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A mercury day if you consider its solar day is 2 mercury years.

Now if you consider just 1 half of mercury, in 1 of those years it is facing towards the sun and in another one of those years it is facing away from the sun.

Mercury is tidally locked. This tidal locking happens when the rotation rate slows down until it reaches 0.

How is it possible for a planet to not rotate and yet have 1 half of it facing towards the sun each year with that half changing every year? It would seem like it would have to rotate 180 degrees in a fraction of a second which I know is not possible.

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  • $\begingroup$ Mercury has days?? This is news to me. $\endgroup$ – Scottie Oct 14 '14 at 15:42
  • $\begingroup$ yes it does have days just like all our other planets $\endgroup$ – Caters Oct 14 '14 at 16:34
  • $\begingroup$ But none of the other planets are tidally locked... How does a tidally locked planet have days? $\endgroup$ – Scottie Oct 14 '14 at 16:37
  • $\begingroup$ @Scottie a tidally locked planet doesn't have (solar) days (and nights). But though it might be news to you, Mercury is not tidally locked, and this is old news. See my answer. $\endgroup$ – Jeremy Oct 15 '14 at 9:49
  • $\begingroup$ @Jeremy: Good to know! Thanks for the update. $\endgroup$ – Scottie Oct 15 '14 at 14:12
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Your premise is incorrect. We used to think Mercury was tidally locked, but since 1965 we now know it is in a 3:2 spin-orbit resonance, which gives it (long) days.

Anything you read that says it is tidally locked is old, or itself using old reference material.

Here are some quick sketches to show the orbit, relative orientation of the planet, and the sidereal and solar days. If you like it as an answer I'll redraw with a graphics tool instead of my iPad Pen app...

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  • $\begingroup$ that is only considering the sidereal day. The one most important to us for discovering earth-like planets is the solar day length. $\endgroup$ – Caters Oct 14 '14 at 20:05
  • $\begingroup$ ...and the question is essentially asking about sidereal days on Mercury, not solar days on exoplanets. $\endgroup$ – Jeremy Oct 14 '14 at 20:11
  • $\begingroup$ But still, even in mercury day to earth day conversion you would use the solar day which is 2 mercury years, not the sidereal day with its 3:2 resonance. $\endgroup$ – Caters Oct 14 '14 at 23:03
  • $\begingroup$ I've added sketches to help explain how it works $\endgroup$ – Jeremy Oct 15 '14 at 7:53
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Like any other planet day/night will not flip immediately in a seconds. It will have its sunrise , noon and then sunset . The 1st year day refers to the time between morning and evening, and the 2nd half year of night (i.e. day in the other half) refers to the time between sunset and sunrise. Something similar happens in earth's noth pole and south.

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