Many, but not all, space stations are in orbit around a planet. An orbit is a clever way to constantly fall towards a planet but never hit the ground.
There are certain preferred orbits.
An equatorial orbit is a noninclined orbit with respect to Terra's equator (i.e., the orbit has zero inclination to the equator, 180° inclination if retrograde). Most civilian satellites use such orbits. The United States uses Cape Canaveral Air Force Station and the Kennedy Space Center to launch into equatorial orbits.
An ecliptic orbit is a noninclined orbit with respect to the solar system ecliptic.
An inclined orbit is any orbit that dock does not have zero inclination to the plane or reference (usually the equator).
A polar orbit is a special inclined orbit that goes over each pole of the planet in turn, as the planet spins below (i.e., the orbit is inclined 90° to the equator). Heinlein calls it a "ball of yarn" orbit since the path of the station resembles winding yarn around a yarn ball. The advantage is that the orbit will eventually pass over every part of the planet, unlike other orbits. Such an orbit is generally used for military spy satellites, weather satellites, orbital bombardment weapons, and Google Earth. The United States uses Vandenberg Air Force Base to launch into polar orbits. Google Earth uses data from the Landsat program, whose satellites are launched from Vandenberg.
ORBITAL PERIODS
How long it takes a space station or ship to make one orbit depends upon how massive the planet is and the altitude of the orbit. The mass of the station doesn't matter. The equations below are for a circular or nearcircular orbit (low eccentricity) and where the mass of the planet is much larger than the mass of the space station (which is always the case unless the station is built out of stellar black holes or something). The equations for elliptical orbits are a bit more complicated.
An orbit with an orbital period exactly equal to the planetary rotation (one planetary "day") is highly prized for communication satellites.
OrbitalRadius = OrbitalAltitude + PlanetRadius
OrbitalAltitude = OrbitalRadius  PlanetRadius
OrbitalPeriod = (2 π) sqrt[ OrbitalRadius3 / (G PlanetMass) ]
OrbitalVelocity = sqrt[ (G PlanetMass) / OrbitalRadius ]
OrbitalRadius = cubeRoot[ (G PlanetMass OrbitalPeriod2) / (4 π2) ]
μ = G PlanetMass
μTerra = 3.99×1014
OrbitalPeriodTerra = 6.28318 sqrt[ (6.37×106 + OrbitalAltitude)3 / 3.99×1014 ]
OrbitalVelocityTerra = sqrt[ 3.99×1014 / OrbitalRadius ]
OrbitalRadiusTerra = cubeRoot[ (3.99×1014 OrbitalPeriodTerra2) / 39.478 ]
where:
OrbitalRadius = distance from station to center of planet (m)
OrbitalAltitude = distance from station to surface of planet (m)
PlanetRadius = distance from center of planet to planet's surface (m) (Terra = 6.37×106 m)
OrbitalPeriod = time it takes station to make one orbit around the planet (sec)
OrbitalVelocity = mean velocity of station in its orbit (m/s)
π = pi = 3.14159...
G = Newton's gravitational constant = 6.673×1011 (N m2 kg2)
PlanetMass = mass of planet (kg) (Terra = 5.98×1024 kg)
μ = standard gravitational parameter
sqrt[ x ] = square root of x
cubeRoot[ x ] = cube root of x
Example
The planet Mars has a mass of 6.4171×1023 kg and a mean radius of 3,390,000 m. What is the orbital period of a station at an altitude of 20,073,000 meters?
OrbitalRadius = OrbitalAltitude + PlanetRadius
OrbitalRadius = 20,073,000 + 3,390,000
OrbitalRadius = 23,463,000 m
OrbitalPeriod = (2 π) sqrt[ OrbitalRadius3 / (G PlanetMass) ]
OrbitalPeriod = (2 3.14159) sqrt[ 23,463,0003 / (6.673×1011 6.4171×1023) ]
OrbitalPeriod = 6.28318 sqrt[ 12,916,671,713,847,000,000,000 / 42,821,308,300,000 ]
OrbitalPeriod = 6.28318 sqrt[ 301,641,220.84628 ]
OrbitalPeriod = 6.28318 17,367.82142
OrbitalPeriod = 109,125 seconds = 1,818.8 minutes = 30.3 hours
For this problem I used the altitude of the Martian moon Phobos, which has an orbital period of 30.312 hours. Our result of 30.3 is close enough for government work.
