# geosynchronous orbit

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First, imagine you have a perfectly rigid tower that extends from Earth to an altitude beyond geosynchronous orbital altitude. An elevator shaft in the center of the tower allows an elevator to travel up through the tower. As the elevator ascends, the force required to maintain constant speed decreases until it reaches geosynchronous altitude, where it remains stationary provided it is stopped and no further vertical force is added to it in either direction, right? Then, if it is moved to a higher altitude and stopped again, the force of the tower itself should work to push it upward (a sort of catapult effect), and this upward motion will be acceleration, right, since the angular speed of the tower is increasing as a function of its radius?

Now, imagine you are observing from a satellite in geosynchronous orbit near the tower but not connected to it. Above you and below you are two other satellites that are orbiting at speeds appropriate to maintain constant altitude. The one below you would appear to be moving faster than you relative to the ground/tower, but the one above you would appear to not be moving slower but rather backwards relative to the ground/tower, right? Does this mean that altitude-maintenance is question of acceleration with descent below geosynchronous altitude but a question of deceleration with ascent above it?

I'm not sure what is confusing me about this, except maybe I can't figure out at what point you would shift from using the ground as a reference frame to using yourself or some other point as a reference frame with the Earth simply rotating below. It seems like a point in geosynchronous orbit is a natural altitude above the ground, e.g. if a tall mountain or tower would extend that high, you would be able to go up it without "detaching" from the reference frame of Earth's rotation. But then by remaining in the same reference frame and ascending even higher, work would actually have to be done to prevent objects from ascending higher.

I guess what this all boils down to is differentiating between net-force of ascent descent and the force of gravity itself, since any satellite experiences "weightlessness" but the phenomenon of weight itself appears to be relative to something else besides gravity directly. E.g. the elevator in the tower loses weight as it ascends toward geosynchronous orbit and gains "negative weight" as it ascends above that. So is weight a phenomenon dependent on both gravity AND speed, and if so speed relative to what?

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Now, imagine you are observing from a satellite in geosynchronous orbit near the tower but not connected to it. Above you and below you are two other satellites that are orbiting at speeds appropriate to maintain constant altitude. The one below you would appear to be moving faster than you relative to the ground/tower, but the one above you would appear to not be moving slower but rather backwards relative to the ground/tower, right? Does this mean that altitude-maintenance is question of acceleration with descent below geosynchronous altitude but a question of deceleration with ascent above it?

Orbital mechanics are weird. But yes, if you're in orbit and you apply your thrusters to try to speed up, you end up in a higher orbit, going more slowly than before. If you decelerate, you drop into a lower orbit and accelerate.

On the other hand, if you accelerate and go into a higher orbit, but continue applying thrusters, you will just move into higher and higher orbits until you just launch yourself out into space. That's what's happening with the elevator that rises above a geosynchronous orbit.

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I think there is confusion here between the velocity of an orbiting object and the orbital period. An object in a higher orbit than another orbiting object (farther from the earth) will have a larger orbital period (take longer to complete an orbit) and will also have a higher velocity. SM

Edited by SMF
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No, it'll have a lower velocity; the higher you are, the slower you need to go to stay in orbit.

$F = \frac{mv^2}{r} = G \frac{m \, m_e}{r^2}$

$v^2 = G \frac{m_e}{r}$

As you can see, velocity decreases with increasing orbital radius.

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Orbital mechanics are weird. But yes, if you're in orbit and you apply your thrusters to try to speed up, you end up in a higher orbit, going more slowly than before. If you decelerate, you drop into a lower orbit and accelerate.

On the other hand, if you accelerate and go into a higher orbit, but continue applying thrusters, you will just move into higher and higher orbits until you just launch yourself out into space. That's what's happening with the elevator that rises above a geosynchronous orbit.

If you are geosynchronous orbit, your altitude makes no difference!

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lemur's scenario is more accurately a geostationary orbit, and there is only one altitude at which you can have a geostationary orbit.

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Cap'n, it would help me to understand if you would label the variables in the formulas. SM

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All right, let me explain a bit. In circular motion, the force required to keep the object moving in the circle is:

$F = \frac{m v^2}{r}$

where m is its mass, v its velocity, and r the radius of the circle.

The gravitational force is, of course,

$F = G \frac{m \, m_e}{r^2}$

where G is the gravitational constant, m is the mass of the object, me the mass of the Earth, and r the distance.

Setting these equal (since gravity is the force keeping the object in circular motion), we see that:

$v^2 = G \frac{m_e}{r}$

so the velocity of the orbit decreases as the radius increases. Of course, the math gets more complicated when you consider non-circular orbits.

