**CLB**, I don't believe you understand what I'm saying. Hopefully, this will clarify things to everyone. (Yes, Mercury does rotate in the manner described by

**DrBashir**.)

I'll use a particular example with fixed numbers, then please reread everything I've said to make sure you see that this is what I've been saying.

Suppose you know that the planet's

*sidereal day *is 10 hours. Suppose you are orbiting the planet in the same direction as the planet's rotation, in the plane of the planet's equator, and you arrive back over the same spot every 6 hours.

**That information allows you do deduce your orbital period exactly. **
If you are orbiting to the East, then by assumption the planet rotates in the same direction, but it appears to pass under you moving to the West. Measure angular speeds to the East as positive. Let x be your angular speed. The sidereal day of the planet has an angular speed of 36 degrees per hour. Transforming your orbital angular speed to rest transforms the angular speed of the planet to negative 60 degrees per hour. Therefore,

36 - x = -60

So, your orbital angular speed must be 96 degrees per hour.

**That means your orbital period must be exactly 3 hours and 45 minutes**, which is the exact amount of time it takes to go around 360 degrees at the rate of 96 degrees per hour.

**So, while of course the rate of rotation that you observe has nothing causal to do with your orbit, how fast you observe the planet to be rotating beneath you constrains the parameters that your orbit can have. **
ETA: I see I have a minor typo in what I wrote before:

Are you taking into account **apparent the **rotation of the planet and the length of its day?

should read

Are you taking into account **the apparent **rotation of the planet and the length of its day?