Most of us know that the "geosynchronous orbit" is special. A satellite placed there will take 24 hours to go around the Earth, and since some point on the Earth's Equator also takes 24 hours to go around its axis, the satellite stays in the same position above the Equator. That orbit is positively
crowded with satellites these days, because it is so usefully special.

Well, I happened to think of another orbit that should be considered special. This orbit is defined as the place where the speed of the orbiting body is the same speed as the rotation of the Earth's equator.
No, it is NOT the geosynchonous orbit. Remember that a geosync satellite is going in a rather large circle around the Earth, while that point on the Equator is going in a much smaller circle around the Earth's axis, both doing this in 24 hours. So the actual velocity numbers are: The satellite is moving at about 3.075 km/sec while the Equatorial point is moving at about 0.465 km/sec.

So, where IS the orbit in which a satellite goes around the Earth at 0.465 km/sec? After researching the matter, I have computed it to be 1,842,596 km from the center of the Earth, or 1,836,218 km from the Earth's surface at the Equator. This is pretty far away, since the Moon is only 384,403 km from the Earth.

OK, now that I've defined it, why do I call it special? The answer has to do with the recent X Prize. That little rocket shot straight up into the air and then fell back. Just suppose we had a more extreme version of SpaceShip One:

This rocket launches from the Equator and goes straight up, EXACTLY straight up! After its initial acceleration has ended, it coasts. Gravity pulls on it and slows it down. We know that if the rocket went fast enough, more than 11 km/sec, it would "escape" the Earth completely, and head off to Mars or some other locale. But this rocket is not going quite that fast. As a consequence, it is SUPPOSED to go slower and slower away from the Earth, until its speed reaches zero, after which the continuing tug of gravity makes it begin actually falling back to the Earth.

However...remember that special orbit? If we plan things just right, then AT the point where gravity has slowed the spaceship to zero velocity away from the Earth, then it will be 1,836,218 km above the Earth, AND IT WILL BE IN ORBIT! Because of the Law of Conservation of Momentum, the spaceship STILL has its original side-ways velocity of 0.465 km/sec that it possessed simply by sitting on the Earth's equator when it was launched!!!
And since that sideways velocity IS the orbital velocity at the altitude where its distance from the Earth stopped increasing, the spaceship must now be in orbit. Q.E.D.

Because it is so far away, it probably won't be as useful a place as I originally thought it might be, when I first got this Idea. So that makes it half-baked, and thus it is presented here for your entertainment and minor enlightenment.

P.S. Synchronous orbits are also called Clarke orbits, after Arthur C. Clarke (of "2001: A space Odessey" book/movie fame), because he was the first guy to write about it. What do you suppose the chances are that I might also be remembered that way, someday? (Maybe somebody else thought of this first...?)

And now, before somebody beats me to it, I will proceed to partially trash this Idea. The problem is the Sun; the computed orbit is fine only if the Sun wasn't there!

See, there is a point between the Earth and the Sun where gravity cancels. This distance is about 1.5 million km from the Earth, which obviously is LESS than the 1.8 mill km specified for Special Orbit #2. So, as soon as the orbit carries the satellite around between Earth and Sun, it breaks free from the Earth and just orbits the Sun. Alas.

However, please remember that while I've described it specifically using Earthly data, the IDEA is not limited to the Earth. Mars, for example, rotates in about 24.5 hours. But Mars has less mass than the Earth, so relative to the Earth, any specified distance-from-world means that the satellite goes slower around Mars than Earth. So Special Orbit #2 would be closer to Mars than to the Earth.

Caveat: Mars is also smaller-diameter than the earth, so the rotational velocity at its surface is also somewhat less than Earth's. This cancels some of the previous effect.

Anyway, Mars is also 50% farther from the Sun than Earth, so the diminished effect of the Sun's gravity COULD allow Special Orbit #2 around Mars to be stable. I don't have all the figures needed to compute that, but at the moment it seems to me a likely conclusion.

Nothing. Well, it orbits the sun, not that that answers the question. I think our [Vern] simply realised the existence of the orbit, and posted it. As he says, it's not feasible round Earth, and would be difficult to attain in any case. Interesting, though.

