After Angella and Alan and Chris left tonight, I was packing away my equipment when Jupiter became visible – having previous been obscured by cloud – so I took some quick spectra before it disappeared again.
Andy
I thought I could use the above to calculate the speed of rotation of Jupiter at the surface but I was wrong.
Surface speed from my data = 1/4 (Doppler shift in practice needs to be counted 4 times) x 300000km/s (speed of light) x change in wavelength (1200A above)/wavelength (5780A)
= 15570 km/s.
Real speed of rotation: Since Jupiter is a gas planet, it does not rotate as a solid sphere. Jupiter’s equator rotates a bit faster than its polar regions at a speed of 28,273 miles/hour (about 43,000 kilometers/hour). Jupiter’s day varies from 9 hours and 56 minutes around the poles to 9 hours and 50 minutes close to the equator. (From coolcosmos.ipac.caltech.edu/ask/200-Which-planet-spins-the-fastest-)
My own calculations =
Jupiter’s circumference = 439,264 km
Rotation period (length of day in Earth days)
Jupiter’s day = 9.8 Earth hours
So surface speed = 439,264/9.8 = 44,822 km/hour = 12.45 km/s
So I am way out!!
Looking up methodology for calculating surface speed my method is wrong (hence incorrect result) – the real way to do it requires high resolution spectrograph and measure the spectrum at the equator. This spectrum will show a tilt due to doppler shift and from that SINGLE spectrum the speed of rotation at the surface can be calculated from the amount of tilt. See https://www.shelyak.com/planets-rotation/?lang=en for more details.
Identifying spectral lines on Jupiter:
I have attempted to do this using diagram as my source from https://www.researchgate.net/figure/Jupiter-Spectrum-The-white-boxes-represent-the-Fraunhofer-lines-and-colored-boxes_fig2_283695178
Not sure whether I have identified the correct lines!