The planet Terra has a mass of 5.98×1024 kg and a mean radius of 6.37×106 m. What is the orbital altitude of a station with an orbital period of 5,559 seconds?
OrbitalRadius = cubeRoot[ (G PlanetMass OrbitalPeriod2) / (4 π2) ]
OrbitalRadius = cubeRoot[ (6.673×1011 5.98×1024 5,5592) / (4 π2) ]
OrbitalRadius = cubeRoot[ (6.673×1011 5.98×1024 30,902,481) / (4 9.86961) ]
OrbitalRadius = cubeRoot[ 12,331,492,891,637,400,000,000 / 39.47844 ]
OrbitalRadius = cubeRoot[ 312,360,186,766,179,210,728.69140725925 ]
OrbitalRadius = 6,785,031 m
OrbitalAltitude = OrbitalRadius  PlanetRadius
OrbitalAltitude = 6,785,031  6.37×106
OrbitalAltitude = 415,031 m = 415 km
For this problem I used the orbital period of the International Space Station. It has a mean orbital altitude of 404.55 km, which is close to our calculated 415 km. I presume the discrepancy is due to the fact that the ISS has an inclination of 51.64°, while Deimos has an inclination of about 1°.
Elliptical Orbital Periods
Just so you know, when it comes to planetary orbits and spacecraft trajectories, none of them are perfectly circular. It is just that so many of them are close enough to being a circle that a science fiction author can get away with using the above equations. We call them Kepler's laws of planetary motion because Kepler found that the equations worked if you assumed the planet orbits were ellipses (which are eccentric circles). Kepler's boss Tycho Brahe was dumped in the dustbin of history because he stubbornly insisted that planet orbits were perfect circles.
And when you get to things like spacecraft transfer orbits, some are not even close to being circular.
What you have to do is use the orbiting object's SemiMajor Axis instead of OrbitalRadius.
Don't panic, it is easy to calculate. As long as you have the object's Periapsis and Apoapsis (in meters), which means the object's closest approach and farthest retreat from the planet it is orbiting. Those numbers are easy to find, for example see Wikipedia's entry for the Moon. Periapsis (called perigee) of 362,600 kilometers and Apoapsis (called apogee) of 405,400 kilometers, right in the data bar on the right. Don't forget to convert the values to meters for the equations, e.g., multiply 362,600 km by 1,000 to convert to 362,600,000 m.
Sometimes an orbit will actually be specified by the periapsis and apoapsis. For instance the Orion bomber's patrol orbit is described as a 190,000410,000 km Terran orbit.
Given periapsis and apoapsis in meters, the SemiMajor Axis is:
SemiMajorAxis = (Periapsis + Apoapsis) / 2
Take the equation for OrbitalPeriod, replace OrbitalRadius with SemiMajorAxis, and you are good to go.
OrbitalVelocity unfortunately is a major pain. You see, the orbiting object moves at different speeds at different parts of its orbit. It moves fastest at periapsis and slowest at apoapsis. Only if the orbit is perfectly circular does the orbiting object always move at the same speed.
If you use the OrbitalVelocity equation replacing OrbitalRadius with SemiMajorAxis, you will get the Mean or Average orbital velocity.
If you want the orbital velocity at a specific point in the orbit, you will specify said point by its distance from the primary. The distance will be somewhere between periapsis and apoapsis, inclusive. Again it will be fastest at periapsis and slowest at apoapsis. The equation is:
OrbitalVelocity = sqrt[ (G PlanetMass) ( (2 / CurrentOrbitalRadius)  (1 / SemiMajorAxis) ) ]
This is the famous Vis Viva Equation, which comes in real handy to calculate deltaV requirements for various missions.
If for some reason you want to draw the orbit, it isn't too hard. As long as have a drawing program that can create an ellipse given a bounding box (The Gimp, Inkscape, Adobe Photoshop, Adobe Illustrator). First you calculate the semimajor axis and the semiminor axis.
