My book says that Geostationary Satellites can only be found in the equatorial plane. I wonder if there would be any other planes about which they could revolve around the earth.
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5Possible duplicate of Is it possible to have a geostationary satellite over the poles? – isanae May 21 '17 at 20:00
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2Note: An orbit that has the precise orbital period (equivalently, precise major axis length) required, but is not necessarily over the equator, is called geosynchronous, not geostationary. It repeats its position wrt. a fixed point on the surface once every day (23.93 hours). – Jeppe Stig Nielsen May 22 '17 at 11:45
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Lets consider what "Geostationary" actually means:
A geostationary orbit is a particular type of geosynchronous orbit, the distinction being that while an object in geosynchronous orbit returns to the same point in the sky at the same time each day, an object in geostationary orbit never leaves that position.
So we need to be above the same point on Earth all of the time.
The Earth is rotating, a point on the surface moving around west to east (ie in the equatorial direction). If your satellite moves in any direction other than this then it will diverge from the path of the point on earth that it wants to stay above.
(In case you weren't aware an orbit needs to have the center of mass of the two objects within it's orbital plane, so having an orbital plane parallel to the equatorial one but higher up wouldn't work).
Lio Elbammalf
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This answer is correct for practical purposes. But it should be noted that it is possible to have a geostationary satellite in any orbit. Higher than the equator or even not parallel to it. It would require unpractical amounts of energy to do so thus nobody does it. – Anonymous Coward May 21 '17 at 21:31
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@JoseAntonioDuraOlmos Could you elaborate? Even if you expend the continuous energy required to fix an object above a fixed point on Earth's surface which is not on the equator, wouldn't the circle that object describes still be parallel to the equator? – trichoplax is on Codidact now May 21 '17 at 21:54
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10@JoseAntonioDuraOlmos An orbit doesn't require additional energy other than to correct orbital drifts and energy loss from friction and air resistance. If it requires additional energy input, it is not an orbit and the object is simply performing a powered hover. – March Ho May 21 '17 at 22:28
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Whats to add is, that although this answer is correct, under some circumstances you can have an geostationary orbit at Lagrangian points. For example, if you are on a planet that always exposed the same side to it's sun, then there is a point between the sun and the planet where the gravitational forces would cancel. A satellite at that position would be geostationary if seen from the planet. https://en.wikipedia.org/wiki/Lagrangian_point – dani May 22 '17 at 08:25
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@everybody After checking some definitions... I was wrong. None of my proposed variations work. An orbit is a gravitationally curved path, not a propulsion curved path. Radovan's proposal has merit though, why would that not be a geostationary orbit? It would be above the same point on Earth all of the time. – Anonymous Coward May 22 '17 at 22:26
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@JoseAntonioDuraOlmos I'm guessing you mean Dani's proposal. In which case we're talking about different planets (since the earth isn't tidally locked with the sun) and also, whilst you may be facing the same point of the planet, you aren't orbiting the planet, you orbit the star. – Lio Elbammalf May 23 '17 at 11:41
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Strictly speaking, the only geostationary orbits are circular equatorial orbits at a very specific altitude above the surface of the Earth.
Satellites orbit on a plane that contains the center of the Earth. A satellite in a non-equatorial circular orbit spends half of each orbit north of the equator and the other half, south of the equator. Such satellites are not geostationary. A satellite at the wrong altitude orbits at a different rate than the Earth's rotation rate. Such satellites are not geostationary. A satellite in a non-circular orbit with the right semi-major axis results in the satellite sometimes moving faster than the Earth's rotation, other times slower. Such satellites might be geosynchronous, but they are not geostationary.
In practice, all geostationary satellites exhibit these behaviors to some extent. There are always errors in the orbit placement, and there are perturbations that inevitably nudge a satellite a bit off of its intended orbit. One of the jobs of the operations center for that satellite is to track its drift from the desired geostationary orbit and occasionally correct for deviations.
Some satellites are geosynchronous (or close to it) but are far from geostationary. A special class of geosynchronous orbits are the highly inclined (by about 63.4 degrees) and highly eccentric tundra orbits. The large eccentricity means the satellite spends a good portion of each orbit in a nearly stationary position near apogee. These address a key issue with geostationary satellites, which is that geostationary satellites aren't particularly useful as communications satellites north of about 60 degrees north latitude due to the low elevation angle at those high latitudes. The tundra satellites have their apogees near the northernmost reach of their orbits, making them nearly geostationary for about 1/3 of their orbits.
A related concept are the Molniya orbits. These too are highly eccentric orbits, but their period is half a day. The nearly hover over two northern spots twice daily. The Soviet Union (and now Russia) made extensive use of these orbits. The satellites spend a good portion of each orbit either high over northern North America (good for spying) or high over Russia (good for communications).
David Hammen
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Let's imagine that a satellite is geostationary over a point at 42 latitude (e.g. not over the equator). In this case, the satellite must always be found at 42 Latitude.
Because the Earth is (approximately) a sphere, we may approximate the earth as a point particle. If we do this, the satellite is not orbiting the centre of the Earth, but instead orbiting a point away from the centre of the Earth (which would be unphysical).
Any other orbit will not be geostationary, since the Earth will move (rotate) under the satellite as it orbits.
