We had a minor geomagnetic storm on Monday just after the major storm on Saturday that everyone saw. This minor storm launched a CME caused by an X-5.8 solar flare on Friday, but despite early estimates it might rival the major storm, it was a glancing blow to Earth’s magnetic field and caused no aurora over much of the Lower-48 States. Many had hoped they would get to see an aurora in Maryland and other mid-latitude locations but the storm was too week to be seen in most states that had enjoyed the Great Storm of May 10-11.

Nevertheless, my DIY magnetometers did show some life for this Kp=6 event as shown below. This time I had three different magnetometers operating. The top numbers are the 3-hour Kp indices. The red trace is from the Fredricksberg Magnetic Observatory. The black trace is from the RM3100-Arduino system. The blue trace is from the Differential Hall Sensoe system. The green trace os from the Differential Photocell Magnetometer. The two dips marked with ‘Sq’ are the diurnal Sq variations, which were recorded by all magnetometers.

We just passed through the biggest ‘solar storm’ in the last 20 years caused by the massive naked-eye sunspot group called AR-3664. Its size was 15 times the diameter of Earth and rivaled the size of the famous Carrington sunspot of September, 1859. Since it first appeared on May 2, it remained inactive until May 9 when it released an X2.2-class solar flare at 10:10 UT.

This enormous and violent release of energy stimulated the launch of six coronal mass ejections of which three merged to become an intense ‘cannibal CME’ that arrived near Earth on May 10 at 16:45 UT. Its south-directed magnetic field was perfect for imparting the maximum amount of energy to our planet’s magnetosphere. For a transit time of about 24-hours, it was traveling at a speed of about 1,700 km/s when it arrived. It sparked a G5-level extreme geomagnetic disturbance with a Kp index of 9 between May 10, 21:00 UT and May 11, 03:00 UT. The image below was taken by Tom Wasiela on May 10, 2024 from Roanoke, Virginia and is courtesy of the SpaceWeather Gallery. Reports suggest that aurora were seen as far south as Florida and Puerto Rico.

On May 9th at 06:54 UT AR-3664 produced an X-3.9 flare. This was followed on May 11 with a fourth major X-5.8 flare at 1:39 UT, which caused an immediate shortwave radio blackout across the entire Pacific Ocean that lasted for several hours. It is expected that the May 11 flare sparked anoher CME that may arrive near Earth on Monday evening May 13.

The last G5 geomagnetic storm that we experienced was way back in October 28 to November 5, 2003. These Halloween Storms caused power outages in Sweden and damaged transformers in South Africa. Despite many recent cautionary comments in the news media about cellphone and satellite outages and power grid problems, as yet none of these have been identified but perhaps in the next few weeks these technological impacts may start to be mentioned as anecdotes begin to surface.

Unfortunately, many areas on the East Coast were covered by clouds during this three-day period and missed the opportunity to see these major aurorae. However, my DIY magnetometer (see my earlier blog on how to build your own $50 magnetometer (located in Kensington, Maryland (latitude 39^{o} N) was able to keep up with the invisible changes going on, and produced a very respectable record of this entire storm period. As a scientist, I am often working with things I cannot directly see with my eyes, so the fact that I had my trusty magnetometer to reveal these invisible changes around me was pretty cool!

This graph shows a side-by-side comparison of the data recorded by my RM3100 magnetometer (black) and the magnetometer at the Fredericksberg Magnetic Observatory (red). I have shifted and rescaled the plots so you can more easily see how similar they are. This is very satisfying because it shows that even a simple home-made magnetometer can perform very well in keeping up with the minute changes in the geomagnetic field. This plot shows the variation in the so-called D component, which is the local magnetic declination angle. Mathematically is is defined by D = arctan(Bx/By). It’s the angle relative to geographic North that your local compass points.

Below is a slightly different graph of the RM3100 data. As you can see in the first part of the above plot between 36 and 63 UT hours, the smooth change is caused by the diurnal Sq current effect that is correlated with the solar elevation angle. During this storm period, it is assumed to have behaved smoothly during the actual storm, so in the graph below I have subtracted it from the magnetometer data. The result is that I have now isolated the changes due to the storm itself. The top row of numbers are the 3-hour Kp index averages from NOAA. The marked times are for EDT in Maryland. Universal Time is 4 hours ahead of EDT.

