So Wikipedia says that the orbit period at 2000km (upper limit of LEO) is about 127 minutes, but when I calculate...

$$T=\sqrt{\frac{4 \pi^2r^3}{\mu}}=\sqrt{\frac{4 \pi^2(2000 \text{ km})^3}{398600 \text{ km}^3/\text{s}^2}}=890 \text{ s}$$

I get 14.8 minutes. Can someone tell me what I'm doing wrong?

  • 1
    $\begingroup$ Here's the notation we use. $\endgroup$ – HDE 226868 Apr 9 '15 at 1:12
  • $\begingroup$ You got the answer, now here's a shortcut. Just replace the 2000 km with any altitude above Earth to get its orbital period. If you want in minutes, just add in minutes to the end of the formula. Or any other time unit you want that Google recognizes. ;) $\endgroup$ – TildalWave Apr 9 '15 at 10:44

You're using the orbit's altitude as the orbital radius. Add the radius of Earth to the orbit altitude to get the actual radius.

Note that an Earth orbit at 2000km radius would be subject to significant lithospheric drag.

| improve this answer | |
  • 3
    $\begingroup$ lithospheric drag ... lol! $\endgroup$ – user8406 Apr 10 '15 at 2:17
  • $\begingroup$ quite a sleeper here! maybe we need a Wiki to start curating these. $\endgroup$ – uhoh Apr 24 '17 at 2:14

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