<digression>
If two satellites were tethered so that one was, say 10kms higher than the geostationary orbit, and the other was 10kms lower than the geostationary orbit, would they remain stable?
</digression>

[contracts], The original idea had to do with the XPrize rocket. As you know, the altitude that a rocket can reach depends on its initial boost. BUT MOSTLY this involves accelerating to something like 8km/sec, because that is the orbital speed close to the Earth. Well, if you can get into an orbit where the orbital speed is only .465 km/sec, then maybe you don't need such a fast rocket! All you need is enough boost to lift straight up, to Special Orbit #2.

Unfortunately, I don't know the calculus needed to compute what the initial speed should be, for a rocket going straight up to end in Special Orbit #2. At such distances from the Earth, the lessening of the pull of gravity significantly affects the rate at which the rocket slows, so the "normal" max-height equation you use for ballistics on Earth does not apply. Sure, it's good enough for SpaceShipOne, which had a max speed of a little less than Mach 3 (roughly 0.33 km/sec), and coasted up past the 100-km mark.

But 6000km above the surface of the Earth, gravity is acting at 1/4 the power that it acts AT the surface of the Earth. If SpaceShipOne had done an identical boost starting at that altitude, it probably would have flown 4 times as high, relative to its starting point, before falling back.

So, what I was thinking was that MAYBE you could reach Special Orbit #2 with a rocket that didn't have to go as fast as 8 km/sec, like they must do to reach ordinary Low Earth Orbit. Of course, now that I know how far away it is, that's less likely, but I'd still like to know what the initial boost-velocity should be! (Might actually be practical for Mars....)

[ling], the two satellites would experience tidal forces, and with the tether also an influence, they would probably both end up 0 km away from geosync.

So your idea is that it might be more economical (energy-wise) to reach a distant orbit than a low orbit?

This begs the question, would it not then be incrementally more effective the further one is from Earth - - making this not "Special Orbit #2" but a [velocity / 'altitude'] trade extrapolated to its maximum expected efficiency?

Yup, but only the math can tell us for sure, regarding the particular Special Orbit #2. I do understand that with a little extra boost, you would reach an even higher altitude, and your Equator-provided sideways velocity would be more-than-enough for an orbit at that height. But you DO first have to apply that little extra boost. Reaching only Spcial Orbit #2 is the most efficient, energy-wise, of that category.

[Vernon], maybe another way to look at the problem is to think "Do I need to add or subtract energy to get my satellite in GeoStationary orbit to go to the special orbit #2?"
My intuition tells me that the outer satellite could fall into the lower orbit, but not the other way around.

I haven't done the math for this yet, but conceptually it sounds flawed. Remember, it takes time to get to altitude. In that time, you're travelling "horizontally" as well as vertically. When you get to "special orbit #2", your original horizontal velocity vector will no longer be parallel with the surface of the Earth. Thus, your velocity in a direction parallel to the surface will be less than the original equatorial velocity. I have a feeling that whatever this actual orbit is, it will still intersect the surface of the Earth.

I have a strong feeling that the additional fuel required to reach this altitude will exceed whatever it would have taken to put the craft into a proper orbit anyway.

I can't properly say "bad science", since I haven't actually done the calculation, but it just doesn't sound right to me.

[Ling] (replying to first of two annos), that is a good point, because one of the ways they SAVE energy getting into LEO is to launch Eastwards, to take advantage of the Earth's rotation. When in any orbit, you DO have to accelerate to reach a higher orbit (where you end up actually travelling at a slower velocity than before, heh heh).

With regard to your second anno, are you sure your PE values take into account the lessening of gravity with distance?

[Freefall], I think the more appropriate word is "tangentially" than "horizontally". And that will be true at any height, provided all your launch-velocity is aimed directly upwards away from the center of the Earth.

Now when you are on a moving (level) train, and you decide to juggle, the balls share your forwards velocity, so they come back down to where you are, to catch them. But on a very high up-and-then-down rocket flight, you are reaching altitudes where you would have to be moving faster at that altitude (larger-radius circle) than the Earth's surface is moving, to keep up with it. So you will end up falling to the ground "behind" your launching site.

