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Aodoi
2001-Dec-27, 09:36 PM
Hey, got into a debate with my father over Christmas dinner on this (yes, we're a family of mostly engineers), figured I have a rare chance to prove him wrong so I'm looking for corroboration.

In several places on this site and others I've seen that a geosynchronous orbit is around 40,000 km above the equator (I think somebody gives it as like 37,500 km) but I can't seem to find anybody who shows the math behind it.

My father maintains the distance of the orbit depends on the mass of the satellite, my brother and I maintain that the mass of the satellite doesn't matter as it cancels out between the gravity attraction and the force needed to keep the satellite in orbit.

Anybody that can post the math or a link to a site that does? Thanks.

DALeffler
2001-Dec-27, 10:51 PM
Rocket and Space Technology (http://users.commkey.net/Braeunig/space/index.htm)

Go git 'im!

Wiley
2001-Dec-27, 11:04 PM
The period of the orbit is essentially independent of the satellite's mass. If we can assume the mass of the earth M is much greater than the mass of the satellite m, the period of the orbit will only depend on the mass of the earth M.

We assume that the centripetal acceleration is due to solely to gravity, this gives

g = (G M)/r^2 = r w^2

where
G is the gravitational constant (6.67e-11 N m^2/kg^2)
M is the mass of the earth (5.98e24 kg)
w is the angular velocity.

We can thus derive the relation

r = (G*M/w^2)^(1/3)

and for geosynchonous orbits w = 2*pi/24 hours or w= 7.29e5 rad/s. And this gives a radius of 4.22e7 m. To get the altitude, we must subtract the radius of the earth R = 6.37e6m. Thus the altitude of a geosynchronous satellite is 3.58e7 m.

Hope this helps

GrapesOfWrath
2001-Dec-28, 12:12 AM
On 2001-12-27 18:04, Wiley wrote:
for geosynchonous orbits w = 2*pi/24 hours or w= 7.29e5 rad/s.

Since you're keeping three digits in your arithmetic, you probably should make that w = 2*pi/ (23 hours 56 minutes)

Aodoi
2001-Dec-28, 01:49 PM
Thanks guys /phpBB/images/smiles/icon_smile.gif

Wiley
2001-Dec-28, 05:18 PM
On 2001-12-27 19:12, GrapesOfWrath wrote:


On 2001-12-27 18:04, Wiley wrote:
for geosynchonous orbits w = 2*pi/24 hours or w= 7.29e5 rad/s.

Since you're keeping three digits in your arithmetic, you probably should make that w = 2*pi/ (23 hours 56 minutes)



I actually used 23 hrs. 56 min, so 7.29e5 rad/s is correct. A period of exactly 24 hrs. gives 7.27 rad/s. /phpBB/images/smiles/icon_wink.gif

GrapesOfWrath
2001-Dec-28, 05:23 PM
Very good! I should have known better than to doubt you.

jkmccrann
2005-Nov-17, 11:43 AM
Does anybody know how many satellites there currently are in geostationary orbit?

Cheers! :)

Damburger
2005-Nov-17, 12:40 PM
If somebody believes that an orbit depends on a satelites mass, then you correct them by explaining that an object in orbit is one that simply moves so quickly the Earth curves away beneath it as fast as it falls the Earth. The object is basically falling around the Earth. Because all things fall at the same rate regardless of mass, mass doesn't affect orbit.

Then it is simply a case of demonstrating to your skeptic that two things of different mass fall at the same rate. Choose your demonstration articles carefully though - a hammer and a feather won't work on Earth due to air resistance.

Footage of the Apollo astronauts doing it on the moon would probably do.

Bob B.
2005-Nov-17, 03:18 PM
Rocket and Space Technology (http://users.commkey.net/Braeunig/space/index.htm)

Go git 'im!
Thanks for referencing my web site, DALeffler. Unfortunately you're given the old URL. It is now located at:

Rocket and Space Technology (http://www.braeunig.us/space/index.htm)

The Orbital Mechanics (http://www.braeunig.us/space/orbmech.htm) page gives everything you need to know. Example problem 1.3 (http://www.braeunig.us/space/problem.htm#1.3) calculates geosynchronous distance.

The mass of the satellite is infinitesimally small in comparison to the Earth, thus it is not considered in the calculation. For a large satellite, like the Moon, then the satellite’s mass must be considered because the center of mass of the system is shifted far enough toward the secondary that the orbital period is effected.

publiusr
2005-Nov-18, 10:53 PM
With a true geostationary orbit you are dead stopped with respect to the Earth below. That is very hard. Geosynch sats wander a bit in a figure-8 over roughly the same spot.

So geosynch and geostationary aren't quite the same.

tlbs101
2005-Nov-18, 11:59 PM
With a true geostationary orbit you are dead stopped with respect to the Earth below. That is very hard. Geosynch sats wander a bit in a figure-8 over roughly the same spot.

So geosynch and geostationary aren't quite the same.

Commsats launched by Zenit Sea Launch have achieved nearly perfect geostationary orbits, but that is because they are launched from the equator.

The only real difference between a geostationary and geosynchronous orbit is inclination. Geostationary orbits could be considered subsets of geosynchronous orbits, where the inclination is 0.0 degrees.

genebujold
2005-Nov-20, 03:17 AM
Hey, got into a debate with my father over Christmas dinner on this (yes, we're a family of mostly engineers), figured I have a rare chance to prove him wrong so I'm looking for corroboration.

In several places on this site and others I've seen that a geosynchronous orbit is around 40,000 km above the equator (I think somebody gives it as like 37,500 km) but I can't seem to find anybody who shows the math behind it.

