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Imagine billions of years have been passed and the Sun has been so big that the Earth is completely swallowed by it. Now the habitable zone reaches Europa, one of the moons of Jupiter.

Let's assume Jupiter completes its orbit around the Sun just like today (e.g. every 4333 Earth days) and Europa also completes its orbit around Jupiter every 3.5 Earth days (just like today).

The orbit of Europa around Jupiter is too short to consider it a year for people living there. So we consider the orbit of Jupiter around the Sun a year.

When Jupiter completes its orbit, Europa has orbited around Jupiter 1238 times. Since the moon is tidally locked to the parent planet, Europa also has rotated 1238 times on its own axis; thus each year has 1238 days for people living on Europa.

Let's assume Europa will have four seasons just like the Earth. Also, let's assume that the frequent total solar eclipses by Jupiter doesn't affect the lives of people living on the moon.

Which will be the best choice for the length of months in Europa so that all 1238 days be in 12 months?

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  • $\begingroup$ Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. $\endgroup$
    – Community Bot
    Mar 19 at 12:20
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    $\begingroup$ With a .1 degree axial tilt, hard to imagine four seasons. $\endgroup$
    – Allan
    Mar 19 at 12:23
  • $\begingroup$ @Community The question is limited to a specific problem. Other parts are just facts and assumptions. $\endgroup$ Mar 19 at 12:23
  • $\begingroup$ @Allan It's an imaginary world. Don't forget the verse "Imagination, life is your creation!" $\endgroup$ Mar 19 at 12:35
  • $\begingroup$ Humans living on Europa might well adopt a ‘month’ of eight or nine local days to preserve a familiar rhythm. $\endgroup$ Mar 20 at 2:22

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This question is marked with an Astronomy tag. Now suppose you insist on having 12 time periods in a year called "months", the answer of Daron will do fine. But that number of 12 is actually Earth-moon system bound, not Jupiter-moon system bound.

Europa's calendar will differ from Earth's calendar

Earth's calendar has 3 independent parameters, that is:

Earth's day (1) is a result of its rotation

Earth's month (2) represents moon orbit, which is completed about 12 times per year.

Earth's year (3) represents orbit around the sun counted in days, on Earth 365.24

Drop the "day"

On Europa, which is tidally locked, the month will be equal to Europa's day, that is 3.5 Earth days, so you'll have only 2 independent parameters. Days are not needed, or you can call your months days.

Assuming the Europa year is the same as a Jupiter year, connected to the sun: the orbit around the sun will result in a year of 4333 earth days, so you'll have 4333/3.5 that is you'll have exactly 1238 Europa months (2) in an Europa year, instead of 12.

Adding a "week" to the calendar

The Earth's week is for convenience, not astronomy-bound. It is about a quarter moon. For your moon Europa, suppose you would introduce a 20 day "week", you'd get 61.9 weeks in a year. Counting e.g. 62 weeks in a year, your calendar would require a leap week of 19 days instead of 20 days, every 10 years.

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  • $\begingroup$ So short months! $\endgroup$ Mar 19 at 14:09
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    $\begingroup$ Absolutely... Only 3.5 Earth days, astronomically, because of the tidal lock of Europa. You could introduce a week, but to make it handy, that week would be longer than the month. I've finished my answer, thanks for the green flag ! $\endgroup$
    – Goodies
    Mar 19 at 14:22
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    $\begingroup$ You're welcome :) btw, Do you mean 20 European days (e.g. months)? Then a year would have 61.9 weeks and not 69.9. $\endgroup$ Mar 19 at 14:47
  • $\begingroup$ Yes, a Europa week would be 20 Europa months (=days). You're right about 61.9, I'll adjust :) $\endgroup$
    – Goodies
    Mar 19 at 15:18
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1238/12 = 103 remainder two.

That means there should be 10 months of 103 days and two months of 104 days.

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  • $\begingroup$ If it is 100 days per month then a common year will have 1200 days. Where are the remaining 38 days? $\endgroup$ Mar 19 at 13:43
  • $\begingroup$ I divided wrong -- whoops! It should say 103 days. $\endgroup$
    – Daron
    Mar 19 at 13:44
  • $\begingroup$ It's a good choice, but there will be 2 remaining days again. So we can have two 104 days long months and the rest are 103 days long months. But there is another problem again. In our calendar, there is just one month with the least number of days (e.g. February). Shouldn't the same rule work here? I mean arranging month lengths so that there is only one month with the least number of days. $\endgroup$ Mar 19 at 13:50
  • $\begingroup$ I made another mistake. The two leftover days do not make a leap year. They just get added to make some months longer. The leap year only comes up if the year is a non integer number of days long. $\endgroup$
    – Daron
    Mar 19 at 13:54
  • $\begingroup$ Leap years would happen for example if the year was 1036.125 days long. Then you get 12 months of 103 days, which makes 1036, and for the extra part every 8 years you add an extra day to some month. $\endgroup$
    – Daron
    Mar 19 at 13:56

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