# Rotation rate of Mercury vs. a day on Mercury

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.

• Hint: take those missing four minutes and multiply them by 365, and see what you get. – Emilio Pisanty Jun 4 '18 at 5:42
• 1460 is what I get – Growler Jun 4 '18 at 5:44
• Moon rotation period is 27.321 days. How often earthrise happens on its far side? – OON Jun 4 '18 at 6:55
• ... and 1460 minutes is, in hours... – Emilio Pisanty Jun 4 '18 at 7:09
• 24.33 hours. Why multiply by days in a year though? – Growler Jun 4 '18 at 13:09

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, $$\frac{1}{T_{day}}=\frac{1}{T_{rot}}-\frac{1}{T_{orb}}$$ 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.
• I think the third fraction should be $\frac{1}{T_{orb}}$. Copy-paste mistake probably. ;) – GRB Jun 4 '18 at 15:14