Recent variable star observations

Just over two weeks ago on the evening of the 30th March 2025 I made a couple of observations of two pulsating variable stars. The moon was one day after new and astronomical twilight ended at about 20:20 UT. The sky transparency was average and nothing to write home about.

Using 10x50 binoculars the first star I looked at was Z Ursae Majoris. This was obviously going through one of its minima in brightness as it was pretty faint:-

Z Uma, 21:13 UT, chart 217.02, =H, mag.8.7

I think, to be honest, I was struggling to estimate the brightness accurately because the star was getting near the limit of my binoculars and so my estimate is 0.6 magnitudes brighter than two other BAA visual observers on this night (their estimates were 9.2 and 9.4 averaging to 9.3; also there was one DSLR estimate of 9.1). However, when I look at the AAVSO data, they have two visual estimates on this night of 8.4 and 8.8 averaging to 8.6 in rough agreement with what I saw so I don't feel too bad.

Using the same chart I looked at RY Ursae Majoris:-

RY Uma, 21:25 UT, chart 217.02, =2, mag. 7.4

This observation seems more in keeping with other BAA observers. 

All text and images © Duncan Hale-Sutton 2025

Partial Solar Eclipse on the 29th March 2025

The good weather over the last few weeks extended to Saturday before last when we had a partial solar eclipse viewable in the UK. From Norwich, the eclipse was to begin at 10:10 UT and maximum coverage would be seen at 11:06 when the magnitude of the eclipse was predicted to be 0.413. The magnitude is a measure of the depth of the eclipse and is the fraction of the Sun's diameter that is occulted by the Moon. The eclipse was then predicted to end at 12.03 UT.

I was able to set up my Celestron NexStar 102 SLT on my HEQ-5 mount well in advance of the start of the eclipse. I intended to project the image from the telescope onto white paper using a 25mm eyepiece and a star diagonal. I used my Samsung Galaxy A21s phone camera to take images and the first image I took was at 10:16:52 UT:-

You can just start to see that the moon is creeping over the western side of the sun. Fortuitously the solar image is just about the right way up with north uppermost and west to the right (this resulted from using the combination of the star diagonal and the projecting the image onto paper). The prominent sunspot group on the left-hand side is AR4046 and to get an idea of what the sunspot activity looked like on this day here is the SDO image (taken at 19:54 UT, courtesy of NASA/SDO):-

 

After deleting poor images and duplicates, I had a total of 18 images from that morning but only until just after maximum eclipse. Rather than show all of these I have picked out six that are approximately 10 minutes apart and show the progression of the moon across the sun. The next image was taken at 10:25:36 UT:-

The chunk taken out of the Sun has increased noticeably but still appears to be increasing from the west. The next image was taken at 10:35:16:-

The same applies here. The moon is slowly encroaching over the sun. The fourth image was taken at 10:47:41:-

Now it appears that not only has the area of the sun that is occulted increased but also the moon appears to be moving northwards. The 5th image was taken at 10:54:42:-

The obscuration is approaching maximum and it is more noticeable that the moon has drifted northwards. The 6th and final image was taken at 11:06:25 and this is within 19 seconds of the eclipse maximum:-

From this image we can measure the percentage area of obscuration. Looking back at my previous discussion of this, I have measured the chord AB to be 994 pixels and the diameter to be 1200 pixels. This gives that the angle w is 1.952 radians (or about 112 degrees) and the observed obscuration is 33% (assuming that the sun and the moon have the same apparent diameter). From the predicted magnitude of 0.413 we can find the predicted obscuration and this is 30%. So pretty much in agreement given the potential errors. 

All text and images © Duncan Hale-Sutton 2025 (except the SDO image).

Final evening observations of Venus

The 2024-2025 eastern elongation of Venus is now over as the planet reached inferior conjunction on the 23rd March and is now appearing in the morning sky. Before this happened I was able to make a few more observations of the planet as its phase shrank to a thin crescent. Firstly, I saw the planet on the 5th March 2025 in clear skies but relatively poor seeing. Venus was at an altitude of about 24 degrees in the West just after sunset. Here is the page from my notebook:-

I made the drawing using my 140mm Orion Maksutov-Cassegrain between 17:35 and 17:52 UT. The phase has shrunk considerably since my last drawing on the 25th January. From the drawing I estimated that the phase was 13% by measurement. This was slightly larger than the predicted phase of 11.2% predicted by the BAA. Notice also how the size of the planet has grown considerably from 29.6 arc seconds on the 25/1/25 to 51.5 arc seconds in this observation. 

