[Petrochemicals (OP)]

If orbit velocity is dependent on height and velocity,  and geosynchronous and geostationary orbits are due to the speed at a given distance, what is their velocity relative to: the surface of the earth, or the earth's position around the sun?

I understand the gravitational attraction of the earth and accelerations at opposites, but a geostationary orbit is not in anyway contrary to the gravitational attraction of earth, yet it does not descend. Does this mean that the orbit velocity is actually measured by the earths position to the sun and that is the measure of the velocity needed ? (and the sun galaxy, galaxy universe)

[Colin2B]

To stay in orbit and counter a gravitational field you have to be moving relative to the body creating the field. Although the satellite is stationary relative to earth surface it is moving at the same angular speed and in the the same direction as earth’s rotation. The reference point you need is earth centre.

[Petrochemicals (OP)]

To stay in orbit and counter a gravitational field you have to be moving relative to the body creating the field. Although the satellite is stationary relative to earth surface it is moving at the same angular speed and in the the same direction as earth’s rotation. The reference point you need is earth centre.

The earths rotation relative too the sun ?

The moving at opposites too the gravitational pull of the earth, the angular momentum created by a force one direction ie down, and the force perpendicular to at right angles  and not circlular orbit  primary force in  relation to that gravity, thus creating an orbit. From your position to the centre of the earth, and a force independant of that gravitational system but within the frame at perpendiculars to it.

So that must mean that the satelite in stationary orbit has the gravitational force on it and its speed is matched by the earths rotation, relative to the sun ?

[wolfekeeper]

If orbit velocity is dependant on height and velocity,  and geo syncronou and geo stationary orbits are due to the speed at a given distance, what is there velocity relative too ? The surface of the earth or the earths position around the sun.

 I understand the gravitational attracti9n of the earth and accelerations at opposites, but a geo stationary orbit it not in anyway contrary to the gravitational attraction of earth, yet it does not decend. Does this mean that the orbit velocity is actually measured by the earths position to the sun and that is the measure of the velocity needed ? (and the sun galaxy, galaxy universe)

The normal quoted numbers are relative to the center of the nearest big body. So a body in low earth orbit is moving about 7.8km/s relative to the center of the earth. But a body in low earth orbit is also moving relative to the sun. So for example, the moon is, in a real sense, in orbit around the sun and moving at about 30km/s around the sun (sometimes a bit faster or slower as it also goes around its orbit around the Earth).

[Janus]

As long as the satellite has a small mass with respect to the Earth (as are any of the ones we have put up), then the non-rotating Earth centered frame is what one measures the sattelite's velocity with respect to.   In fact, in the formula V= sqrt(GM/R) for orbital velocity, R is the distance between the centers of the orbiting mass and the primary body.

This changes as the mass of the satellite increases. At some point(when this is depends on just how accurate of an answer you require), you have to start including the mass of the satellite into the calculation.  At this point you no longer use the Earth's center as the focus of the orbit, but rather the "barycenter",  which is the center of  mass of the two bodies combined.
You also consider both bodies as orbiting this point. (this is not to say that small satellites don't orbit a barycenter, but that is is so close to the center of the Earth as to make no practical difference.) 
An example is the Earth-Moon pair. the Moon is massive enough and far enough away that the Earth-Moon barycenter is some 4267 km from the Earth's center and 2111 km below it surface.  The Earth's center orbits around this point with the same period as the Moon does.   (With many astronomical resources, it is common to use the E-M barycenter as the reference upon which our orbit around the Sun is based rather than the Earth's center.)
The Sun-Jupiter barycenter is just above the Sun's surface.(The earliest detection of extra-solar planets were based on this. The planets detected this way were so massive that they caused a large enough shift in the position of its star as it orbited the barycenter that we were able to measure it. This is also how the first black hole candidate was discovered. )

[evan_au]

what is their velocity relative to: the surface of the earth, or the earth's position around the sun?

You can look at the relative force on a satellite from the Earth, the Sun, the Moon, Jupiter, etc.

As Janus says, at the orbit of the Moon (380,000km from the center of the Earth), the tug of the Sun and the Earth is similar.

Because of the inverse square law, at geosynchronous orbit (42,000 km from the center of the Earth), the tug of the Earth is about 80 times stronger than tug of the Sun. So you can almost ignore the effect of the Sun on the location and speed of geosynchronous orbit.

In fact, if you transported the Earth into interstellar space, the radius of geosynchronous orbit would not change, even though the impact of the Sun is now zero (it is merely another star).

