I'm not asking about velocity, I'm talking about overall speed. Shortest time from perigee to perigee. This orbit would be a low, circular orbit, but I can't really find any specifics on any world records.
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$\begingroup$ Around Earth, or other bodies too? (Something smaller and denser like a neutron star could be orbited in just a few seconds, I heard somewhere.) $\endgroup$– Camille GoudeseuneDec 3, 2020 at 21:26
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$\begingroup$ See this question $\endgroup$– UweDec 3, 2020 at 21:51
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2$\begingroup$ The JAXA satellite Tsubame was recognized for achieving the lowest altitude, of only 167.4 kilometers. That broke the old mark of 224 km set by the European Space Agency. $\endgroup$– Joe JobsDec 3, 2020 at 22:10
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2 Answers
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See this question
The fastest possible orbit around Earth is just the last full orbit just before reentry caused by the drag of the atmosphere. At 160 km an orbit will last for one day only. Only 1 hour 27 min 32 sec is needed for an orbit.
But an orbit at 400 km needs 1 hour 32 min 24 sec, only 5 minutes longer. If you need about 90 minutes anyway, there is no use of being 5 minutes faster if the spacecraft is destroyed a short time after. At 400 km you may stay for about several years.
By the way, an elliptical orbit has one perigee (farthest point) and one apogee (closest point). So every point of a circular orbit may be apogee as well as perigee, or we may say there is no apogee and no perigee. "Shortest time from perigee to perigee" is not a useful definition of the period of circular orbit.
But a circular orbit at 400 km height and an elliptical orbit from 390 to 410 km have the same period 1 hour 32 min 24 sec. Same story for 380 to 420 km. So you don't need to insist on circular orbits. Same story for 380 to 420 km.
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The time for the fastest orbit is exactly the same as dropping an object from the same height down a pipe through the center of the Earth (in vacuum) and waiting for it to come back. It is a standard problem in intro calculus based physics courses.
Fortunately the result is simple. The period is 2pi times the square root of (radius of the Earth / g). Just add the altitude you want to the radius, like the absolute minimum for an orbit, and calculate. Use meters and the answer is in seconds.
If the altitude is more than a couple hundred km the values will trend too fast because once you leave the mass of the Earth, the value of g is decreasing and not accounted for in this simple result.
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1$\begingroup$ I'd recommend using the more accurate and generally-useful Orbital Period formula . The provided calculation of the fall time of an object through the Earth makes the assumption that the Earth is of uniform density, when it actually increases by a factor of about 5.5 between crust and core. $\endgroup$– notovnyDec 5, 2020 at 13:50