SemiMajorAxis = (Periapsis + Apoapsis) / 2
SemiMinorAxis = sqrt[ SemiMajorAxis2  ( SemiMajorAxis  Periapsis )2 ]
Chose a convenient scale for your drawing program, like 1,000,000 meters equals one pixel. Draw the upper and lower sides of the box (red lines in diagram above) which are twice the length of the SemiMajorAxis in scale. Draw the left and right sides of the box (green lines) which are twice the length of the SemiMinorAxis in scale. That is the bounding box.
Draw a horizonal center line so it is equidistant from the top and bottom edges of the box. Draw a vertical line (blue in diagram above), and move it so it is one Periapsis scale length away from the right edge. Where these two lines cross is the location of the planet.
Move the box so the crosshairs are on the planet image. Use the "draw ellipse" function of the drawing program such that the ellipse fits in the bounding box. That is the orbit. You can erase the bounding box now, you don't need it any more.
THE HILL SPHERE
It is greatly desired that satellites and space stations stay in stable orbits, because corporations and insurance companies become quite angry if hundred million dollar satellites or expensive space stations with lots of people are gravitationally booted into The Big Dark.
A good first approximation is ensuring that the orbiting object stays inside the parent's Hill Sphere. This is an imaginary sphere centered on the parent planet (the planet or moon the satellite is orbiting). Within the sphere, the planet's gravity dominates any satellites.
For first approximation you have three players: the space station (e.g., SupraNew York), the planet or moon it is orbiting (e.g., Terra), and the object the planet is orbiting (e.g., Sol) otherwise known as the planet's "primary".
The point is that Sol cannot gravitationally capture SupraNew York as long as all of the space station's orbit is inside Terra's Hill Sphere.
You can calculate the approximate radius of a planet's Hill Sphere with the following equation:
r ≈ a cbrt( m / (3 M) )
where:
r = Radius of Hill Sphere (kilometers)
a = Distance between the planet and its primary (kilometers)
m = mass of the planet (kilograms)
M = mass of the primary (kilograms)
cbrt(x) = cube root of x (the ∛x key on your calculator)
Actually you can use any desired unit of distance for r and a as long as you use the same for both. The same goes for units of mass for m and M.
This equation assumes that the planet is in a nearcircular orbit. If it has some weird eccentric orbit the Hill Sphere link has the more complicated equation. The above equation also assumes that the mass of the station or sattelite is miniscule compared to the object it is orbiting. It further assumes that the mass of the primary is quite a bit bigger than the mass of the planet.
In practice, for long term stability, the station should not orbit its planet further than onehalf the Hill sphere radius. No further than onethird the Hill sphere radius if you are ultracautious.
EXAMPLE
What is the Hill sphere radius of Terra?
In this case, the planet is Terra, the primary is Sol, the mass of Terra (m) is 5.97×1024 kg, the mass of Sol (M) is 1.99×1030 kg, the distance between Terra and Sol (a) is 149.6 million kilometers (149,600,000 km).
r ≈ a cubrt( m / (3 M) )
r ≈ 149,600,000 cubrt( 5.97×1024 / (3 1.99×1030) )
r ≈ 149,600,000 cubrt( 5.97×1024 / 5.97×1030 )
r ≈ 149,600,000 cubrt( 0.000001 )
r ≈ 149,600,000 0.01
r ≈ 1,496,000 kilometers
So the ultimate Hill sphere radius is 1.496 million kilometers from Terra's center. Luna is at 0.384 million kilometers, safely inside the Hill sphere. The implication is that all of Terra's stable satellites have an orbital period of less than seven months. SupraNew York would do well to stay inside.
The long term stable radius is 0.748 million kilometers from Terra's center (1/2 Hill sphere). The ultracautious radius is 0.499 million kilometers (1/3 Hill sphere).
If you were interested in Lunar satellites, the planet would be Luna, the primary would be Terra, and a would be the distance between Terra and Luna.
Any object (like a spaceship) which enters a planet's Hill sphere but does not have enough energy to escape, will tend to start orbiting the planet. The surface of the Hill sphere is sometimes called the "zerovelocity surface" for complicated reasons.
