Solar Activity, 26-31 July 2020
In late July 2020, the solar disk displayed simultaneously two active regions and a gigantic solar prominence.
All images below were taken with a Lunt LS60THa/B1200CPT solar telescope on a SkyWatcher EQ6R-Pro mount with ZWO ASI174mm camera and SharpCap Pro for capturing images and application of flats. I used a Tele Vue 5x Powermate for close-ups. I stacked the images using AutoStakkert 2, inverted them using the free software ImPPG, and processed ABE, curves, contrast and sharpening and applied colour using PixInsight. Below, the image of AR2767 was created from the best 75% of 200 frames; the other images from the best 20% of 1000 frames. Images captured in monochrome, with false colour applied to some during processing.
The below image, taken at 07:53:33 UT on 31 July, shows the full solar disk. Active region AR2768, which recently appeared on the north-western limb, is to the left above middle, a huge prominence is visible on the south-western limb (bottom left) and AR2767 is soon to disappear on the eastern limb (middle right).
The below close-up, captured five days earlier, at 07:51:31 UT on 26 July, shows active region AR2767 in detail.
A close-up of AR2768 is below:
A close-up of the prominence is below:
I had no idea of the size of the prominence, so I asked on an astronomy forum for a way of estimating it. Whilst waiting for a reply, I uploaded the image to Affinity Photo and brought up the ruler tool. I measured the full width of the solar disc as 28.12 cm, representing the diameter of the Sun, 1.3927 million km. Next, I assumed that the prominence was a straight line which represented the hypotenuse of a right-angle triangle. Applying the ruler tool to the tip and root of the prominence I obtained two further measures of 2.57 cm and 1.69 cm. Then using Pythagoras' theorem I calculated a length 3.07 cm for the prominence, representing a true length of 152,000 km.
To put my estimate into perspective, the Earth has an equatorial diameter of 12,756 km, only one-twelfth of the length of the prominence. Subsequently, a member of the forum provided an overlay (see below) which provides a means of estimating the length of a prominence; it produces a figure in agreement with my calculation.
John Hughes