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In the latest episode of Last Week Tonight (June 3rd), John Oliver says a day on Mercury is 1,407 earth hours (58 days and 15 hours), which is correct.

Neil D. Tyson says

The time you gave is the rotation rate of Mercury. That's not the same thing as a day. A day on Mercury would be from sunrise to sunrise. And on Mercury that lasts three times longer than the number (1,407 earth hours) than the number you just gave.

I know it's a comedy show, but isn't rotation rate and a day the same? I don't see how John Oliver is actually wrong. Earth rotates once every 23 hours, 56 minutes and 4.09053 seconds, the approx. time in a day.

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    $\begingroup$ Hint: take those missing four minutes and multiply them by 365, and see what you get. $\endgroup$ – Emilio Pisanty Jun 4 '18 at 5:42
  • $\begingroup$ 1460 is what I get $\endgroup$ – Growler Jun 4 '18 at 5:44
  • $\begingroup$ Moon rotation period is 27.321 days. How often earthrise happens on its far side? $\endgroup$ – OON Jun 4 '18 at 6:55
  • $\begingroup$ ... and 1460 minutes is, in hours... $\endgroup$ – Emilio Pisanty Jun 4 '18 at 7:09
  • $\begingroup$ 24.33 hours. Why multiply by days in a year though? $\endgroup$ – Growler Jun 4 '18 at 13:09
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The rotation period is defined in the inertial reference frame (with respect to the "fixed stars") However in addition to this rotation the planet is orbiting. This makes the day slower than the rotation period. If we neglect the eccentricity and the axial tilt the day length can be written as, \begin{equation} \frac{1}{T_{day}}=\frac{1}{T_{rot}}-\frac{1}{T_{orb}} \end{equation} The extreme case would be $T_{rot}=T_{orb}$ when the planet is always turned with the same side to the parent body. From the pov of the observer on such a planet the star will always be in the same spot in the sky, without any day-night changes. The example is the Moon with Earth being the parent body. We also expect exoplanets in the habitable zone of the red dwarves to be tidally locked to those stars and experiencing the same synchronicity.

As for the Mercury we obtain about 176 Earth day which is indeed about three times larger. It also is very close to the two orbital periods. For some time it was even believed that it is tidally locked to the Sun just as the Moon to the Earth.

This becomes even more interesting when you take into account the rather significant eccentricity of the Mercury orbit. When it's near the perihelion its orbital angular velocity exceeds its own rotation. That means that for some time the Sun in the Mercurian sky goes in the reverse direction. For some placrs this means that the Sun may go back below the horizon shortly after the dawn and then rise back.

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  • $\begingroup$ I think the third fraction should be $\frac{1}{T_{orb}}$. Copy-paste mistake probably. ;) $\endgroup$ – GRB Jun 4 '18 at 15:14
  • $\begingroup$ @GRB Rather writing latex from the phone. Thank god if no autotypo... uh... I mean autocorrect insanity. Thanks! $\endgroup$ – OON Jun 4 '18 at 15:15

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