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 

Kepler's Equation

In my previous description of a program that I wrote to calculate binary star orbits I mentioned Kepler's Equation which can be stated as:-

E - e sin E = M                . . . (1)

(see section 43 on page 84 of Practical Astronomy with your Calculator by Peter Duffett-Smith). E is the eccentric anomaly, M is the mean anomaly and e is the eccentricity of the orbit. From E we can calculate the true anomaly ν. For a pair of binary stars, the mean anomaly is essentially how far round star B has gone round star A assuming that it orbited at a constant rate. If we count the number of years since the epoch of periastron (the closest approach for which data is recorded) and divide this by the orbital period then the remainder is the fraction of an orbit that has been completed since the last periastron. Multiply this by 2π and you get the mean anomaly in radians. The true anomaly is the angle taking into consideration that star B moves in an ellipse about star A, A is at a focus of this ellipse and star B does not move at a constant rate.

As you can see equation (1) is not easy to solve for E if you know M and e and that is why people resort to numerical solutions (see page 85 for Duffett-Smith's book). I wondered if there was an approximate solution which didn't rely on a numerical solution. Firstly, if we write 

Δ = E - M                        . . . (2)

Then equation (1) becomes

Δ = e sin E                     . . . (3)

If you look at page 122 of Duffett-Smith's book you can see plots Δ of versus M for various values of e. Note that the modulus of Δ is always less than e which is less than 1. This comes from equation (3); since -1 ≤ sin E ≤ 1 this implies -e ≤ e sin E ≤ e (e > 0), so from (3) -e ≤ Δ ≤ e. 

This lead me to an idea for an approximate solution to (1). Substituting for E = M + Δ in equation (3) we get 

Δ = e sin (M + Δ) = e (sin M cos Δ + cos M sin Δ)                . . . (4)

If Δ is much less than 1 then sin Δ ≈ Δ and cos Δ ≈ 1 (to first order). Thus (4) becomes

Δ ≈ e (sin M + Δ cos M)

Rearranging we get

Δ ≈ e sin M / (1 - e cos M)            . . . (5)

I have found that this solution does pretty well if e ≤ 0.2 (the maximum difference between this solution and the true estimate of Δ is less than 2.1% over all values of M).

All text and images © Duncan Hale-Sutton 2025

Observation of Mars on the 2nd February 2025

Last Sunday night we had some clear but very cold weather. I had been thinking for a while that I ought to get my Maksutov-Cassegrain out to have a look at Mars as it is so high up in the sky at the moment. Rather than spend an hour trying to set the telescope up properly I just decided to use the hand controller to keep Mars in view. This is what I had been doing with my Venus observations and it is so much quicker to get going. It is just as well because only ten minutes after I had started everything froze up (including the correcting objective lens)!

My main priority was to do a drawing of Mars rather than try to photograph it. I looked out some coloured pencils before I started and chose those which I thought would match the colours I was expecting. It was still quite a difficult thing to do. Fortunately, I worked out that I could perch on some step ladders next to the telescope so I could draw and look down the eyepiece at the same time. Holding a torch, a notebook and various pencils whilst the air temperature was below freezing is no easy task. However, I succeeded and here is the page out of my notebook:-

Mars reached opposition on January 16th and it has now moved away from us decreasing in size from about 14.6 arc seconds to 13.6. Its phase has correspondingly reduced but at 99% this was hardly noticeable. My Orion OMC 140 has a focal length of 2 metres which together with a 9mm eyepiece gives a magnification of about 222x. Even at this magnification Mars appears to be pretty small! I did try and include a x2 Barlow lens but this didn't really improve matters and it seemed that a brighter image at 222x was a better option.

The main features to be seen were the northern polar cap (which was delineated by a dark area close to its southern edge) and a large area of darker colour to the south. Not knowing exactly what I was looking at I have found a useful tool produced by Sky and Telescope - this is their Mars profiler. You can use the tool to set the date and time of your observation and to choose what telescope set-up you have. Using the date and time of 2nd February 2025 at 19:15 UT and Mirror Reversed (I was using a Maksutov-Cassegrain with a star diagonal) you get a projected map of what I was looking at. Unfortunately, it doesn't show it as a sphere so features at the edge of the map will appear foreshortened. 

What I saw does sort of match up with this projected map. The central red circle on the map is the part of Mars that points directly at Earth and this is placed over an apparently featureless red area called Tharsis which is actually a vast volcanic plateau which is home to the largest volcanoes in the solar system. With more resolution it would be possible to see white clouds associated with the tops of these mountains. The darker area to the south of Tharsis is probably Solis Lacus and shows up well on my drawing. To the left and right of this are other dark areas of Mare Erythraeum and Mare Sirenum. There is a bit of a gap between these dark areas and another dark area hinted in my drawing called Niliacus Lacus. It is to the far left of my drawing. The gap is called Chryse Planitia.

For a further explanation of the Mars profiler look at this article in Sky and Telescope. Note that the profiler also gives the position angle (PA) of the north pole and the central-meridian longitude (CM) which for my observation were 345 and 89 degrees respectively. The PA is measured from North through East and explains why the northern polar cap appears to be tilted 15 degrees westwards in my drawing. The CM is a longitude measurement on Mars and shows which longitude is pointing directly at Earth.

 All text and images © Duncan Hale-Sutton 2025