On the 13th March at 18:16 UT I captured an image of Venus using my Tasco 18-36x spotting scope:-

 

This terrestial telescope has an aperture of about 50mm. This picture was taken at 36x and I used my Samsung phone to capture it. At this time the phase of Venus was about 5% and the planet was at an elongation of 19 degrees.

The final drawing I made of Venus was on the 17th March and by then the phase appeared to have shrunk to just 1%:-

Here I used my NexStar 102 SLT set up at the front of the house which is west facing. Venus was by this date not very far above the horizon after sunset (a mere 5 degrees in the west). The low altitude caused the seeing to be very poor but it made for a very beautiful sight. The crescent of the lit side was so thin that the atmosphere caused the image to flicker in different colours, like diamonds on a necklace. What's more, after careful observation, it was possible to see that the cusps of the crescent had extended further round the planet than would be expected for solid body with no atmosphere and I have tried to show this in my drawing. The cusps on such a body would be expected to end diametrically opposite each other where the terminator meets the edge of the planet. Instead the light carries on past this point causing appearance of the cusps to curve back on each other. I was very pleased to see that Paul Abel at the BAA produced a very similar drawing a day later.

By comparing what I could see at 73x with 5 phase diagrams ranging in phase from 1 to 5% in steps of 1% I estimated the phase to be 1% and this is smaller than the 3% predicted by the BAA. 

Finally, on the 18th March at 18:12 I captured my last image of Venus using my spotting scope at 36x again:-

 

 All text and images © Duncan Hale-Sutton 2025

Conjunction of the Moon, Venus and Mercury (and some geese!)

On the 2nd March, in the period of continued settled weather, we had a nice conjuction of the Moon, Venus and Mercury:-

 

This picture was taken at 18:14 UT about 20 minutes after sunset. The Moon is just 2 days old and a thin crescent. Venus is the bright object to the right of the Moon. The planet Mercury, which never strays very far from the Sun, can just about be made out to the lower right and just above some geese that happened to fly into my view! I used my Sony DSC-RX100 and an exposure of 1/40s, f/3.5 at ISO 1600. The equivalent SLR focal length was 58mm.

On this date Mercury was at an elongation of 16 degrees east and was magnitude -1.0 (compared to -4.8 for Venus). It reached greatest elongation on March 8th.

All text and images © Duncan Hale-Sutton 2025

Observation of Mars on the 1st March 2025

For the last week and a half we have had some very settled mild weather dominated by high pressure to the south of us. This has enabled me to leave my HEQ-5 telescope mount outside covered up with a plastic bag and to use it whenever I was able. Ten days ago on the 1st March I had another go at drawing Mars. This planet is now retreating from us and its diameter has shrunk from 13.6 arc seconds on the 2nd of February to 10.9 arc seconds on the 1st of March. As a consequence it is getting harder to make out the detail. Here is the page out of my notebook:-

At about eight in the evening Mars was quite high above the horizon and roughly due south. There was quite a bit of thin cloud/haze about but the seeing was pretty good and the image steady. I was using my Orion OMC140 Maksutov Cassegrain again with a 9mm orthoscopic eyepiece that gives a magnification of about 222x. I began with a pre-printed template and drew in the details using coloured pencils. The above image has washed out the original colours, so I have just reproduced the drawing below where I have adjusted the saturation and colour temperature:-

 

It seemed to me that the northern polar cap has either shrunk in size since my last observation or the cap has tilted away from us. To my eyes the southern cap area looked less dark than before and I actually thought frozen white parts were visible but this may not be the case. At this time the central-meridian longitude that was pointing towards us was 220 degrees. According to the Sky & Telescope Mars Profiler the dark area to the south is likely to be Mare Cimmerium. 

I have been in touch with the BAA Mars Section director Richard McKim to see if my drawings are of any use to the BAA but it seems I need to be able to resolve more detail before any science can be made from them. To this end I should really be using my x2 barlow lens as well to get the magnification up to 444x but I think I had been reluctant to do this because the image appears so much dimmer. The other alternative is to wait until another opposition in 2 years time when Mars will once again be a bit bigger (and brighter). Still, I have made a start! He did like the colours I was using which he said did match what he expected. I may try and do a disc drawing of Jupiter as further practice.

