Just had 2 evenings when the ISS was visible twice on successive passes. Times for appearance were 94 minutes apart.

If I use the quoted value of g at 400 km above the earth of 0.89m/s/s I can calculate orbit time of 91 minutes which is pretty close BUT it makes no allowance for the earth's rotation in the 90 minutes which is not insignificant - about 22deg. For the ISS to 'catch up' with the earth's rotation surely it should need another 6 minutes or so. Where is the flaw in my thinking?


Where is the flaw in my thinking?

This is a good question, but there are a few flaws in your thinking.

If I use the quoted value of g at 400 km above the earth of 0.89m/s/s I can calculate orbit time of 91 minutes.

Your calculation of the ISS' orbital period is a bit off. The ISS is currently (Dec. 20) in a 404 x 415 km orbit, or a mean altitude of 409.4 km. That corresponds to an orbital period of 92.75 minutes. Even using a 400 km orbit, you should have computed an orbital period of 92.56 minutes.

This of course makes your perceived problem even worse. With this correction, there's but a ~1.25 minute difference between the 92.75 minute long orbit and the 94 minutes between passes.

Times for appearance were 94 minutes apart.

I suspect you subtracted the start times of the two passes, and that the first pass was short in duration while the second pass was a nice long one. The difference between consecutive start times can vary from about 94 minutes to 99 minutes. The difference between zenith times on consecutive passes has a much smaller variance, from a bit under 96 minutes to a bit over 97 minutes, with a mean of about 96.75 minutes. This corresponds to about four minutes in excess of the orbital period.

For the ISS to 'catch up' with the earth's rotation surely it should need another 6 minutes or so.

That would be the case if the ISS was in an equatorial orbit. An equatorial observer of a vehicle with a 92.75 minute prograde equatorial orbit would see successive passes separated by about 99 minutes. All passes would be more or less the same, with the vehicle moving from west to east and passing straight overhead.

The ISS is instead in an orbit inclined by about 51.6°. This changes the geometry considerably. An observer might see the first of two consecutive passes starting in the south and disappearing in the southeast, while the second pass might start in the southwest and disappear due east. The zenith passage of the two events corresponds to a 360°+15.5°change in true anomaly rather than the 360°+25° change for the equatorial orbiter.


Flaw in your thinking (I wouldn't really call it like that, it's an honest question) is that when the station is observable during two consecutive orbits, it would be observable during the first pass eastward from your stationary location, and on the second pass westward. You see, the station's ground track moves westward (about 2,505 km or 1,556 miles at the equator) as the Earth rotates towards the east. So the Earth's rotation is, in a sense, caching up with the first observation's ground track.

Now, if the second pass will go closer to zenith than the first one did (makes an apparent movement arc more directly overhead), then you'll be able to also see it a bit earlier in its orbital position that you did the first time around. Depending on your location relative to the station's first pass ground track, this slight variation in observable times could also go the other way around (like it seems it happened in your case), and you happen to have caught two consecutive passes when the first pass was closer to you (it would also be observable for a longer time, the station would appear brighter and it would have an eastward azimuth compared to the second pass).

Additionally, you probably also slightly miscalculated the station's orbital period, might be due to rounding input values, which in your case only increased the perceived difference in station passes timetable. Station's average orbital altitude is 420 km above mean sea-level, higher than your input of 400 km, so its true orbital period would also be slightly longer than 91 minutes. Let's do the calculations again:

If orbital period (in seconds) is given by $T = 2\pi\sqrt{a^3/\mu}$ (derived from Kepler's third law saying that the square of the orbital period of a planet is proportional to the cube of the semi-major axis of its orbit), $a$ is semi-major axis of the orbit (6798.1 km from Earth's center), and $\mu$ is standard gravitational parameter (398,600.4418 km3s−2 for Earth), then:

$$T = 2\pi\sqrt{314,168,505,637.141\ km^3/398,600.4418\ km^3s^{-2}}$$

$$T = 2\pi\sqrt{788,179.02\ s^{-2}} = 6.28318531 \times 887.7945\ s = 5,578.18\ s$$

Or 92.97 minutes. So the westward drift you observed on the second pass caused that the station was from your stationary position observable about one minute later than on the first pass, i.e. the station must have passed more directly overhead to your location on the first pass than it did on the second. Note that part of this might also be due to observational conditions, such as natural obstacles limiting your horizon towards the first and the second pass differently, and if your observations were just after dusk, it would be slightly darker on the second pass and you might be able to spot the station sooner because of that, too. So there's many different factors sliding this observational time up to a few minutes forward or backward in stations orbital position relative to your stationary location.


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