But the goal here has nothing to do with ending up in some position relative to a particular point on the surface of the Earth, as IS the goal of satellites in Clarke orbits. The goal here is just to let your Equatorial tangential velocity become an orbiting velocity. And THAT will work perfectly, even if it takes more energy than I orginally supposed, to reach the proper altitude. But then , I DID say in the original text that the Idea was half-baked, didn't I? Now you know precisely why!

It seems like I remember hearing that it takes a larger rocket to get to geosyncronous orbit than it takes to get to low earth orbit. Therefore I would expect that it would be even harder to get to special orbit #2.

Regarding a satellite in geosyncronous orbit tethered to one in special orbit #2: one goes around the earth every day, the other takes about 288 days. Both will be pulled out of orbit immediately.

This, is what I used to think orbit was, back when I was 8. I thought that the spaceshuttle just blasted off and sat there not moving, in orbit. I don't get how a space elevator would not just drag a suposed connected station down.

I get the idea of orbit, you just fire a supposed object at a velocity and angle around the planet, and it will continue falling around, until it falpls to the ground. I like to think of the universe as like a giant video game run by gods. Silly yes, but helps immensely in realizing and theorizing about the universe itself.

I imagine the universe has 3 dimensions, and from my 3D modeling and modding expertise, I know there are X, Y, and Z axis's in a 3 dimensional world. Now, in a human computer, the world has a set limit on how small and accurate objects can be in the infinite 3d space. Because, if it did not, the computer would end up in a loop, endlessly calculating an infinite number. But, in real life, I also believe that the universe is infinite. Suppose it is, then what? Then, obviously the ammount of accuracy in an supposed orbit is, indeed, infinite. You can just get smaller and smaller.

You can use a computer to calculate the orbit, but eventually, there comes a point in calculation, where your finitely capable computers cannot calculate the infinite smallness of the orbit. At this point, it means that, if you make the best, most perfectly calculated orbit you can make, it will eventually, over time, fall to earth.

In fact, as soon as the obit has begun, gravitational forces start pulling and degrading the orbit, but, the degradation is so small, you cannot detect it, or you can, but it is so small, it does not matter. It starts out at a perfect angle. Now, a small number of degration happens, it multiplies over time, soon, it becomes a great number, then you have a problem.

This can always be corrected as soon as it is managable. A small thruster on an orbiting object cannot always fix the problem, because it is so small, that the tiniest of an adjustment could fix it, but over do it. It can, later on, because it is big enough, that the computer can make an adjustment, and not overshoot.

Just my theory and speculations. :D

I suppose that, way out in space, way, WAY, out, where there is no matter filling the area, and no forcesa acting on anything, that you could possibly fix an object in space, and have it stway there forever, or untilk acted upon by another object.One of newqtons laws states that an inanimate object will stay in place unless acted upon by anothewr object, or force. Newtons laws, know all :D .

[EP] Gravitational forces won't pull the satellite back to earth. What will is interaction with gases; this reduces the speed of the satellite and hence the diameter of the orbit decreases.

A stable orbit isn't about perfect calculations. If you underestimate the speed necessary, the diameter of the orbit will be smaller than you intend. If you overestimate, the orbit will be larger.

[vern] //When in any orbit, you DO have to accelerate to reach a higher orbit (where you end up actually travelling at a slower velocity than before, heh heh//

Not slower velocity, merely fewer revolutions per day. Your new higher-altitude orbit will have a larger circumference, which will take longer to traverse, but as you yourself said, you had to accelerate to get there.

It would be a very simple matter to balance out the sun's gravity with manuevering rockets. Same thing for other large masses nearby, ie, the moon. Very simple to balance it out.

[gardnertoo], sorry but it really is true that object in higher orbits have lower orbital velocities than objects in lower orbits. Yes, you do have to accelerate to get to a higher orbit, but the initial velocity you gain during that acceleration is lost in the climb to the higher orbit, along with some extra!

[Desert Fox], the "simple" you suggest becomes non-simple as soon as your rockets run out of fuel....

Well, [Vernon], I did a quick search to prove I was right. I found out that no, I was NOT right. The outer planets travel slower through space than the inner planets. (Guess it's a good thing I'm not an astronaut.) In my gut, it seems that the amount of energy it would take to get all the way out to your special orbit would be more than enough to put yourself into a much lower orbit with higher speed, but maybe I'm wrong about that too. Keep the croissant, in any case.