Bob
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To justify “Because the Earth is (approximately) a sphere, we may approximate the earth as a point particle” you should also state your assumptions about mass distribution. Whether balanced out or symmetric or uniform, whatever. – James Waldby - jwpat7 May 21 '17 at 18:47
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Previous answer is correct. However, IIRC, if your launch site is at high latitude so cannot practicably access equatorial orbit, you may mimic the coverage using several satellites sharing a very elliptical orbit. This lets each spend a significant proportion of each orbit in same narrow band of sky, so a fixed antenna should still work.
Downside, such orbit may be much more sensitive to geoid variation, MasCons etc when 'low', then Lunar position when 'high' so need more frequent tweaks than a genuine GeoSat's, so burn through more station-keeping fuel...
Nik
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7The principal reason the Russians use Molniya orbits is not lack of access to launch sites. It's because, as seen from the ground in many parts of Russia, a geostationary satellite would be too low on the horizon to be of any practical value. – Solomon Slow May 21 '17 at 15:38
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Which are yiu referring to as a “previous answer”? The order they are presented in will vary. – JDługosz May 22 '17 at 07:01
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This answer only works if you take the definition of geostationary orbit as an orbit that allows for the satellite to be constantly directly above a point on the planet.
Imagine you place some perpendicular poles down on some points on earth. One on the equator and one at anywhere else (let's say Australia ). Now you trace out the path traveled by the tip of the pole as the earth rotates. You should get two circles.
The circle by the pole of the equator should be centered at the center of earth, which is allowed as the force of gravity is canceled out by the centrifugal force of the satellite as it orbits (might not be accurate but should get the point across ). The path in which the satellite travels is in line with the force of gravity that acts on it. (The path is the circle traced out, while imagine the force of gravity as the radius of the circle.)
On the other hand, the pole at Australia would trace out a circle which is centered below the centre of the earth. In this case the direction of the force of gravity will not be aligned with the centrifugal force caused by the satellite's motion. This means that if some object were to travel in this orbit it would either have to have some other forces supplied, like rocket boosters, or it will fall to an orbit where all forces are canceled out.
Just tell me if my explanation is flawed or if I need to add pictures to clarify my idea.
height of pole is not specified here because it involves knowing the geostationary orbit altitude.(which you need the mass of the planet and speed in which the planet rotates to find out). For this case just assume that it is a very long pole that stretches into space.
See Jian Shin
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What centrifugal force? It's best not to invoke centrifugal force to explain orbits. This concept only works for circular orbits, and even then, it only works in a weird rotating frame in which the orbiting object is stationary. – David Hammen May 21 '17 at 14:59
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@David yeah I did try to explain it in other ways but I couldn't. And also an ideal geostationary orbit is circular. I was trying to show the idea of forces not balancing out for a satellite being geostationary and not on top of the equator. The "you are constantly falling but you are moving sideways si fast you miss earth" is more accurate I agree, but it simplifies to the idea of centrifugal force too as it is technically opposing the centripetal force of gravity. – See Jian Shin May 21 '17 at 15:10
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«definition of geostationary orbit as an orbit that allows for the satellite to be constantly directly above a point on the planet.» yes, that’s what Geo-Stationary means. It stays put relative to points on the Earth’s surface. – JDługosz May 22 '17 at 06:58
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@JDługosz couldn't believe that I didn't state this out. Yes my answer is a no. For a normal planet the geostationary orbit can only exist within the equatorial plane (whatever plane the planet revolves in.) Additionally some planets have geostationary orbits within the planet itself due to the speed in which it rotates. – See Jian Shin May 22 '17 at 07:04
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@SeeJianShin wouldn’t it fly apart if the surface were moving faster than orbital speed? – JDługosz May 22 '17 at 07:09
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@JDługosz I can't find reports on this, but you just need a planet that is relatively big and spins relatively fast. I might do some calculations on this one as I yet to find example planets. For reference, earth should spin around 20 times it's speed to have a geostationary orbit under the surface. – See Jian Shin May 22 '17 at 07:25
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If a planet were spinning that fast, the loose soil and rocks, water, air, etc. on the surface would fly away, since it would be moving at a speed greater than orbital velocity. (Note: s/it's/its/) – JDługosz May 22 '17 at 07:57
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@JDługosz allowing for my calculations , Beta Pictoris B seems to have a geostationary orbit well within the planet itself. As I said, the planet doesn't have to be very fast or very big, it just needs to be big and fast , like 8 hour day fast and 7 times Jupiter mass big. – See Jian Shin May 22 '17 at 08:36
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I get a period of 2.37 hours for an orbit that skims the surface of Beta Pictoris b. The planet takes 8.1 hours to rotate. By Kepler’s 2nd law, a synchronous orbit would be a semimajor axis of 2.27, or the altitude will be 1 and a quarter times the planet’s radius — not less than the planet’s radius. – JDługosz May 22 '17 at 09:00
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@JDługosz kepler's second law, law of equal areas? Also if the orbit is synchronous, the orbital period should be equal to the planet's time for a complete rotation. – See Jian Shin May 22 '17 at 09:07
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2nd is square/cube; you’re thinking of 3rd. Yes, I know the synchronous orbit on that body would be 8.1 hours— that's what we're talking about and I’ve not forgotten the meaning of geosynchronous since we started this conversation. – JDługosz May 22 '17 at 09:17
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@JDługosz I did my calculations again and found my error. The geostationary orbit is actually around 265000km, well above the surface. Thanks for clearing it up. But I still remember seeing some example planets that have geostationary orbits outside of its hill sphere and within its body. Maybe I'll start a new thread on this. – See Jian Shin May 22 '17 at 10:15