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Lowest Stable Low Earth Orbit?

The idea

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What is the theoretically smallest possible distance between the surface of the Earth and a satellite in stable orbit around the Earth?

Does the answer depend on knowing how high a satellite must be before the atmosphere is sufficiently attenuated? Presumably the atmosphere doesn't actually stop at a particular, exact height?
 
The idea said:
What is the theoretically smallest possible distance between the surface of the Earth and a satellite in stable orbit around the Earth?
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Zero. My coffee cup is in ultra-low earth orbit as I write this.

If you're not talking about geostationary orbit, then I was under the impression that any orbit radius can be stable as long as the orbit speed is correct, so any arbitrarily low orbit is fine as long as your satellite is travelling fast enough. Of course, I could be talking complete crap here.
 
Reading this article it seems that even geostationary satellites carry fuel to keep them on station and they're about 22,000 miles up.

I suppose if you had sufficient fuel and did away with that pesky atmosphere, you could orbit one just high enough to clear Mt. Chimborazo which is the farthest point from the center of the earth. Everest is only the highest spot from mean sea level. ;)
 
darn. iconoclast beat me to it. i wanted to point out i have been maintaining a low orbit since registering at this forum.:D
 
I would have to correct you here, Iconoclast and Bug girl. An object resting on the surface (any surface ;)) is NOT in orbit. If you throw the coffe-cup or make a jump, on the other hand, you will be in free fall, and therefore in an orbit. It will be an orbit you cannot follow for vary long, though, since it will intersect the surface of Earth after a short while, putting you back in the "resting on surface" mode.

Hans
 
MRC_Hans said:
I would have to correct you here, Iconoclast and Bug girl. An object resting on the surface (any surface ;)) is NOT in orbit.
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Well, if we're going to get picky. According to that wheelchair guy, it's not possible for two objects to touch each other, therefore my coffee cup is not actually resting on the desk, it's hovering just above it.
 
The idea said:
What is the theoretically smallest possible distance between the surface of the Earth and a satellite in stable orbit around the Earth?

Does the answer depend on knowing how high a satellite must be before the atmosphere is sufficiently attenuated? Presumably the atmosphere doesn't actually stop at a particular, exact height?
Click to expand...


As others have pointed out, it could be at any height if the velocity was high enough..

I read a short story once about some space travelers who landed on a planet, and began to observe that there were these holes passing through everything in a straight line, just a few feet off the ground. As they finally began to put 2 and 2 together, they realized their vehicle was parked in the path of a billiard ball sized moon...
 
Re: Re: Lowest Stable Low Earth Orbit?

Diogenes said:
As others have pointed out, it could be at any height if the velocity was high enough..
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I think you'd have to be sufficiently high up in the stratosphere.

Down here in the Troposphere, the air molecules get in the way. Due to the uneven heating of the atmosphere (even across a relatively small area), the air density will undoubtedly effect the projectile.

(Watch out for those billiard ball sized moons, they sting...)
 
Yahweh is right, you have to be outside the atmosphere. Anything orbiting inside will eventually have its kinetic energy wasted away and come crashing down. And no, things resting on the surface of the Earth are not "in orbit". Static friction moves them along.

In any case, there is no truly stable orbit anywhere, because gravitational radiation constantly bleeds from the system, albeit at a very slow rate.
 
Re: Re: Lowest Stable Low Earth Orbit?

Diogenes said:



As others have pointed out, it could be at any height if the velocity was high enough..

I read a short story once about some space travelers who landed on a planet, and began to observe that there were these holes passing through everything in a straight line, just a few feet off the ground. As they finally began to put 2 and 2 together, they realized their vehicle was parked in the path of a billiard ball sized moon...
Click to expand...
I love that story! Asimov (the editor, but not the author) claimed that the velocity was wrong for that radius orbit, however, so that situation would be physically impossible. I've been meaning to check that calculation for a while now but never got around to it.
 
Re: Re: Re: Lowest Stable Low Earth Orbit?

QuarkChild said:
I've been meaning to check that calculation for a while now but never got around to it.
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OK, I finally just broke down and did it...

Someone can check this, but using v^2/R = M G /R^2 for the equation of a circular orbit around a planet of mass M, I found that using Mars' radius and mass the orbit described would require a velocity of 1140 m/s. (I can't remember whether Mars had a breathable atmosphere in the story...if so, that would break the sound barrier many times over.)

That's certainly fast enough to leave neat holes in the hillsides, but it might be too fast to be reasonable. The satellite would then orbit Mars every 10 Earth-hours. Earth's moon takes about a month to orbit Earth.
 
Something's got to be wrong there, QC. Consider that a satellite in low-earth orbit circles the Earth in about 90 minutes. Consider also that Phobos (Mars' innermost moon), at a distance of about 6000 km from the surface of Mars, orbits Mars about 3 times each Earth day.
 
Skeptoid said:
Something's got to be wrong there, QC. Consider that a satellite in low-earth orbit circles the Earth in about 90 minutes. Consider also that Phobos (Mars' innermost moon), at a distance of about 6000 km from the surface of Mars, orbits Mars about 3 times each Earth day.
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I took the circumference of Mars and divided by the velocity I found above to get the time of orbit in seconds (3.6 million). Then I converted to hours. Is there a different way to find the time of orbit?
 
I just redid the velocity calculation and 3551m/s instead, yielding a time of 3.3 hours for the orbit, which according to skeptiod is still oddly high.

I give up.
 
3.3 hours? Why does that sound wrong? Phobos takes several hours and is 6000km higher (is it three times a Mars day or Earth day?), that seems to make sense.

There was another story SF short story on this, I fail to remember the author, but it does have an Asimov scent ;):

This was during the cold war, and the author has it in a future where the USA and USSR are at the umptheenth fringe war on Earth, but the more or less independent moon colonies of the two countries keep an amiable peace. So a comitee goes to investigate, and they haven't been in the office of the local commander for very long, when a siren sounds, and everyone hits the floor. Then there is a wizzling and slammming of bullets through the walls, people get up, run around and patch holes, and sit down to continue business.

It turns out that they HAD tried to fight, but on the airless moon, the high-velocity riffle bullets tended to enter low orbits, so they soon discovered that if you fired a gun and missed, you would eventually hit your own backside! ..And so, peace ensued :).

Hans

Edited for grammar and some of the typos...
 
QuarkChild said:
I just redid the velocity calculation and 3551m/s instead, yielding a time of 3.3 hours for the orbit, which according to skeptiod is still oddly high.

I give up.
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Actually, that seems much more reasonable. Consider that Mars' escape velocity is 5020m/s.

Edit-- I thought your previous result to be oddly slow. :)



Hans, a Martian day is 24.66 Earth hours.
 
QuarkChild said:
I just redid the velocity calculation and 3551m/s instead, yielding a time of 3.3 hours for the orbit, which according to skeptiod is still oddly high.

I give up.
Click to expand...
Okay, using the equation for orbital velocity, V<sub>o</sub> = (GM/R)<sup>1/2</sup> (which is equivalent to the equation you used), I get 3554m/s for a small mass orbiting at Mars' surface. But I'm getting 1.66 hours per orbit. I think you forgot to double the radius when you calculated Mars' circumference.
 
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