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Cap'n, thanks. This is really fun. I had been thinking about it starting with my super cannon on top of mount Everest (and no air resistance). A low power shot makes too small an orbit so the projectile hits the earth. With just the right increased velocity the projectile hits me in the back. So the velocity had to increase. An even higher velocity makes an elliptical orbit, but my error was assuming the mean velocity over the orbit was greater, forgetting that gravity decreases with distance, and not taking into account the energy required just to increase the altitude of the projectile. SM

Edited by SMF
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First, imagine you have a perfectly rigid tower that extends from Earth to an altitude beyond geosynchronous orbital altitude. An elevator shaft in the center of the tower allows an elevator to travel up through the tower. As the elevator ascends, the force required to maintain constant speed decreases until it reaches geosynchronous altitude, where it remains stationary provided it is stopped and no further vertical force is added to it in either direction, right? Then, if it is moved to a higher altitude and stopped again, the force of the tower itself should work to push it upward (a sort of catapult effect), and this upward motion will be acceleration, right, since the angular speed of the tower is increasing as a function of its radius?

Now, imagine you are observing from a satellite in geosynchronous orbit near the tower but not connected to it. Above you and below you are two other satellites that are orbiting at speeds appropriate to maintain constant altitude. The one below you would appear to be moving faster than you relative to the ground/tower, but the one above you would appear to not be moving slower but rather backwards relative to the ground/tower, right? Does this mean that altitude-maintenance is question of acceleration with descent below geosynchronous altitude but a question of deceleration with ascent above it?

I'm not sure what is confusing me about this, except maybe I can't figure out at what point you would shift from using the ground as a reference frame to using yourself or some other point as a reference frame with the Earth simply rotating below. It seems like a point in geosynchronous orbit is a natural altitude above the ground, e.g. if a tall mountain or tower would extend that high, you would be able to go up it without "detaching" from the reference frame of Earth's rotation. But then by remaining in the same reference frame and ascending even higher, work would actually have to be done to prevent objects from ascending higher.

I guess what this all boils down to is differentiating between net-force of ascent descent and the force of gravity itself, since any satellite experiences "weightlessness" but the phenomenon of weight itself appears to be relative to something else besides gravity directly. E.g. the elevator in the tower loses weight as it ascends toward geosynchronous orbit and gains "negative weight" as it ascends above that. So is weight a phenomenon dependent on both gravity AND speed, and if so speed relative to what?

When analyzing satellites, you should never be using the Ground as a reference frame but should be using the ECI frame or Earth center inertial frame, which does not rotate with the Earth. . It is also the best frame for considering what happens with your tower. You simply consider the effect of gravity on the tower at a given height vs. the effect of traveling around the Earth center in a circle.

The Ground frame is not an inertial one, so analyzing things from this frame adds all kinds of unnecessary complications.

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If you are geosynchronous orbit, your altitude makes no difference!

A geosynchronous/geostationary orbit has only one altitude for its solution.

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The Ground frame is not an inertial one, so analyzing things from this frame adds all kinds of unnecessary complications.

Maybe, but it is the frame in which I experience gravity as weight and acceleration of falling objects. From this thread, it seems that I should have used "geostationary" instead of "geosynchronous" (not sure what "geosynchronous means now though); but my point was that if the ground continued up to that altitude (e.g. if a mountain went that high), you would become completely weightless at that point. Yet that doesn't mean that you have left Earth's gravity well. Likewise, if you were some distance from the top of the mountain, you could presumably take of with a hang-glider (assuming sufficient air density, which must by physically impossible with such low gravity) and accelerate to an non-decaying orbital speed. So the issue that really puzzles me is the relationship between gravity and falling/weight, because it doesn't seem to be a direct one. I.e. if you jumped off the mountain without the hang-glider you would fall, but if you use the hang-glider to accelerate, you would orbit weightlessly. If you would ascend to an altitude above the top of the mountain (above geostationary altitude), you would still orbit and not escape gravity, but you would have to fly away from the mountain in the opposite direction as Earth's rotation. But if you would fly the same speed away from the same mountain in the same direction as Earth's rotation, you would be leaving the gravity well. Granted this is frame-confounding in a sense, but in the context of standing on a mountain with atmosphere rotating at the same rate as the planet and therefore stationary relative to the mountain (i.e. no wind), it seems like your weight would change depending on your direction and speed. Then the bigger issue is what causes weight if not gravity only? Is it gravity combined with relative motion in some way that causes objects to fall? I.e. How does the ground and atmosphere slow down below orbital speed in the first place enough to cause things to fall downward into the gravity well?

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So the issue that really puzzles me is the relationship between gravity and falling/weight, because it doesn't seem to be a direct one.