In practice, geosynchronous satellites have to stay within an imaginary "box" in space, so that the simple satellite TV antennas on the roof  don't need to be adjusted every year. The predictable gravitational tugs of the Sun, Moon and Earth's not-quite-spherical shape, combined with the variable solar wind means that geostationary satellite operators have to use planned bursts from onboard rockets to keep the satellite within its specified box.

https://en.wikipedia.org/wiki/Geostationary_orbit#Orbital_stability

[evan_au]

the earth's position around the sun?

Geostationary orbit is about the satellite having a constant location in the sky from a given point on the Earth's surface (independent of the position of the Sun), and is mostly used for communications satellites. This orbit sits above the Earth's equator, and orbits the Earth once in 24 hours (relative to the Sun), just as the Earth spins once in 24 hours (measured relative to the Sun). This orbit also goes once around the Earth once in 23 hours and 56 minutes (relative to the stars), just like the Earth spins once in 23 hours and 56 minutes (relative to the stars).

Another type of orbit is a Sun-synchronous orbit, which is mostly used for spy satellites. It aims to have a constant angle of illumination of the Earth's surface, using light from the Sun. This orbit typically passes near polar regions, and orbits once every 90 minutes.

See: https://en.wikipedia.org/wiki/Sun-synchronous_orbit

[Petrochemicals (OP)]

the earth's position around the sun?

Geostationary orbit is about the satellite having a constant location in the sky from a given point on the Earth's surface (independent of the position of the Sun), and is mostly used for communications satellites. This orbit sits above the Earth's equator, and orbits the Earth once in 24 hours (relative to the Sun), just as the Earth spins once in 24 hours (measured relative to the Sun). This orbit also goes once around the Earth once in 23 hours and 56 minutes (relative to the stars), just like the Earth spins once in 23 hours and 56 minutes (relative to the stars).

But subtract the sun from the equation and the geo stationary orbits are merely a satelite of the earth with no angular velocity, that for some reason does not decend, right?

[Colin2B]

But subtract the sun from the equation and the geo stationary orbits are merely a satelite of the earth with no angular velocity, that for some reason does not decend, right?

No, look through the answers above.
The satellite and the earth are both rotating with the same angular velocity, this results in a centripetal acceleration that balances the pull of gravity.
Earth's surface experiences a similar centripetal acceleration which causes the equatorial bulge and results in a small reduction in effective g at the surface. So g is slightly less at the equator than the poles.

[Janus]

the earth's position around the sun?

Geostationary orbit is about the satellite having a constant location in the sky from a given point on the Earth's surface (independent of the position of the Sun), and is mostly used for communications satellites. This orbit sits above the Earth's equator, and orbits the Earth once in 24 hours (relative to the Sun), just as the Earth spins once in 24 hours (measured relative to the Sun). This orbit also goes once around the Earth once in 23 hours and 56 minutes (relative to the stars), just like the Earth spins once in 23 hours and 56 minutes (relative to the stars).

But subtract the sun from the equation and the geo stationary orbits are merely a satelite of the earth with no angular velocity, that for some reason does not decend, right?

Look at it this way.  Take the Earth and put it out in interstellar space and stop its rotation.  Put a satellite up at a height of 35,667 km above the equator.  Accelerate it to ~3.07 km/sec to the East, parallel to the surface and the equator.  It is now in an orbit that will take  just a tad under 24 hrs to complete( I'll explain why it  is not exactly 24 hrs later), returning to the same spot above the equator every orbit.
Now start the Earth rotating back to its normal rate of 1 rotation per day.  This has no effect on the orbit of the satellite, but now the same point of the Equator stays directly under the satellite and you have a geostationary orbit.  From the ground it looks like it is just hanging motionless above the ground, but it is really just orbiting the Earth in in the dame direction and with the same period as the Earth rotates.   If you really need to use an outside reference system to measure these motions against, you can use the stars in the sky. Motions judged against this reference are labeled as being "sidereal".  Thus both the Earth and satellite have the same sidereal periods.  This brings up the issue of why you want to have your satellite orbit the Earth in just under 24 hrs.  This is because the sidereal period of the Earth's rotation is just under 24 hrs.  The 24 hr day is based on the Solar day of the Earth's rotation, or the time it takes for the Earth to complete one rotation with respect to the Sun.  Since the Earth is also orbiting the Sun, the Solar day is just a bit longer than the sidereal day (the time it takes the Earth to rotate once with respect to the stars.)  This is important to know because when you use a formula like  T= 2 pi sqrt(r^3/GM) to find the period of a satellite, the answer you get will be the sidereal period*, and thus if you want it to match the rotation of the planet, you need to use the sidereal period of rotation for the planet.

*Or at least really, really close. Since the stars in the sky are themselves part of the rotating galaxy, it won't be an exact match, but since it takes hundreds of millions of years for the galaxy to complete even one rotation, the difference is too small to be of any practical importance.

