I suppose it should be a simple thing, but i can't solve it....
Suppose i'm on the equator of a planet which have a day of X hours.
From there i see a satellite raising each Y hours.
What's the formula to find the satellite orbital period?
 I'm thinking to a simplified, theoretical system, we can think at the satellite orbit as a perfectly circular, equatorial orbit, ecc...
 found that formula, i should put Y<0 for satellites with a period longer than X, right?
Help: Angular velocity, and "rendezvous" formula?

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Re: Help: Angular velocity, and "rendezvous" formula?
My gut feeling says is should look something like P = Y + Y/(XY)
Where Y is positive when the satellite rotates in the same direction as the planet and negative when it's in a retrograde orbit.
Let's say X = 24 and Y =1 then the result for P = 1 + 1/23 which is a little more than one hour, an expected result: the satellite has to 'catch up' with the viewer.
For Y = 1, we get P = 1 + 1/25 = 1  1/25 which is a little less than one hour. Also as expected.
For Y = 24 (e.g. geostationary), we get no P, because of the division by zero, which is to be expected as well.
Does it work when Y > X. Let's try:
if Y = 72, we get:
P = 72 + 72/48
No it doesn't Well. Maybe someone else has an idea.
Where Y is positive when the satellite rotates in the same direction as the planet and negative when it's in a retrograde orbit.
Let's say X = 24 and Y =1 then the result for P = 1 + 1/23 which is a little more than one hour, an expected result: the satellite has to 'catch up' with the viewer.
For Y = 1, we get P = 1 + 1/25 = 1  1/25 which is a little less than one hour. Also as expected.
For Y = 24 (e.g. geostationary), we get no P, because of the division by zero, which is to be expected as well.
Does it work when Y > X. Let's try:
if Y = 72, we get:
P = 72 + 72/48
No it doesn't Well. Maybe someone else has an idea.
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