Precession of polar satellites

Assuming a satellite in Earth-polar orbit of about 20,000 km radius,
would the orbit precess and, if so, how fast? Are there simple,
approximate equations for precession in general?

In article ,
Bill Bogen wrote:
Assuming a satellite in Earth-polar orbit of about 20,000 km radius,
would the orbit precess and, if so, how fast?

If the orbit is *exactly* polar, the precession rate is zero, to a first
approximation.

Are there simple,
approximate equations for precession in general?

The first-order approximation is

precessionrate = -3/2 * J2 * (R^2 / p^2) * n * cos(i)

where J2 is a constant related to Earth's flattening (about 0.001), R is
the radius of the Earth (about 6378km), p is the "parameter" of the orbit
(= a*(1-e^2)), n is the mean motion of the orbit (=sqrt(mu/a^3)), and i is
the inclination. (And "a" is the semimajor axis of the orbit, equal to
its radius for a circular orbit, "e" is the eccentricity, 0 for a circular
orbit, and "mu" is about 398600km^3/s^2, Earth's mass times the universal
gravitational constant.) The result is in radians per second if you've
been consistent with the other units.

So not entirely simple, but not too hard to calculate.
--
MOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer
pointing, 10 Sept; first science, early Oct; all well. |

(Henry Spencer) wrote in message ...
In article ,
Bill Bogen wrote:
Assuming a satellite in Earth-polar orbit of about 20,000 km radius,
would the orbit precess and, if so, how fast?

If the orbit is *exactly* polar, the precession rate is zero, to a first
approximation.

Are there simple,
approximate equations for precession in general?

The first-order approximation is

precessionrate = -3/2 * J2 * (R^2 / p^2) * n * cos(i)

where J2 is a constant related to Earth's flattening (about 0.001), R is
the radius of the Earth (about 6378km), p is the "parameter" of the orbit
(= a*(1-e^2)), n is the mean motion of the orbit (=sqrt(mu/a^3)), and i is
the inclination. (And "a" is the semimajor axis of the orbit, equal to
its radius for a circular orbit, "e" is the eccentricity, 0 for a circular
orbit, and "mu" is about 398600km^3/s^2, Earth's mass times the universal
gravitational constant.) The result is in radians per second if you've
been consistent with the other units.

So not entirely simple, but not too hard to calculate.

Assuming the orbit is exactly polar (inclination = 0) but rather
eccentric (non-circular), then the _plane_ of the orbit does not
precess but are the axes of the orbit affected by the lumpiness of the
Earth? IOW, if the perigee of the orbit starts at, say, 45 degrees
above the Earth's equator, is the perigee eventually tugged down
toward the bulge of the equator? Or does the Earth's lumpiness change
the shape of the orbit in other ways?

In article ,
Bill Bogen wrote:
Assuming the orbit is exactly polar (inclination = 0) but rather
eccentric (non-circular), then the _plane_ of the orbit does not
precess but are the axes of the orbit affected by the lumpiness of the
Earth?

Precession of an elliptical orbit *within* the orbit plane is another
story entirely.

For an exactly polar orbit (that's 90deg, not 0deg), the precession rate
of the orbit plane is (to a first approximation) zero.

However, the equatorial bulge also makes an elliptical orbit precess
within its plane. *That* effect is zero at a different inclination, about
63.5deg, which is why there are a bunch of satellite orbits at around that
inclination (starting with Sputnik 1!).

The first-order-approximation formula for that one is

precessionrate = 3/2 * J2 * (R^2 / p^2) * n * (2 - 5/2 * sin^2 i)

with the same symbols as before.

IOW, if the perigee of the orbit starts at, say, 45 degrees
above the Earth's equator, is the perigee eventually tugged down
toward the bulge of the equator?

This one too is a steady precession at essentially a constant rate, the
orbit rotating within its plane.

Or does the Earth's lumpiness change the shape of the orbit in other ways?

There are smaller effects from lumpiness other than the equatorial bulge,
and also more subtle effects (again generally small) from the bulge itself
when you go beyond first approximations. Finally, for a significantly
elliptical orbit, the higher altitude at apogee means that lunar and solar
perturbations are not negligible. (For example, they cause periodic
oscillations in orbit eccentricity, which can push the orbit perigee down
into the atmosphere "temporarily", long before the orbit would actually
decay.)
--
MOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer
pointing, 10 Sept; first science, early Oct; all well. |

Henry Spencer wrote:
However, the equatorial bulge also makes an elliptical orbit precess
within its plane. *That* effect is zero at a different inclination,
about 63.5deg, which is why there are a bunch of satellite orbits at
around that inclination (starting with Sputnik 1!).

How did they know? Presumably the Russians had no computers then.
And of course they had no previous satellite orbits to study (except
the moon's).

And why did they care? Wasn't Sputnik 1 in a roughly circular orbit,
anyway?
--
Keith F. Lynch - - http://keithlynch.net/
I always welcome replies to my e-mail, postings, and web pages, but
unsolicited bulk e-mail (spam) is not acceptable. Please do not send me
HTML, "rich text," or attachments, as all such email is discarded unread.

