What's the lowest (practical) altitude an object can be and still be considered in a stable orbit.

This is probably a stupid question as I don’t know a lot about these things so forgive my ignorance…

Inspired by this post: http://boards.straightdope.com/sdmb/showpost.php?p=20451604&postcount=26

I had not really thought about it until I read this post. I guess it makes sense. So two questions:

  1. How did they determine the best height/speed for the ISS?
  2. What’s the lowest practical height for a stable orbit (I’m guessing air resistance will enter the equation at some point)?
    Thanks

https://www.quora.com/What-is-the-lowest-possible-stable-Earth-orbit

But in a perfect vacuum, around a perfect non-rotating sphere of huge mass, could a perfect sphere of negligible mass maintain a stable orbit a couple inches above the surface?

Since ISS was assembled by modules lifted by the shuttle and serviced by Soyuz, both altitude and orbital inclination were a compromise based on the performance of those two vehicles. Launch vehicle payload performance varies based on latitude and launch inclination (angle from the equator). Also, the higher the orbital altitude the lower the payload. The final selected ISS orbital altitude and inclination considered the limitation of those launchers.

This depends on the definition of “stable” and whether the orbit is circular or elliptical. The final Mercury Atlas mission flown by Gordon Cooper used a 88 x 143 nautical mile orbit (163 x 265 km), and stayed up 34 hours.

The Saturn V S-IVB 3rd stage used a parking orbit around the earth of 115 nautical miles (185 km) for several hours before igniting the engine for the moon.

The KH-7 spy satellite reportedly could orbit with a perigee as low as 76 miles: (122 km): KH-7 GAMBIT

The 2nd American satellite, Vanguard 1, was launched in 1958 and is still in orbit even though communication was lost in 1964. The orbit is 2073 x 659 nautical miles (3840 x 659 km), and the expected orbital life is 240 years: Vanguard 1 - Wikipedia

Stability also depends on the size of the satellite. At any given height, a big beefy satellite will last longer than a little CubeSat, because air resistance is much more significant for the smaller satellite.

And that Quora answer is misleading: There are plenty of satellites in geosynchronous orbit, and while those would still eventually decay, it would take longer than anyone’s standard of “unstable”. They still need station-keeping thrusters, but that’s to keep them over one particular spot, not to keep them up at all.

In principle, yes. There’s an amusing '50s-era sci-fi short story premised on this: The Holes Around Mars, by Jerome Bixby.

Depends upon your time scale as well. Given enough time the Earth’s orbit around the Sun is not stable, and both it and the moon, plus any remaining satellites, will be ejected into the freezing wastes.

This question has been answered but raises another in my mind. What about a rotating sphere? I’ve heard of something called “frame dragging”, but I have no idea what it means.

I consider myself to be in an extremely low geosynchronous orbit, but I doubt that’s technically true. It livens up a dull morning just thinking about it though.

Thanks all. For the purposes of this OP I was defining stable as a practical time limit, as in serving their purpose for quite a few years (like most satellites) but I realize the orbits will eventually decay.

So based on the above answer, is the orbiting height of any particular satellite based on its size and there is no standard geosynchronous orbit height?

Not Bixby’s best work, that. (Trivia moment–his most famous work is the ultimate source of a phrase used here on the dope.)

I not too long ago read a SF novel with another especially low orbiting satellite (can’t remember which one, possibly a Neil Asher book?) One scene in the book takes place at an atmosphere-free, geologically dead planet where some past ancient civilization had carved a miles deep and wide trench around the circumference of the planet. A large city/satellite had been built (by a modern civilization) that orbited in that trench below the surface of the planet.

No. For a mass of a satellite much less than the planet, the satellite mass cancels out for the period of an orbit, leaving only the radius and the mass of the planet.

The bigger satellite has more mass, thus more energy, and the square/cube law suggests the drag will not have increased as much, so overall the larger satellite will not decay as quickly, as the drag will take longer to take effect.

I’m not convinced that this is really a metric of orbital stability, rather one of orbital decay. A stable orbit can decay, and an object in an orbit where there is no decay due to drag or other energy loss can be unstable.

