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Length of Day
Posted: 29.07.2004, 18:47
by ajtribick
How do you go about calculating the length of day on a planet which rotates once every D days and takes Y days to orbit its sun, given that the rotation of the planet is a) directed opposite to the orbital motion or b) directed with the orbital motion?
Posted: 29.07.2004, 19:39
by maxim
Out of my head:
There's an offset of one day for one complete orbit, so
O = 1/(Y/D) = D/Y is the per local day offset, so a day length is
D+O = D+D/Y for case a) and
D-O = D-D/Y for case b).
Wonder if that could be correct
maxim
Posted: 29.07.2004, 19:43
by granthutchison
Calculate the two angular velocities (degrees/sec) associated with the sidereal rotation period D and the sidereal orbital period Y. Subtract the orbital angular velocity from the rotational angular velocity for direct motion, add the two values together for retrograde motion. Convert back to a rotation period.
Example:
Earth's sidereal rotation period = 86164.098904s => 0.004178074 deg/s
Earth sidereal orbital period = 31558149.5s => 0.000011407 deg/s
0.004178074 deg/s - 0.000011407 deg/s = 0.004166666 deg/s
0.004166666 deg/s => solar day of 86400s (+ rounding errors in 3rd decimal place)
86400s = 24 hours (Q.E.D.)
Grant
Posted: 30.07.2004, 15:21
by cpotting
The formula that I derived some time ago is:
P = 1 / ( 1/a - 1/b)
where
P = length of the solar day
a = siderial period (star-rise to star-rise)
b = revolutionary period (length of a year)
I have found that this formula also works for sydonic periods where
P = sydonic period
a = inferior planet's revolutionary period
b = superior planet's revolutionary period
I developed it some time ago while working out some stuff for a game I was playing (GDW's Traveller), but I never really had a practical use for it until now.
Grant, or anyone out there, I'd be interested in knowing if you think this works or if there are any holes in it (other than orbital eccentricities, etc.)
Posted: 30.07.2004, 15:25
by cpotting
cpotting wrote:The formula that I derived some time ago is:
P = 1 / ( 1/a - 1/b)
Hmmm. I just took a closer look at Grant's reply, and it would appear that his method and my formula are the same. And with a little fiddling, can also be expressed as
P = (b*a)/(b-a)
which is even easier and less prone to rounding errors by calculators and spreadsheets.