(the section about launch site inclinations has been moved here)

MEO: yellow
HEO: orange
Hill Sphere is about four times the radius of the orange sphere
Orbits around Terra (geocentric) are sometimes classified by altitude above Terra's surface (which is 6.37×103 km from Terra's center):
 Low Earth Orbit (LEO): 160 kilometers to 2,000 kilometers. At 160 km one revolution takes about 90 minutes and circular orbital speed is 8 km/s. Affected by inner Van Allen radiation belt.
 Medium Earth Orbit (MEO): 2,000 kilometers to 35,786 kilometers. Also known as "intermediate circular orbit." Commonly used by satellites that are for navigation (such as Global Positioning System aka GPS), communication, and geodetic/space environment science. The most common altitude is 20,200 km which gives an orbital period of 12 hours.
 Geosynchronous Orbit (GEO): exactly 35,786 kilometers from surface of Terra (42,164 km from center of Terra). One revolution takes one sidereal day, coinciding with the rotational period of Terra. Circular orbital speed is about 3 km/s. It is jampacked with communication satellites like sardines in a can. This orbit is affected by the outer Van Allen radiation belt.
 High Earth Orbit (HEO): anything with an apogee higher than 35,786 kilometers. If the perigee is less than 2,000 km it is called a "highly elliptical orbit."
 Lunar Orbit: Luna's orbit around Terra has a pericenter of 363,300 kilometers and a apocenter of 405,500 kilometers.
 UltraCautious Hill Sphere: 496,540 kilometers from surface of Terra (498,670 km from center)
 Long Term Stable Hill Sphere: 744,820 kilometers from surface of Terra (748,000 km from center)
 Ultimate Hill Sphere: exactly 1,489,630 kilometers from surface of Terra (1,496,000 km from center)

It shows how three space stations in geostationary orbit spaced at 120° can provide coverage over all of Terra
Geosynchronous Orbits (aka "Clarke orbits", named after Sir Arthur C. Clarke) are desirable orbits for communication and spy satellites because they return to the same position over the planet after a period of one sidereal day (for Terra that is about four minutes short of one ordinary day).
A Geostationary Orbit is a special kind of geosynchronous orbit that is even more desirable for such satellites. In those orbits, the satellite always stays put over one spot on Terra like it was atop a 35,786 kilometer pole (remember: 42,164 km from center of Terra). For complicated reasons all geostationary orbits have to be over the equator of the planet. In theory only three communication satellites in geostationary orbit and separated by 120° can provide coverage over all of Terra.
All telecommunication companies want their satellites in geostationary orbit, but there are a limited number of "slots" available do to radio frequency interference. Things get ugly when you have, for instance, two nations at the same longitude but at different latitudes: both want the same slot. the International Telecommunication Union does its best to fairly divide up the slots.
The collection of artificial satellites in geostationary orbit is called the Clarke Belt.
Note that geostationary communication satellites are marvelous for talking to positions on Terra at latitude zero (equator) to latitude plus or minus 70°. For latitudes from ±70° to ±90° (north and south pole) you will need a communication satellite in a polar orbit, a highly elliptical orbit, or a statite. Russia uses highly eccentric orbits since those latitudes more or less define Russia. Russian communication satellites commonly use Molniya orbits and Tundra orbits.
About 300 kilometers above geosynchronous orbit is the "graveyard orbit" (aka "disposal orbit" and "junk orbit"). This is where geosynchronous satellites are moved at the end of their operational life, in order to free up a slot. It would take about 1,500 m/s of delta V to deorbit an old satellite, but only 11 m/s to move it into graveyard orbit. Most satellites have nowhere near enough propellant to deorbit.
GEO Space Station
"Okay, T.K., look at it this way. Those three hundred people in LEO Base can get back to Earth in less than an hour if necessary; we'll have lifeboats, so to speak, in case of an emergency. But out there at GEO Base, it's a long way home. Takes eight hours or more just to get back to LEO, where you have to transfer from the deepspace passenger ship to a StarPacket that can enter the atmosphere and land. It takes maybe as long as a day to get back to Earth from GEO Base— and there's a lot of stress involved in the trip."