Not Bixby’s best work, that. (Trivia moment–his most famous work is the ultimate source of a phrase used here on the dope.)

I not too long ago read a SF novel with another especially low orbiting satellite (can’t remember which one, possibly a Neil Asher book?) One scene in the book takes place at an atmosphere-free, geologically dead planet where some past ancient civilization had carved a miles deep and wide trench around the circumference of the planet. A large city/satellite had been built (by a modern civilization) that orbited in that trench below the surface of the planet.

Crap, stupid double post. The first one (I thought) didn’t go through when the forum went down during posting–got an error message and everything.

I suggest looking at the wikipedia entries for geosynchronous orbit and lists of orbits. Or play Kerbal Space Program, where you can get intimately familiar with all kinds of orbits.

If the orbit is elliptical, then “height of the orbit” varies, depending on where the satellite is in the orbit, and whether it is moving towards or away from the apogee or perigee. As an extreme example, the Infrared Space Observatory was in a highly elliptical orbit with a perigee of 1,000 km (621 mi) and an apogee of 70,600 km (43,868 mi).

If the orbit is circular, geosynchronous, and around Earth, then the standard height of the orbit is 42,164 km (26,199 mi).

You can orbit at basically any height you want, but the lower you are, the faster you have to move in order to stay in the same altitude. Objects in low-earth orbit (like the ISS) typically move about 20,000 km/hr (12,400 mph), or about half the escape velocity, while objects in a geosynchronous orbit only need to move half that speed (10,800 km/hr or 6,900 mph). The moon, by comparison, orbits Earth at a leisurely 3,683 km/hr (2,288 mph).

The higher up you want to orbit, the more expensive it is to get up there. It takes a lot of fuel to claw your way out of a gravity well. Low-earth orbit is the easiest, and if you have a lot of mass (like the ISS), it’s easier to just stick to that altitude.

Ben Bova wrote a story about extremely-low orbiting debris from a conflict on the moon that still returned to weak havoc on lunar bases.
This site has some commentary on the notion: Shoot the Moon | ScienceBlogs

Most satellites are, broadly speaking, at one of two different orbital heights. If your satellite’s orbit doesn’t matter much, and it just has to be Up There somewhere, then you go for Low Earth Orbit: Basically, just high up that it’ll last long enough to do whatever it is you want it to do (which will vary some depending on the satellite and what you want it to do, but they’re all pretty close). You don’t go any higher because it’s more expensive, and there’s no point.

Alternately, some satellites have special requirements for their orbits. The most common special requirement is that you want your satellite to stay more or less above the same point on the surface: Maybe you want to spy on that site, or maybe you want to transmit TV signals to there, or whatever. For that, you want geosynchronous height, which is much higher than LEO. There is only one geosynchronous height, with no variation by type of mission or spacecraft.

It’s also possible to have a mission that calls for an orbit somewhere in between LEO and GEO, or for one out beyond GEO, but those are fairly rare.

You’re standing on the equator? … at any other location you’d be rotating around the axis of Earth’s rotation, not the center of gravity … work the vectors over morning coffee, your kids will be amazed …

Nitpick: you’re confusing geosynchronous with geostationary. All geostationary orbits are geosynchronous, but not all geosynchronous orbits are geostationary. A geosynchronous orbit is just one whose orbital period is equal to earth’s rotational period. This can be achieved by orbiting in a (nearly) perfect circle at 42k km, or by orbiting in a variety of eccentric orbits where the perigee is lower than 42k and the apogee is higher. Now, all these eccentric orbits will have the same semi-major axis, so I suppose you might argue that all geosynchronous orbits have the same height if one equates height with semi-major axis, but I don’t believe one would ordinarily say that a tundra orbit is the same height as a geostationary orbit when a satellite in a tundra orbit spends the majority of its time at significantly higher altitudes.

But obviously standing on the equator is not remotely similar to being in orbit anyway. As discussed, there’s a unique geostationary distance of 42km, or 36km above the surface.

Theoretical orbital speed at the equatorial surface is 8km/s, or 17 times faster than the speed of the earth’s rotation.
https://en.wikipedia.org/wiki/Orbital_speed#Tangential_velocities_at_altitude