 All text and images © Duncan Hale-Sutton 2025

Real use of approximate solution to Kepler's Equation

I wanted to try out the 2nd order approximate solution to Kepler's Equation in my binary star program to see how well it compared to the numerical solution. To this end I have replaced the code in the program from the line 

X = M + Zsin M/(1 - Zcos M)

to the line

Lbl 2

(inclusive) with the following code

J = Z sin M

K = 1 - Z cos M

X = M + (-K + √(K² + 2J²))/J

Previously I had a prediction for the binary star Xi Bootis using data from my Norton's 2000.0 Star Atlas. I now want to use the same data to compare the results I had before with the results from the modified program. I get the following:-

 

Numerical                    2nd order solution

Year : Sep. : PA            Sep. : PA

2024.0 : 4.85 : 290.4    4.84 : 290.3

2034.0 : 3.90 : 270.5    3.91 : 270.7

2044.0 : 3.08 : 238.7    3.09 : 239.3

2054.0 : 2.63 : 191.3    2.63 : 191.6

2064.0 : 2.16 : 125.5    2.16 : 125.5

2074.0 : 2.76 : 53.5      2.78 : 52.9

2084.0 : 4.23 : 21.3      4.24 : 21.0

2094.0 : 5.47 : 5.2        5.47 : 5.1

2104.0 : 6.36 : 354.4    6.36 : 354.5

2114.0 : 6.92 : 346.0    6.92 : 346.0

2124.0 : 7.19 : 338.5    7.19 : 338.5

2134.0 : 7.20 : 331.2    7.20 : 331.2

2144.0 : 6.96 : 323.8    6.96 : 323.8

2154.0 : 6.50 : 315.5    6.49 : 315.5

2164.0 : 5.83 : 305.6    5.82 : 305.5

2174.0 : 4.98 : 292.7    4.98 : 292.7

I think the approximate solution does pretty well considering that the eccentricity of these stars orbit is 0.512 which is probably at the limit of what this solution can cater for. The largest error in separation is 0.02 arc seconds but mostly they are within 0.01. The largest error in PA is 0.6 degrees and this occurs in 2044 and 2074. 

If I had the situation where it wasn't practicable to seek a numerical solution then I think this approximate one would do very well.

 

All text and images © Duncan Hale-Sutton 2025

A further refinement to an approximate solution to Kepler's Equation

In my previous post I described the importance of Kepler's Equation in orbital dynamics and came up with an approximate solution which covers the cases where the eccentricity of the orbit is less than or approximately equal to 0.2. I wondered if we could do slightly better than this and come up with a solution for larger values of the eccentricity.

If we go back to equation (4) of that last post and consider a second order approximation to sin Δ and cos Δ of the form sin Δ ≈ Δ and cos Δ ≈ 1 - Δ²/2 then equation (4) becomes

Δ ≈  e (sin M (1 - Δ²/2) + Δ cos M )

Rearranging this we obtain a quadratic in Δ

½ (e sin M) Δ² + (1 - e cos M) Δ - e sin M = 0                . . . (6)

(I have assumed equality for now). If we write this in the form of

a Δ² + b Δ + c = 0

with a = ½ (e sin M), b = (1 - e cos M) and c =  - e sin M then the two solutions are 

Δ = (- b ± √(b² - 4ac))/2a

that is

Δ = (- (1 - e cos M) ± √((1 - e cos M)² + 2e² sin² M))/e sin M                . . . (7)

Note that (1 - e cos M)² + 2e² sin² M is the sum of two squares and so is always greater than or equal to zero. This means that the solutions in (7) are real. Unfortunately, the solutions are not defined when M = 0, ±π, ±2π as then sin M = 0 and they may become unstable as M tends to any of these values.

A bit of further analysis shows that only one of these two solutions is viable and that is

Δ ≈ (- (1 - e cos M) + √((1 - e cos M)² + 2e² sin² M))/e sin M                . . . (8)

I have compared this solution to the true values of Δ over all values of M and for e = 0.2 the maximum deviation is less than 0.1%. For e = 0.4 the deviation is less than 0.8% and e = 0.6 the deviation is less than 3.1%. This shows that this approximation can probably be successfully used where e ≤ 0.5.

All text and images © Duncan Hale-Sutton 2025