There are a couple of issues with this thread:

1. You are mixing up a lot of different concepts, lemur. The end result is that you are confused.

2. The answer to the quoted question depends on what you mean by "weight" and what you mean by "falling".

Regarding orbits:

One challenge is getting past the seemingly paradoxical fact that to transfer a vehicle from a circular orbit to a higher circular orbit, the vehicle's velocity needs to be increased twice and yet at the end of the process the vehicle's velocity is smaller than it was originally. The resolution of this apparent paradox is that you are extending your Earth-bound common sense to regimes where it doesn't apply. Orbital mechanics is a bit counterintuitive at times. Some suggested reading material: The wikipedia article on Hohmann transfers, http://en.wikipedia.org/wiki/Hohmann_transfer_orbit (google "Hohmann transfer" for more), and the wikipedia article on the vis viva equation, http://en.wikipedia.org/wiki/Vis-viva_equation.

Regarding weight:

This term has multiple meanings. One meaning is the magnitude of the gravitational force vector exerted on some body. This is the definition of weight that you will see in most introductory physics texts and in almost all aeronautics engineering texts. In this sense, the "weight" of an astronaut on the International Space Station is about 90% of the astronaut's weight on the ground. Another definition is the value read by an ideal spring scale, so some call this "scale weight". In terms of Newtonian mechanics, this is the magnitude of the net non-gravitational force acting on some body. In terms of general relativity, this is the magnitude of sum of all real (non-fictitous) forces acting on some body. Some introductory physics texts and most general relativity texts use this definition of "weight". In this sense, the "weight" of an astronaut on the International Space Station is zero.

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FYI, the little green "+" sign on the lower right of someone's post is another way of expressing this.

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Swansont:

I did it. Thanks. I gave you one also.

SM

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Regarding orbits:

One challenge is getting past the seemingly paradoxical fact that to transfer a vehicle from a circular orbit to a higher circular orbit, the vehicle's velocity needs to be increased twice and yet at the end of the process the vehicle's velocity is smaller than it was originally. The resolution of this apparent paradox is that you are extending your Earth-bound common sense to regimes where it doesn't apply. Orbital mechanics is a bit counterintuitive at times. Some suggested reading material: The wikipedia article on Hohmann transfers, http://en.wikipedia...._transfer_orbit (google "Hohmann transfer" for more), and the wikipedia article on the vis viva equation, http://en.wikipedia....s-viva_equation.

The Hohmann transfer was most helpful (I don't get the vis viva equation). It seems that acceleration within an orbit initially causes the orbit's shape to change from circular to elliptic, so for example if you would accelerate from the top of the mounting at geostationary altitude, you would eventually gain altitude as a result of your own inertia, because your orbital shape became more elliptical. Then, just by your own inertia, you would later descend to the altitude of the mountain again. I assume you would experience this as the mountain receding downward from you, since you wouldn't feel yourself shift into assent if you were moving only on your own inertia. I suppose the only way you could remain at the level of the mountain top would be to thrust toward the mountain in some way. I wonder if your sense of vertical/horizontal orientation relative to the ground would become confounded.

Regarding weight:

This term has multiple meanings. One meaning is the magnitude of the gravitational force vector exerted on some body. This is the definition of weight that you will see in most introductory physics texts and in almost all aeronautics engineering texts. In this sense, the "weight" of an astronaut on the International Space Station is about 90% of the astronaut's weight on the ground. Another definition is the value read by an ideal spring scale, so some call this "scale weight". In terms of Newtonian mechanics, this is the magnitude of the net non-gravitational force acting on some body. In terms of general relativity, this is the magnitude of sum of all real (non-fictitous) forces acting on some body. Some introductory physics texts and most general relativity texts use this definition of "weight". In this sense, the "weight" of an astronaut on the International Space Station is zero.

The zero-weight (i.e. "weightlessness") is how I think of weight. I don't understand exactly what the 90% weight refers to. The problem I have is distinguishing between the force of gravity and the rate of acceleration toward the floor (when the floor isn't on the ground). Things get very strange when I think about being in an elliptical orbit where "falling" is increasing your altitude part of the time. Likewise, if you think about a very fast sustainable-altitude orbit that could occur if air-friction wasn't an issue (e.g. 10m above the surface of the moon). In such an orbit, gravity gets expressed completely as speed parallel to the ground. Yet, as you slow down below that orbital speed, gravity gets increasingly expressed as weight (i.e. rate of falling toward the ground). So depending on the speed of the ground, you could fall faster or slower and weigh more or less. I guess I have this figured out, but it still confuses me to think about gravity as a separate concept from rate of falling toward the ground.

e.g. would you say we are actually as heavy as we would be if the Earth wasn't rotating, and that Earth's rotation lightens us a certain percentage - or would you say that our weight is accurate in the frame of the planet as it rotates and that it would change if the rotation-speed changed? I guess it's just a question of choosing a frame, isn't it?

Edited by lemur

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