ISS has an inclination of about 56 degrees but because his orbit is not exactly a whole part of 24 hours, ISS will the next day be above other places. So after a while he has 'visited' most of the planet (exept north and south pole) but how long does this take?

Or with other words, when ISS travels above your house how does it take before he will cross it again?

I hope it is clear, but perhaps I mixed things up?

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    $\begingroup$ That depends on the size of the house ;) $\endgroup$ Commented Dec 23, 2016 at 18:32
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    $\begingroup$ Depends on how you define "visit". Does the ISS only have to be visible from the ground to have visited a location? Or does it need to be directly overhead (i.e. the ISS only visits a track ~100 m wide)? $\endgroup$
    – Hobbes
    Commented Dec 23, 2016 at 19:05
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    $\begingroup$ Well if ISS is really going directly overhead to all places and that would be within a lifetime than it is ok. But perhaps that takes too long, I don't know. $\endgroup$
    – Marijn
    Commented Dec 23, 2016 at 19:28
  • $\begingroup$ Looks like "one lifetime" only gives you a 63% chance - see below. $\endgroup$
    – uhoh
    Commented Dec 24, 2016 at 4:29

1 Answer 1


Let's consider different interpretations of what "travel over" a place means to get a ballpark idea.

The ISS is literally "over" a given place:

The ISS is about 108.5 meters wide, at an altitude (recently) of about 400 km, and the Earth's equatorial radius (where the coverage is most sparse) is 6378 km. So on the surface that projects down to a 102 meter wide "footprint" on the surface at the equator. If each equator-crossing is at an angle of 51.6 degrees with respect to the equator, it paints a section of the equator that is 102 meters / cos(90-51.6) ~ 130 meters.

Since the Earth's equator is about 40 million meters in circumference, it would take an absolute minimum of 308,000 crossings to paint the equator. At a period of 93 minutes, that would be about 19,900 days or a minimum of 57 years.

That lower limit is unrealistic. Since the ISS isn't tightly coupled to the Earth's surface in a repeat ground-track orbit one could try a statistical analysis assuming that over the decades the passes were distributed randomly. In that case it turns out only about 63% of the equator's circumference would be painted in the first 57 years, and something like 63% of the leftover would be painted in the next 57 years, etc. (see python script at the end)

The ISS passes within 1 degree of the zenith:

That's about 7 kilometers, so the equator would be 63% painted after 5,700 orbits (about one year).

The ISS passes within 45 degrees of the zenith:

That's about 800 kilometers, so the equator would be 63% painted after 50 orbits (three days).

Remember this is just statistical, so there could be much longer periods where some areas go without seeing the ISS, but only very careful (and pointless) planning could prevent any one given place on the equator from not having the ISS within 45 degrees of the zenith within a given year.

note: for points above the equator, but within 51.6 degrees North and South latitude, the track's motion is increasingly parallel to the ground, so the probabilities become higher. Equator coverage is the hardest.

Here's a quick python calculator to confirm the "about 63%" statistical estimate:

import numpy as np

total_meters = int(6378 * 1000 * 2 * np.pi)
n_orbits     = int(total_meters / 130)
equator      = np.zeros(total_meters, dtype=bool)
positions    = (total_meters * np.random.random(n_orbits)).astype(int)

for pos in positions:
    equator[pos:pos+130] = True  # ignore wrapping - can use np.put()

print equator.sum().astype(float)/len(equator)

print "FYI: 1-exp(-1) = ", 1 - np.exp(-1)

gives for example:

FYI: 1-exp(-1) =  0.632120558829
  • $\begingroup$ You should add a calculation for "appears above the observers (visible) horizon". Is that the same as ~80degrees from the zenith? $\endgroup$
    – Innovine
    Commented Dec 24, 2016 at 7:52
  • $\begingroup$ @Innovine be my guest! But you'd better re-read the question first. $\endgroup$
    – uhoh
    Commented Dec 24, 2016 at 8:16
  • $\begingroup$ "when it travels over your house" is open to interpretation. I'd count a few tens of degrees above the horizon as "over my house". I read the question again, according to your suggestion, three times actually, and I still am unable to locate the definition of "over your house". Perhaps you would like to quote the relevant part since I have had such trouble locating it. $\endgroup$
    – Innovine
    Commented Dec 24, 2016 at 8:48
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    $\begingroup$ @Innovine so most Earth houses are much smaller than 100m, and the literally "over" section covers that. The other two are there for reference only, to make sure people understand that 1) the frequency increases dramatically as soon as you open up the constraint or interpretation a bit, and 2) no matter which you choose, it's not really a fixed interval but more like a probability. Adding a "barely visible" point doesn't enhance this further, and there is nothing anywhere in the question about seeing the ISS or it being visible, so it would be sort of a no sequitur. $\endgroup$
    – uhoh
    Commented Dec 24, 2016 at 9:25
  • $\begingroup$ @Innovine I just did a quick calculation, at 10 degrees above the horizon it would actually be over a house that's roughly 2,000 kilometers from your house. I think that's stretching it a bit much. $\endgroup$
    – uhoh
    Commented Dec 24, 2016 at 10:02

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