In article ,
Keith F. Lynch wrote:
... *That* effect is zero at a different inclination,
about 63.5deg, which is why there are a bunch of satellite orbits at
around that inclination (starting with Sputnik 1!).

How did they know? Presumably the Russians had no computers then.
And of course they had no previous satellite orbits to study (except
the moon's).

The effects of the equatorial bulge are relatively easy to study
theoretically, using just pencil and paper, and the bulge is prominent
enough that you can get a first-approximation measurement of it with
geodetic surveys. And indeed, you *do* need to understand its effects to
calculate the Moon's orbit... although analysis of the Moon's motion is
horribly difficult and so it doesn't give you reliable feedback about the
size of the bulge etc.

The first really precise measurements of the bulge came from satellites,
as did essentially all our knowledge of Earth's lesser gravitational
irregularities, but the general nature of the main effects of the bulge
was understood long before Sputnik 1.

And why did they care? Wasn't Sputnik 1 in a roughly circular orbit,
anyway?

228x947 km, which is significantly elliptical. Apparently they chose that
inclination to make orbit prediction simpler. Keeping track of all the
fiddly details without a computer is definitely a hassle, even if you can
understand each of them individually with just pencil and paper, so a
cheap way of getting rid of one complication has its attractions.
--
MOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer
pointing, 10 Sept; first science, early Oct; all well. |

"Keith F. Lynch" wrote in message ...
Henry Spencer wrote:
However, the equatorial bulge also makes an elliptical orbit precess
within its plane. *That* effect is zero at a different inclination,
about 63.5deg, which is why there are a bunch of satellite orbits at
around that inclination (starting with Sputnik 1!).

How did they know? Presumably the Russians had no computers then.
And of course they had no previous satellite orbits to study (except
the moon's).


I think the bulge started being detected during the French revolution
(just read the Nov 2002 American Scientist review of a book about the
establishment of the meter).

And why did they care? Wasn't Sputnik 1 in a roughly circular orbit,
anyway?


May not have been as important for the first satellite itself, but
when you start wanting long flights...Sputnik 1 was probably being
used to check the predictions so that the 2nd satellite would go where
they wanted it.

/dps

(Henry Spencer) wrote in
:

In article ,
Keith F. Lynch wrote:
... *That* effect is zero at a different inclination,
about 63.5deg, which is why there are a bunch of satellite orbits at
around that inclination (starting with Sputnik 1!).

How did they know? Presumably the Russians had no computers then.
And of course they had no previous satellite orbits to study (except
the moon's).

The effects of the equatorial bulge are relatively easy to study
theoretically, using just pencil and paper, and the bulge is prominent
enough that you can get a first-approximation measurement of it with
geodetic surveys. And indeed, you *do* need to understand its effects
to calculate the Moon's orbit... although analysis of the Moon's
motion is horribly difficult and so it doesn't give you reliable
feedback about the size of the bulge etc.

The first really precise measurements of the bulge came from
satellites, as did essentially all our knowledge of Earth's lesser
gravitational irregularities, but the general nature of the main
effects of the bulge was understood long before Sputnik 1.

More to the point, at least with respect to apsidal rotation, you don't
*need* to know the size of the equatorial bulge *at all* to know the
critical inclination where the effect goes to zero - it happens to be
independent of the size of the bulge. The critical inclination is equal to
asin(sqrt(0.8)). The astute observer will note that 1) there are actually
two roots to this equation, one at 63.4 degrees and the other at 116.6
degrees, and 2) the solution is general, not Earth-specific. However, it
does assume no higher-order bulges that can cause other effects, so it
probably wouldn't work on the moon or some other "lumpy" body.

--
JRF

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check "Organization" (I am not assimilated) and
think one step ahead of IBM.

I was always told that the inclination of the orbit of Sputnik was
determined by the flight path that provided the maximum ground
tracking within the USSR. That is why Mir II was supposed to fly at
65 degrees, as well.


(Henry Spencer) wrote in message ...
And why did they care? Wasn't Sputnik 1 in a roughly circular
orbit,
anyway?

228x947 km, which is significantly elliptical. Apparently they chose that
inclination to make orbit prediction simpler. Keeping track of all the
fiddly details without a computer is definitely a hassle, even if you can
understand each of them individually with just pencil and paper, so a
cheap way of getting rid of one complication has its attractions.

In article ,
Explorer8939 wrote:
...Apparently they chose that
inclination to make orbit prediction simpler....

I was always told that the inclination of the orbit of Sputnik was
determined by the flight path that provided the maximum ground
tracking within the USSR.

The "simpler prediction" account comes from one of the Vanguard guys, but
I don't know how solid his information was.
--
MOST launched 30 June; first light, 29 July; 5arcsec | Henry Spencer
pointing, 10 Sept; first science, early Oct; all well. |