Hocksmith paused, and seeing no response from the doctor, added gently, "We can get by with a simple firstaid dispensary at LEO Base, T.K., but not at GEO Base. I'm required by my license from the Department of Energy as well as by the regulations of the Industrial Safety and Health Administration, ISHA, to set up a hospital at GEO Base."
He finished off his drink and set the glass down. "If building this powersat and the system of powersats that follow is the biggest engineering job of this century, T.K., then the GEO Base hospital's going to be the biggest medical challenge of our time. It'll be in weightlessness; it'll have to handle construction accidents of an entirely new type; it'll have to handle emergencies resulting from a totally alien environment; it'll require the development of a totally new area of medicine— true space medicine. The job requires a doctor who's worked with people in isolated places—like the Southwest or aboard a tramp steamer. It's the sort of medicine you've specialized in. In short, T.K., you're the only man I know who could do the job... and I need you."
Stan and Fred discovered that it took almost nineteen minutes just to get to Charlie Victor, Mod Four Seven. There were a lot of hatches to go through and a lot of modules to traverse. "Fred, if we don't find some faster way to move around this rabbit warren, a lot of people are going to be dead before we reach them," Stan pointed out, finally opening the hatch to Mod Four Seven.
Fred was right behind him through the hatch. "I'll ask Doc to see Pratt about getting us an EffMu."
"What's that?"
"Extra Facility Maneuvering Unit. A scooter to anybody but these acronymhappy engineers."
Transporting was easy in zerog, but getting through all the hatches while continuing to monitor his condition and maintain the positivedisplacement IVs was difficult. It required almost a half hour to bring the man back to the med module.
Lagrangian points are special points were a space station can sit in a sortof orbit. Lagrange point 1, 2, and 3 are sort of worthless, since objects there are only in a semistable position. The ones you always hear about are L4 and L5, because they have been popularized as the ideal spots to locate giant space colonies. Especially since the plan was to construct such colonies from Lunar materials to save on boost delta V costs. The important thing to remember is that the distance between L4 — Terra, L4 — Luna, and Terra — Luna are all the same (about 384,400 kilometers). Meaning it will take just as long to travel from Terra to L4 as to travel from Terra to Luna.
For a more exhaustive list of possible Terran orbits refer to NASA.
It is also possible for a satellite to stay in a place where gravity will not allow it. All it needs is to be under thrust. Which is rather expensive in terms of propellant. Dr. Robert L. Forward noted that solar sails use no propellant, so they can hold a satellite in place forever (or at least as long as the sun shines and the sail is undamaged). This is called a Statite.
If the planet has an atmosphere and the station orbits too low, it will gradually slow down due to atmospheric drag. "Gradually" up to a point, past the tipping point it will rapidly start slowing down, then burn up in reentry. Some fragments might survive to hit the ground.
The "safe" altitude varies, depending upon the solar sunspot cycle. When the solar activity is high, the Earth's atmosphere expands, so what was a safe altitude is suddenly not so safe anymore.
NASA found this out the hard way with the Skylab mission. In 1974 it was parked at an altitude of 433 km pericenter by 455 km apocenter. This should have been high enough to be safe until the early 1980's. Unfortunately "should" meant "according to the estimates of the 11year sunspot cycle that began in 1976". Alas, the solar activity turned out to be greater than usual, so Skylab made an uncontrolled reentry in July 1979. NASA had plans to upgrade and expand Skylab, but those plans died in a smoking crater in Western Australia. And a NOAA scientist gave NASA a savage I Told You So.
The International Space Station (ISS) orbited at an even lower at 330 km by 410 km during the Space Shuttle era, but the orbit was carefully monitored and given a reboost with each Shuttle resupply mission. The low orbit was due to the Shuttle carrying up massive components to the station.
After the Shuttle was retired and no more massive components were scheduled to be delivered, the ISS was given a big boost into a much higher 381 km by 384 km orbit. This means the resupply rockets can carry less station reboost propellant and more cargo payload.
If the planet the station orbits has a magnetic field, it probably has a radiation belt. Needless to say this is a very bad place to have your orbit located, unless you don't mind little things like a radiation dosage of 25 Severts per year.
There are known radiation belts around Terra, Jupiter, Saturn, Uranus and Neptune.
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