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How fast would ISS need to travel at the current altitude (LEO 250mi) to be geosynchronous with say 36 degree N lattitude?

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    $\begingroup$ The question is premised on a misconception. Without continuous thrust (which is entirely impractical) there's only one possible speed for circular orbit at a given altitude. $\endgroup$ – Russell Borogove Jul 26 '17 at 20:02
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    $\begingroup$ And also on the misconception that a geostationary orbit is possible about anything but the equator. $\endgroup$ – David Hammen Jul 27 '17 at 1:27
  • $\begingroup$ @DavidHammen OP did say "-synchronous" rather than "-stationary" at least, though it still doesn't work with any of the the other constraints. $\endgroup$ – uhoh Jul 27 '17 at 3:53
  • $\begingroup$ @uhoh - OP also say "with say 36 degrees N latitude", which doesn't make sense. $\endgroup$ – David Hammen Jul 27 '17 at 4:31
  • $\begingroup$ @DavidHammen I know, the consistency has a damping range of about +/- 3 words. $\endgroup$ – uhoh Jul 27 '17 at 4:33
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If the ISS were to decrease speed to try to be geosynchronous it would start to fall in it's orbit. The only way to keep it in its current orbit and be moving slow enough to be geosynchronous you would have to apply thrust backward against its current direction and constantly apply thrust up away from the Earth to counter the ISS from falling in altitude.

The speed at geosynchronous orbit is 3 km/sec, the speed of the rotation of the Earth at the equator is 0.46 km/sec, geosynchronous orbit is at 35,786 km and the ISS is at 400 km, the current speed of the ISS is 7.66 km/sec so the speed the ISS would need to travel to match the ground would only be slightly greater than 0.46 km/sec.

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  • $\begingroup$ The edits completely changed the meaning of the answer... $\endgroup$ – FKEinternet Jul 26 '17 at 20:22
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Actually, the ISS would have to slow down by an order of magnitude in order to take 24 hours to orbit the Earth at its current altitude where it completes an orbit every 1.5 hours - 16 times a day - with an orbital velocity of 4.76 mi/sec. Thus, to keep it "geosynchronous" at its current altitude, you would have to slow it to 1071 mph (4.76 mi/sec / 16 * 3600 - current speed divided by orbits per day times seconds per hour). At that speed, it wouldn't be in orbit any more, so you'd have to be expending a LOT of energy as upward thrust to keep it from falling straight down from the sky like a brick.

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  • $\begingroup$ I swear I caught myself and corrected mine before I read yours, I felt dumb enough getting it reversed, the ground speed threw me off. $\endgroup$ – Brooks Nelson Jul 26 '17 at 20:23
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    $\begingroup$ @BrooksNelson Welcome to life ;) $\endgroup$ – FKEinternet Jul 26 '17 at 20:25
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AFAIK it's not possible for it to be Geosynchronous at it's current altitude. Geosynchronous orbits are only possible at a much higher altitude. At any given altitude there is only one speed which produces a stable orbit. Any faster and you'll spiral outward and any slower and you'll spiral inward. There's only one altitude where the correct orbital speed results in a 24 hour orbital period which is what produces a Geosynchronous orbit.

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  • $\begingroup$ Sans atmospheric drag, orbits do not spiral in or out. They instead form conical sections (circles, ellipses, parabolas, and hyperbolas). $\endgroup$ – David Hammen Jul 26 '17 at 23:59

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