, Uranus, Neptune, and
Pluto frequently disagree in the third or even the second
significant digit. Such a large disagreement between sources
surprises me. Is there any any explanation for it?

Here are the figures I'm currently using:

Mercury 57,910,000 km = 0.3871 AU
Venus 108,200,000 km = 0.7233 AU
Earth 149,597,870 km = 1 AU
Mars 227,940,000 km = 1.5237 AU
Jupiter 778,330,000 km = 5.203 AU
Saturn 1,429,400,000 km = 9.5388 AU
Uranus 2,870,990,000 km = 19.1914 AU
Neptune 4,504,300,000 km = 30.0611 AU
Pluto 5,913,520,000 km = 39.5294 AU

URL:http://nssdc.gsfc.nasa.gov/planetary...ble_ratio.html

Saturn 1,433,530,000 = 9.58 (three times the eccentricity of Earth)
Uranus 2,872,460,000 = 19.20 (nearly three times the eccentricity)
Neptune 4,495,060,000 = 30.05 (less eccentric but completes one orbit in
163 years)
Pluto 5,869,660,000 = 39.24 (fourteen times as eccentric, one orbit
completes in 247 years)

These web pages were updated this month. I suspect the variance is due to
the difficulty that orbit eccentricity and long periods present in
accuracy.

My guess.

David A. Smith

"Jeff Root" wrote in message
om...
I notice that figures on various web pages and in books for
the mean radius of the orbits of Saturn, Uranus, Neptune, and
Pluto frequently disagree in the third or even the second
significant digit. Such a large disagreement between sources
surprises me. Is there any any explanation for it?

Here are the figures I'm currently using:

Mercury 57,910,000 km = 0.3871 AU
Venus 108,200,000 km = 0.7233 AU
Earth 149,597,870 km = 1 AU
Mars 227,940,000 km = 1.5237 AU
Jupiter 778,330,000 km = 5.203 AU
Saturn 1,429,400,000 km = 9.5388 AU
Uranus 2,870,990,000 km = 19.1914 AU
Neptune 4,504,300,000 km = 30.0611 AU
Pluto 5,913,520,000 km = 39.5294 AU

It is always possible that I've made gross errors, as well,
so please point out to me whatever you see. :-)

-- Jeff, in Minneapolis

Subtract 1 from my e-mail address above for my real address.

I doubt that what follows will fully account for the discrepancies but it is
something to keep in mind.

The mean orbital radius is often quoted as being equal to the semi-major
axis, but this is, in fact, just a very good approximation for orbits with
small eccentricity. The full solution is given by ...

Mean Orbital Radius = a(1 + (e^2)/2)

were a = semi-major axis and e = eccentricity

(It is difficult to find a reference for this on-line, but I did find one
once. If you would like, I could see if I can dig it up.)

David A. Smith replied to Jeff Root:

URL:http://nssdc.gsfc.nasa.gov/planetary...ble_ratio.html

I was sure I used the fact sheet pages on that site for the
individual planets, just a couple of months ago. But they all
have more data on them than I remember. They have exactly what
I was looking for-- and eventually found most of, scattered
across four or five different sites, including one in German--
all in one place. Maybe I was looking at a mirror site, and
the mirror needed re-aluminizing.

Thank you!

-- Jeff, in Minneapolis

..

John Zinni replied to Jeff Root:

The mean orbital radius is often quoted as being equal to the
semi-major axis, but this is, in fact, just a very good
approximation for orbits with small eccentricity. The full
solution is given by ...

Mean Orbital Radius = a(1 + (e^2)/2)

where a = semi-major axis and e = eccentricity

I'm lousy at most aspects of math, but had noticed and wondered
about what seemed to be a discrepancy between statements about
the meaning of "mean" in this case. I didn't follow up on that
puzzlement. Now I will.

(It is difficult to find a reference for this on-line, but I
did find one once. If you would like, I could see if I can dig
it up.)

I would like to know. If you can't easily locate the URL, it
might be enough if you were to suggest a search term or two in
addition to the obvious ones in your first reply. Thank you!

-- Jeff, in Minneapolis

..

"Jeff Root" wrote in message
m...
John Zinni replied to Jeff Root:

(It is difficult to find a reference for this on-line, but I
did find one once. If you would like, I could see if I can dig
it up.)

