Mercury’s Shadowy North Pole

[muriel_volestrangler]

6. 58 days is the sidereal rotation period

ie the time for a star (effectively at infinite distance) to come back to the same position in the sky. But since Mercury's orbit period is 88 days, it takes longer than either for the Sun to get back to the same position in the sky, ie midday to midday - by the time Mercury has rotated once, relative to the stars, it's gone 2/3rds through its orbit, so it would have to rotate another 2/3rds to get the Sun in a similar position, ie another 39 days, and then a bit more to make up for how much further it went in that orbit, etc. The end result is the solar day is 176 days long. And for the purposes of the heating from the Sun, that's the day we're interested in.

Here's the NASA figures:

http://nssdc.gsfc.nasa.gov/planetary/factsheet/mercuryfact.html

--------------------------- Mercury Earth -- Ratio (Mercury/Earth)
- Sidereal orbit period (days) 87.969 365.256 0.241
Sidereal rotation period (hrs) 1407.6 23.9345 58.785
------- Length of day (hrs) 4222.6 24.0000 175.942