(posted before on physics stackexchange but was told to come here)

I'm reading this book which has a post-apocalyptic setting. At one point you look at the earth from the view of a 1000-year-old space station with a bunch of frozen astronauts on board. It was an interesting scene, but got me thinking -- doesn't the ISS have to perform orbit corrections semi-regularly (every couple months), just to avoid crashing back down to Earth?

So, how far away would you need to put a space station so it stays in orbit for 1000 years? Imaginary bonus points for $N$ years. Assume no fuel is left on board for orbit corrections.


2 Answers 2


tl;dr: Park your ISS-like space station above 700 km and there is a good chance it will only lose 100 m/s in 1,000 years due to atmospheric drag at least (and 2000 km for a million years). However, there are other problems

This is a really interesting question! Just for example, the LAGEOS satellites are about 6,000 km above the Earth's surface and are expected to re-enter the atmosphere in 8 million years or so. But they are spherical and dense, whereas a space station may be non-aerodynamically shaped and have a low density.

Let's look at the current TLE for the ISS from https://celestrak.org/NORAD/elements/stations.txt

ISS (ZARYA)             
1 25544U 98067A   19203.81086311  .00000606  00000-0  18099-4 0  9996
2 25544  51.6423 184.5274 0006740 168.1171 264.4057 15.50995519180787

The value for B-star is see this 18099-4 which is 0.18099E-04 which is 1.8099e-05 which is pretty big, as it should be for a hollow space station with big solar panels. It has units of inverse Earth radii (see this from this).

Wikipedia's BSTAR gives the following equation for acceleration due to drag:

$$a_D = \frac{\rho}{\rho_0} B^* v^2$$

where $\rho_0$ is a reference density and is about 0.1570 kg/m^2/Earth radii and $v$ is velocity presumably in m/s.

Calculating time to reentry requires some calculus, so let's just estimate the time it takes to lose 100 m/s in velocity.

$$\Delta t = \frac{\Delta v}{\frac{dv}{dt}} = \frac{\Delta v}{a_D} $$

If we then set $\Delta t$ to 1000 years or $\sim 1000 \times \pi \times 10^7$ seconds, we get $a_D \sim 3E-09$ m/s^2 for that 100 m/s loss.

Putting that back into the first equation and using the ISS' $B^*$, we get

$$\rho = \frac{a_D \rho_0}{B^* v^2} $$

The funky units (Earth radii-based) work out and the atmospheric density is about 8E-13 kg/m^2 based on an orbital velocity of about 7000 m/s

What altitude is that? It depends greatly on the Sun's activity. The plot below puts it as low as 380 km during a solar minimum, but up at 700 km during an active Sun.

This link found in this answer puts 8E-13 kg/m^3 closer to 600 km for an active Sun.

That link shows 8E-16 (for 1 million years) at about 2,000 km.

So park your ISS-like space station above 700 km and there is a good chance it will only lose 100 m/s in 1,000 years due to atmospheric drag at least, and above 2,000 km for 1,000,000 years. However, there are other problems due to a big patch of space junk in the 600 to 1000 km neighborhood.

Found in this answer, from https://en.wikipedia.org/wiki/Scale_height Wertz et al. SSC12-IV-6, 26th Annual AIAA/USU Conference on Small Satellites.

enter image description here

  • $\begingroup$ Only fixed a spelling error! I didn't touch anything else, promise :). $\endgroup$ Commented Jul 23, 2019 at 13:53
  • 1
    $\begingroup$ of course, as it looses that 100m/s, it will actually go faster. $\endgroup$
    – user20636
    Commented Jul 23, 2019 at 15:13
  • $\begingroup$ @JCRM yes good catch, that's exactly right! $\endgroup$
    – uhoh
    Commented Jul 23, 2019 at 15:20
  • $\begingroup$ @MagicOctopusUrn that one is hard-wired. I will never, ever be able to get that word right. It's spelled incorrectly in permanent sharpie in my brain for some reason. $\endgroup$
    – uhoh
    Commented Jul 23, 2019 at 15:21
  • 1
    $\begingroup$ perhaps "lose 100m/s of orbital energy" - of course we both know that's what you mean $\endgroup$
    – user20636
    Commented Jul 23, 2019 at 15:21

First, let's ask ourselves "why do orbits decay?". For multiple reasons.

  1. Atmospheric drag.
  2. Solar gravity acting on satellites.
  3. Solar activity

Satellites in geostationary orbit can stay in orbit for billions of years, such as the EchoStar XVI which orbits around 35000 km above Earth. Safe from the atmosphere, far enough to not decay from solar flares, and relatively close to the Earth to avoid solar gravity overtaking the satellite and decaying the orbit. Obviously these satellites won't be functional, but they'll still be there, like the space station you mentioned in your question.

So the answer to your question is: In order for a satellite to stay in orbit for at least 1000 years, it needs to be in a geostationary orbit.

  • $\begingroup$ What about lunar gravity acting on satellites? If you are high enough so that Earth's atmosphere and Earth's inhomogeneity don't significantly affect a satellite's orbit, won't the Moon start to have some effect? $\endgroup$
    – Anthony X
    Commented Jul 23, 2019 at 22:12
  • $\begingroup$ @Anthony X It will but the satellite will still remain in orbit. This is the exact reason geostationary satellites have thrusters. The moon's gravity will effect the satellite's orbit shape, so they have thrusters to make sure it remains circular (or whatever shape they desire). $\endgroup$
    – Star Man
    Commented Jul 23, 2019 at 22:17
  • $\begingroup$ @AnthonyX in case of the Geostationary orbit, sun and moon have their influence. It is what shifts the inclination from 0° to about 17° and back again over a 53 year period. Regarding StarMan's post, Geostationary Orbit is one solution, but not the only one. Basically, an orbit can be determined for any duration. Russian nuclear reactors in space, for example, are made to stay in orbit for around 1500 years, after which the radiation they'll leave behind is weak enough for safe re-entry. $\endgroup$
    – Infrisios
    Commented Jul 24, 2019 at 7:59

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