I would like to know. If you can't easily locate the URL, it
might be enough if you were to suggest a search term or two in
addition to the obvious ones in your first reply. Thank you!

Hi Jeff

I will give you a brief history (if you will bear with me) of how I found
this solution because I think it illustrates well what is going on.

The question of "What is the mean 'r' for an elliptic orbit" came up in
conversation with a friend in relation to a course we were both taking at
the time. "No problem!" I thought to myself (more than a little cocky!), all
you need to do is express the equation of an ellipse, with one focus at the
origin, in polar form, integrate from 0 to 2Pi and divide by 2Pi (search on
something like "mean of a continues function" for this). Well, if you do
this, what you come up with is 'b' (that is the semi-MINOR axis) and this
contradicted all references I had that the mean orbital radius is 'a' (the
semi-MAJOR axis). After re-doing the integration about 4 or 5 times and
coming up with the same answer each time, we came to the realization that
the mean 'r' for a plain old ellipse and for an elliptic orbit are not the
same. The reason for the discrepancy is Kepler's Second Law. That is, for a
plain old ellipse, equal weight is given to all values of 'r' but for an
elliptic orbit, we know from Kepler's Second Law that at large 'r' the
orbital speed is less that at small 'r' (the object hangs out at large
values of 'r' for more time than it does at small values of 'r', so large
values of 'r' need to be more heavily weighted than the small values of 'r'
hence lengthening the mean 'r'). This was what prompted the search for the
actual derivation of the value of the mean 'r'.

I found this paper. It looks like a problem that was posed in a 2nd or 3rd
year math class. It states the problem and gives the solution in the first
part. The problem the students had to solve was the rather nasty definite
integral involved (If, as you say, you are "lousy at most aspects of math"
than the solutions to the definite integral given by three of the students
is likely to look like gobbledygook).
http://www.math.wisc.edu/~robbin/angelic/anIntegral.pdf

Enjoy!

John Zinni

In article , Jeff Root wrote:
I notice that figures on various web pages and in books for
the mean radius of the orbits of Saturn, Uranus, Neptune, and
Pluto frequently disagree in the third or even the second
significant digit. Such a large disagreement between sources
surprises me. Is there any any explanation for it?

Here are the figures I'm currently using:

Mercury 57,910,000 km = 0.3871 AU
Venus 108,200,000 km = 0.7233 AU
Earth 149,597,870 km = 1 AU
Mars 227,940,000 km = 1.5237 AU
Jupiter 778,330,000 km = 5.203 AU
Saturn 1,429,400,000 km = 9.5388 AU
Uranus 2,870,990,000 km = 19.1914 AU
Neptune 4,504,300,000 km = 30.0611 AU
Pluto 5,913,520,000 km = 39.5294 AU

I"ll bet the reason you see different tabulated values is simply
that the orbits aren't constant! The planets perturb each others' orbits,
especially Jupiter. So the precise elements you get depends on the
time you take the "snapshot".

I once wrote a planetarium program that used osculating orbital elements
dated from the beginning of 1996. Within a few years, the outer planets'
positions were different by up to several arc minutes from the
ones my program computed using fixed Keplerian orbits. I'd imagined
that the perturbations would be too small to notice, but they really weren't.

Cheers

Stuart

(Jeff Root) wrote in news:964c68dd.0310261015.13804bf0
@posting.google.com:

I notice that figures on various web pages and in books for
the mean radius of the orbits of Saturn, Uranus, Neptune, and
Pluto frequently disagree in the third or even the second
significant digit. Such a large disagreement between sources
surprises me. Is there any any explanation for it?

Here are the figures I'm currently using:

Mercury 57,910,000 km = 0.3871 AU
Venus 108,200,000 km = 0.7233 AU
Earth 149,597,870 km = 1 AU
Mars 227,940,000 km = 1.5237 AU
Jupiter 778,330,000 km = 5.203 AU
Saturn 1,429,400,000 km = 9.5388 AU
Uranus 2,870,990,000 km = 19.1914 AU
Neptune 4,504,300,000 km = 30.0611 AU
Pluto 5,913,520,000 km = 39.5294 AU

It is always possible that I've made gross errors, as well,
so please point out to me whatever you see. :-)

-- Jeff, in Minneapolis

Subtract 1 from my e-mail address above for my real address.
.

Dunno. JPL will have the best data. Their digital orrery is used to get
spacecraft to most of those planets so must be accurate.

Llanzlan