When a satellite is in geostationary orbit, it is supposed to stay above one spot of the earth and rotate around the earth at the same rate the earth spins. But how exact does that equality need to be? Let's say I pointed a laser such that it hit a geostationary satellite exactly (perhaps a GPS satellite). How long would it take for the satellite to shift out of position such that the laser would no longer hit the satellite?
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asked Mar 12, 2017 at 6:48
David says Reinstate MonicaDavid says Reinstate Monica
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2$\begingroup$ There are several good questions and answers already here in SXSE that relate to GEO satellite drift, perturbations, and station keeping. See for example How much “wobble” does a typical geostationary satellite experience? and How much is a geostationary satellite expected to deviate from the geostationary orbit?. $\endgroup$– uhohCommented Mar 12, 2017 at 8:38
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5$\begingroup$ And just to be clear, GPS satellites are not geostationary; they're in 12-hour orbits that are tilted with respect to the equator. $\endgroup$– Dave TweedCommented Mar 12, 2017 at 13:25
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1$\begingroup$ Pointing lasers at the sky is not recommended and may give you large fines if you manage to hit an airplane that flies thru the laser beam $\endgroup$– FerrybigCommented Mar 12, 2017 at 17:07
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1 Answer
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Generally, the GEO satellites are to keep their desired position above ground within +/- 0.05 deg (both N and E) which translates into a 70 km 2D projection corridor at GEO altitude. The laser pointer case, is dependent on the beam width. A laser pointer of the kind you get in stores has beamwidth of 1-3 milliradian, i.e. about 0.05 to 0.17 deg. This translates roughly into a circle of diameter 35- 105 km at GEO altitude. The satellite should stay within it for a long time. Calculations can be updated for other cases, like the one for LLCD in this answer which shows 3.5 urad beamwidth for the corresponding laser.
Lets look at how it does drift from nominal though. Looking broadly at the perturbation scenario the term geostationary orbit as derived from Kepler's 2BP and its relation with the number 42164.2 km is not as precise in real-life. The real orbit radius is not equal to that number generally. The satellite drifts with time, owing to several sources of perturbation including equatorial bulge, SRP, third-body effects and Earth's precession and nutation. Most are long term variation while some short term variation do exist (considerable longitude drift can be seen within 6 months, semi-major axis might drift by tens of kilometers in a matter of few days. Here are some trends to give a perception of drifting of a satellite in GEO.
Graph 1 : Shows the drift of semi-major axis for a satellite placed at nominal R of 42164.2 km propagated for about 6 months shows an increase of about 21kms.
Graph 2 :Due to asymmetry of Earth (More elliptical at equator) Longitude drift occurs. Longitude can greatly vary with time.The graph shows how over a propagation period of 160 days, the satellite at a nominal longitude of 125 deg can drift eastward till 105 deg.
Graph 3 :The change of longitude-drift can be observed with longitude drift-time trend which is essentially linear. The slope depends upon nominal longitude of the satellite.
Graph 4 :The longitudinal drift rate for a nominal longitude at a point, in turn, is given by this graph and varies in form of a parabola.
Considering the asymmetry there are four equillibrium points- two stable (at 75.3°E and 104.7°W) and two unstable (at 165.3°E and 14.7°W) equilibrium points. Any geostationary object placed between the equilibrium points would (without any action) be slowly accelerated towards the stable equilibrium position, causing a periodic longitude variation.
Inclination changes due to 'wobbling' motion of the Earth with 0.85deg/yr rate between +/-15 deg maximum with a period of 26.6 years. Also, eccentricity, argument of perigee and RAAN varies considerably. This complex variation can better be perceived by drawing relative motion in cartesian coordinates of the satellite in orbit as well as the observer on Earth.
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Source:
[Book] Li, Hengnian. Geostationary satellites collocation. New York: Springer, 2014.
answered Mar 12, 2017 at 7:45
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$\begingroup$ Can you explain the meaning and significance of each of these plots? They aren't helping me understand your answer at the moment. Right now it looks like you've just pasted a bunch of plots to make the answer look good, without explaining how the plots specifically enhance or support your explanation. Thanks! $\endgroup$– uhohCommented Mar 12, 2017 at 8:30
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1$\begingroup$ @uhoh Tried updating with some more information to clariify. Check the relevance. Else, the answers mentioned in your comment are great, as they are. Can delete if the answer is not necessary. $\endgroup$ Commented Mar 12, 2017 at 10:15
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4$\begingroup$ Please make clearer which parts of this answer are your own work, and which have been copy/pasted from the named book. Use "quote" formatting to indicate words that are not your own. $\endgroup$ Commented Mar 12, 2017 at 13:59
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$\begingroup$ For reference that +/- 0.05 deg is about +/- 5.56 km at the Earth's mean radius (ground-ish). $\endgroup$– Jason CCommented Mar 12, 2017 at 14:44
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$\begingroup$ I've just asked Why would the orbit of a satellite in GEO climb by 21 km in altitude over six months by itself? $\endgroup$– uhohCommented Jan 10, 2019 at 6:12