My father maintains the distance of the orbit depends on the mass of the satellite, my brother and I maintain that the mass of the satellite doesn't matter as it cancels out between the gravity attraction and the force needed to keep the satellite in orbit.

Anybody that can post the math or a link to a site that does? Thanks.

Actually, your father's right, although you're only going see a difference of a mm or so between a small satellite and something as large as the ISS.

From wiki: "A geostationary orbit (abbreviated GEO) is a circular orbit directly above the Earth's equator (0 latitude)."

The problem with this entry is that he assumed the satellite's pull on the Earth was negligeable, and for the purpose of our satellites, it is.

But a moon is also a satellite, you must remember that any additional mass of the satellite increases the pull between the satellite and the planet it's orbiting.

That increased gravitational attraction must be countered with both increased distance (orbital altitude) and velocity in order to keep the satellite in a geostationary orbit.

A geostationary orbit is approximately 35,786 km above mean sea level in the plane of the equator.

If you had one Earth oribiting another, and knowing that the gravitational attraction varies with the square root of the distance between them, radial velocities, centripetal accelerations...

Even without going through all the math it's clear that the altitude would be considerably higher due to the mutual gravitational pull being twice as large.

ngc3314
2005-Nov-21, 03:32 AM
Actually, your father's right, although you're only going see a difference of a mm or so between a small satellite and something as large as the ISS.

From wiki: "A geostationary orbit (abbreviated GEO) is a circular orbit directly above the Earth's equator (0 latitude)."

The problem with this entry is that he assumed the satellite's pull on the Earth was negligeable, and for the purpose of our satellites, it is.

But a moon is also a satellite, you must remember that any additional mass of the satellite increases the pull between the satellite and the planet it's orbiting.

That increased gravitational attraction must be countered with both increased distance (orbital altitude) and velocity in order to keep the satellite in a geostationary orbit.

A geostationary orbit is approximately 35,786 km above mean sea level in the plane of the equator.

If you had one Earth oribiting another, and knowing that the gravitational attraction varies with the square root of the distance between them, radial velocities, centripetal accelerations...

Even without going through all the math it's clear that the altitude would be considerably higher due to the mutual gravitational pull being twice as large.


This distinction enters in Newton's generalization of Kepler's third law, in which the period squared or orbital radius cubed differ by the factor (m+M)/M from the vanilla version for infinitesimal satellite mass (in which M is the greater mass and m the smaller one - there is obviously symmetry depending on which one you measure from). It also factors into the definition of the astronomical unit, which is not exactly the semimajor axis of the Earth's orbit. For the really pedantic, it is the semimajor axis of the orbit of the same period by a fictitious massless object orbiting the Sun (example here (http://neo.jpl.nasa.gov/glossary/au.html)).

GOURDHEAD
2005-Nov-21, 03:43 PM
When one says that a geocynchronous orbit is circular, within what tolerances of circularity is this true? Since the plane of the elliptical orbit of the moon is not coincident with the plane of the Earth's equator, is there a cumulative effect tending to drag objects out of geosynchronous orbit? Since the plane of the Earth's equator is not coincident with that of its elliptical orbit around the sun, does a similar effect apply? If such effects exist, how can they be minimized? Answers to these questions are quite important to those planning to invest in space elevators.

01101001
2005-Nov-21, 03:53 PM
When one says that a geocynchronous orbit is circular, within what tolerances of circularity is this true? [...] Answers to these questions are quite important to those planning to invest in space elevators.
Wikipedia: Geosynchronous orbit (http://en.wikipedia.org/wiki/Geosynchronous_orbit)


Circular geosynchronous orbits at the equator are known as geostationary orbits. A perfect stable geostationary orbit is an ideal that can only be approximated. In practice the satellite will drift out of this orbit (because of perturbations such as the solar wind, radiation pressure, variations in the Earth's gravitational field, and the gravitational effect of the Moon and Sun), and thrusters are used to maintain the orbit in a process known as stationkeeping.
How much less of a bother this will be if you can just run more fuel up a tether!

hhEb09'1
2005-Nov-21, 04:32 PM
Thanks for referencing my web site, DALeffler. Unfortunately you're given the old URL.Thread resurrection warning: DALeffler's post was from four years ago :)

When one says that a geocynchronous orbit is circular, within what tolerances of circularity is this true? Since the plane of the elliptical orbit of the moon is not coincident with the plane of the Earth's equator, is there a cumulative effect tending to drag objects out of geosynchronous orbit?The biggest perturbation of geosynch orbits is the non-circularity of the earth equatorial gravity field, IIRC

Bob B.
2005-Nov-21, 07:38 PM
Thread resurrection warning: DALeffler's post was from four years ago :)
D'oh :doh:

aurora
2005-Nov-21, 07:45 PM
Actually, your father's right, although you're only going see a difference of a mm or so between a small satellite and something as large as the ISS.


Not that the original post was from 2001, and the original poster has not posted anything since June of 2004, and so may not read your response.

ToSeek
2005-Nov-21, 11:19 PM
Does anybody know how many satellites there currently are in geostationary orbit?

Cheers! :)

This site (http://www.satsig.net/sslist.htm) lists 348.

genebujold
2005-Nov-22, 01:40 AM
This distinction enters in Newton's generalization of Kepler's third law, in which the period squared or orbital radius cubed differ by the factor (m+M)/M from the vanilla version for infinitesimal satellite mass (in which M is the greater mass and m the smaller one - there is obviously symmetry depending on which one you measure from). It also factors into the definition of the astronomical unit, which is not exactly the semimajor axis of the Earth's orbit. For the really pedantic, it is the semimajor axis of the orbit of the same period by a fictitious massless object orbiting the Sun (example here (http://neo.jpl.nasa.gov/glossary/au.html)).

Or merely the distance between their two geometric centers...