[evan_au]

the geo stationary orbits are merely a satelite of the earth with no angular velocity, that for some reason does not descend, right?

A satellite of the Earth needs some angular velocity around the Earth, or it would crash straight into the Earth at high velocity (or "descend", as you put it).

A point X on the surface of the Earth at the equator completes a journey of 40,000 km around the Earth in 24 hours, as the Earth rotates. It has a velocity of about 1,000 km/h.

A satellite in geostationary orbit  above X is about 4 times farther from the center of the Earth, but also completes one orbit in 24 hours. So the satellite has a velocity that is about 4 times faster than X, or about 4,000 km/h.

Both the surface of the Earth, and a satellite in geostationary orbit have considerable angular velocity.

(Sorry for rounding everything off so savagely - I have a plane to catch...)

[Petrochemicals (OP)]

But subtract the sun from the equation and the geo stationary orbits are merely a satelite of the earth with no angular velocity, that for some reason does not decend, right?

No, look through the answers above.
The satellite and the earth are both rotating with the same angular velocity, this results in a centripetal acceleration that balances the pull of gravity.
Earth's surface experiences a similar centripetal acceleration which causes the equatorial bulge and results in a small reduction in effective g at the surface. So g is slightly less at the equator than the poles.

Rotating at the same angular velocity measured from 2here ? Relative to the sun ? See evans quot3d b3low

A satellite in geostationary orbit  above X is about 4 times farther from the center of the Earth, but also completes one orbit in 24 hours. So the satellite has a velocity that is about 4 times faster than X, or about 4,000 km/h.

Both the surface of the Earth, and a satellite in geostationary orbit have considerable angular velocity.

(Sorry for rounding everything off so savagely - I have a plane to catch...)

Again is this m3asured relative to the sun ? If at geosync orbit is at 13000 es from the centre of the earty and orbits in 90 minutes its 27000 km an hour. If at 52oookm geostationary thats 6800 km an hour. This may no5 account for angular accel3ration, but given that the radius is larger at slower speed, angular velocity will be even less.

These are all in reference tothe sun position though. Subtract the sun and your stationary in geo stationary, subtract the sun and you are relative to the galaxy etc.



Giv3n thar

[Colin2B]

Rotating at the same angular velocity measured from 2here ? Relative to the sun ? See evans quot3d b3low

Sorry, i should have made that clearer.
It is relative to rotating around earth’s axis, so movement relative to centre of earth.
Centre of earth is where gravity is acting from.
As has been said earlier, Sun doesn’t have a significant effect on small mass satellites in near earth orbit. The rotation of the satellite around the sun takes a year, so the rotation around earth centre once a day is the biggest influence. Remember, the objective is not to prevent the satellite falling into the sun, but to stop it falling to earth.
The posts by Janus are very comprehensive and if you read them along with Evan’s you have a lot more detail than the summary i gave.

[evan_au]

Rotating at the same angular velocity measured from where ? Relative to the sun?

You don't need to have a point of reference to know whether you are rotating, or how fast.

Even if you were in deep intergalactic space, and couldn't see any galaxies...
- or in a galactic dust cloud, so you couldn't see any stars

A rotating body is not in an inertial frame of reference, and there are hints that give this away:
- The Coriolis force, which causes low pressure weather patterns to rotate in one direction (and the opposite sense in the other hemisphere)
- The Focault Pendulum, which allows you to measure the rotation of the Earth, even if you are locked in an underground bunker, and can't see the Sun or stars.
- A gyroscope, such as used in inertial navigation systems (and satellite pointing systems)

See: https://en.wikipedia.org/wiki/Foucault_pendulum

Note: If the Earth were not rotating, "Geostationary Orbit" would be at "infinite" radius (and not very useful for communications).

[Petrochemicals (OP)]

But surely this is all relative to the suns position and the earths.the centre of gravity is orbiting the sun.

If deep in space as evan says, the rotation of the earth and orbit of the satelite around the earths centre would be nil. Only would it be judged  to be a twin orbit if the gravitational centre where orbiting around a third influence. The rotation of the atmosphere is itself measured by the sun/earth position. If Relativity is to be judged as correct then you have to include the entire gravitational pull of the universe to justify why the geostationary orbit will not decend ? That would give Relativity a few conflicts of interest?

I agree that the bary c3ntre must be what is calculated about, and dependant upon the distance from. Angular momentum by way of motion and  velocity are tye factors, but it must be calculated from the orbit of the earth about the sun.

If so geosyncronous orbits must be slower when contrary to the earths mo5i9n about the sun than when in concurrance with the orbit, due to the added velocity of ear5h about the sun.