Kepler's Third Law is for satellites around a central point. For them it asserts a ratio, (Y*Y) / (a*a*a), where Y is the time for orbit and a is the semimajor axis. Being a ratio - if you know the values for one satellite, you can bootstrap this for all the other satellites and you give nary a thought about the value of the ratio itself. And then... there's Venus, currently satelliteless.
Let's start by doing this for Earth, in days and kilometers.
For Earth, we have the Moon. So plug in the sidereal(!) month squared and the moon's semimajor. 746.473159798921 / 5.67997922950732e+16. Earth's Kepler is then 1.314218115307562e-14.
For Venus, multiply that by Venus' weight in Earth masses: 1.071087763975663E-14.
Let's discuss a Venuslike with an arbitrarily-large Sphere Of Influence. Grant it a satellite running at 949928 km semimajor. It should make its circuit in 95.8 days.
Again, this is the satellite's sidereal month. Where the sun hits at any given point depends on its synodic month, just like here. If this is an ecliptic-prograde satellite, that month takes quite a bit longer than that. Specifically, 1/(1/m - 1/y): where y is the Venus year and m is our satellite's month. 1/y is 0.0044513687959047 so a 95.8 day orbit returns to its Venerean spot in 167 days. Where the "month" and year are at "retrograde" (prograde to Venus herself), I'd add these bottom numbers: 67.16 days.
(This far down the System, and with Venus' gravity extending the range it has, we can get a lower month/year ratio than you'd expect over on Earth. Depends how far from the planet you're willing to go. Sphere Of Influence is a... thing. Past that we're at L1/L2, where Lagrange takes over from Kepler, and the month is the year.)
Now, this was all a thought-experiment, for benchmarks. I don't see the magic of this particular semimajor (much less period), for Venus' satellites. (Wait for it . . .) Farms, though, might go for a semimajor like it, timing their pericytheria for the Umbra and maximising their sunlight outbound.
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