Jump to content

  •  

CNers have asked about a donation box for Cloudy Nights over the years, so here you go. Donation is not required by any means, so please enjoy your stay.

Photo

Calculating orbital period of Ganymede

  • Please log in to reply
2 replies to this topic

#1 juggle5

juggle5

    Lift Off

  • -----
  • topic starter
  • Posts: 8
  • Joined: 17 Sep 2017
  • Loc: Seattle WA

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. 

 

jupitermontage.jpg

 

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

 

ganymede.png

 

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


  • roelb likes this

#2 roelb

roelb

    Viking 1

  • -----
  • Posts: 862
  • Joined: 21 Dec 2013
  • Loc: Belgium, Antwerp

Posted 14 June 2018 - 03:46 PM

nice jobwaytogo.gif



#3 juggle5

juggle5

    Lift Off

  • -----
  • topic starter
  • Posts: 8
  • Joined: 17 Sep 2017
  • Loc: Seattle WA

Posted 15 June 2018 - 12:35 AM

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




CNers have asked about a donation box for Cloudy Nights over the years, so here you go. Donation is not required by any means, so please enjoy your stay.


Recent Topics






Cloudy Nights LLC
Cloudy Nights Sponsor: Astronomics