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# Calculating orbital period of Ganymede

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### #1 juggle5

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Posted 14 June 2018 - 03:43 AM

This spring, I've sketched Jupiter and its moons most nights I've been observing.  I thought it would be an interesting exercise to figure out what the orbital period of one of the moons based on my logs.

I picked Ganymede since it usually seemed to be farther out than the other moons, so would have a longer period.  Other moons would be possible, but it would be better to observe on consecutive nights to estimate shorter periods more easily.  I don't think I had two nights in a row of clear skies.

I made the following observations.  All were made in Bellevue, WA, with an Omni XLT 150 reflector.

I cheated a little and used Stellarium to figure out which moon was which.  From here, I have the following data:

Days since 4/22 |  Location of G | "Side"

--------------------------------------------
0                        |   G west of J    | 1
32                      |   G east of J     |  -1
35                      |   G near J         |  0
38                      |   G west of J    |  1
40                      |   G east of J     | -1
45                      |   G west of J    | 1
50                      |   G west of J    | 1

I converted east/west location to just a -1, 0, 1 scale ("Side") since I didn't have much confidence in how consistent the sizes of things were in my sketches.  (Since planets are so small, I find it's easy to draw them bigger than they actually appear.)

On 5/27 (day 35), since Ganymede was quite close to Jupiter, I approximated this as halfway through its orbit, so its motion could be described by the equation

x(t) = sin(2 pi (t - 35) / P)

where P is the period in days.

From here, I experimented graphing the function above for various values of P.  I found that 7.0 < P < 7.4; values outside this range meant that the predicted location of Ganymede was inconsistent with my observations (that is, it would have been on the wrong side).  Taking the average, I estimated P = 7.2 days (plus or minus 0.2 days).  Here's the estimated location using P = 7.2

Checking on Wikipedia, the actual value is 7.155 days, surprisingly close!

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### #2 roelb

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Posted 14 June 2018 - 03:46 PM

nice job

### #3 juggle5

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Posted 15 June 2018 - 12:35 AM

Thanks.  It was a fun thing to try while the clouds were out.

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