# How to calculate day length of a tidelocked double planet?

• I
• AotrsCommander
In summary, the conversation discusses the calculation required for determining the day length or rotational period of a double planet where both planets are tidally locked to each other. The speaker mentions using Kepler's Third equation, but is unsure if adjustments need to be made for double planets. They also ask for confirmation on whether the rotation period is equal to the orbital period for tidally locked planets. The expert confirms that this is correct and also mentions that if the planets were not tidally locked, a different equation would be used.

#### AotrsCommander

Quick question: what's the calculation required for working out the day length (i.e. rotational period) of a double planet, where both planets are tidelocked to each other? Do just need to make adjustments to Kepler's Third?

I've got a spread sheet set up to allow me to calculate orbital periods via Kepler's Third, but I'm not sure what adjustments (if any) I need to make to the equation for double planets orbiting each other or whether this is even the right formula to use in this instance (i.e for rotational period of the two bodies.)

Currently set up for calcualtions as follows for orbital periods:

##{T^2}=4π\frac{r^3}{G(M1+M2)}##

(Assuming circular orbits, because that's a good enough abstraction for what I'm doing.)

I am half-thinking that I don't need to do anything and that time for the rotatation period is equal to the orbital period, but I would appreciate a confirmation.

(Secondary question, more hypothetically, if they weren't tide-locked, what equation would you use?)

AotrsCommander said:
I am half-thinking that I don't need to do anything and that time for the rotatation period is equal to the orbital period, but I would appreciate a confirmation.
Of course. If the planets are tidally locked, their rotation periods are equal to their revolution period.

Right, thanks. (It's been a few months since I last looked at the this, so I wasn't entirely sure.)