This was, indeed, a very powerful storm that lasted about 42 hours. This places it among a handful of exceptional geomagnetic storms that includes the great Carrington Storm of August 28 to Septemer 5, 1859.

Why is this important? Well, in the grand scheme of things it may not matter much, but as an astronomer it is still a lot of fun to have access to the invisible universe from the comfort of my suburban home. I will let geophysicists have all the fun deciphering all the bumps and wiggles and what they tell us about our magnetic field and solar storms!

Meanwhile, my gear is primed and ready to go to detect this Monday’s next storm. Some predict that it may be even bigger then the one we just experienced. It’s interesting how the Carrington Storm was actually two major storms separated by a few days, with the CME from the first storm also canibalizing several other CMEs that were also enroute.

Like the introduction of hand calculators into the classroom in the 1970s, ChatGPT offers enormous promise but currently suffers from a variety of negative expectations. Some of the arguments against students using calculators in the 1970s classroom are being used today against ChatGPT. I think there are some applications that work very well if you consider ChatGPT to be an intelligent ‘hand calculator’ in the math and physical science classroom. Here is an example I came up with without much effort!

I. Student Pre-Requisite Knowledge

The following example requires students to work with Scientific Notation, calculating the volume of spheres and shells, and working with the mass:density:volume relationship.

II. Exoplanet Interior Modeling

Astronomers have discovered over 5000 exoplanets orbiting other stars. We call these ‘exoplanets’ so that they don’t get confused with the ‘planets’ in our solar system. From a careful study of these exoplanets, astronomers can figure out how long they take to orbit their stars, their distance from the star, their diameters and their masses. How do they use this information to figure out what the insides of these exoplanets look like? This activity will show how a simple knowledge of mass, volume and density provides the clues!

III. Mass, Density and Volume

Mass, volume and density are related to each other. If two things occupy the same volume but have different masses, the less-massive one will have the lower density.

Density = Mass / Volume.

Example 1: A Prospector had his sample weighed to be 20 grams, and its volume calculated by water displacement and found to be 4 cubic centimeters. If pure gold has a density of 19.3 gm/cc, is his sample actually gold or is it iron pyrite (density 5.0 gm/cc)?

Answer: Density = 20 gm/4 cc = 5 gm/cc so it’s iron pyrite or ‘Fools Gold’.

Example 2: A basic principle of physics is that light things of low density float on top of denser things. Why do you have to shake a bottle of salad dressing before you use it?

ChatGPT Query: There are five different liquids mixed together in a bottle. After 10 minutes they sort themselves out. The liquids are: Olive oil ( 0.92 g/cc ), water ( 1.0 g/cc), molassis ( 1.4 g/cc) , vinegar ( 1.0006 g/cc), honey (1.43 g/cc). From bottom to top, how will the liquids separate themselves?

IV. Designing Mercury with a One-Component Interior Model

Mercury was formed close to the sun where only iron and nickel-rich compounds could condense into a planet. Let’s model Mercury and see what we discover. The actual mass of Mercury is 0.055 times Earth.

Step 1 – Use the formula for the volume of a sphere V=4/3 pR^{3} and with a known radius for Mercury of Rm = 2.43×10^{6} meters to get the volume of the planet of V = 6×10^{19} m^{3}.

Step 2 – Calculate the mass of Mercury for various choices of density. Give the predicted mass for Mercury in multiples of Earth’s mass of 5.97×10^{24} kg.

Step 3: Test your knowledge: For a density of 5000 kg/m^{3} and a radius of 2.43×10^{6} meters, what is the mass of Mercury for these selected values? Give your answer to two significant figures.

Mase(Earth units) = 3.0×10^{23 }kg /5.97×10^{24} kg = 0.05 times Earth

Use ChatGPT to generate data for plotting. Enter this question into the window:

ChatGPT Query: A sphere has a radius of 2.43×10^6 meters. What is the mass of the sphere if its density is 5000 kg per cubic meter? Express your answer in units of Earth’s mass of 5.97 x 10^24 kg. Give your answer to two significant figures.

Repeat the ChatGPT query four times to generate a mass estimate for densities of 4000, 4500, 5000, 5500 and 6500 kg/m^{3}. Plot these points on a graph of mass versus density and draw a line through the values. Which density gives the best match to the observed mass of Mercury of 0.055 Mearth? (Answer: about 5500 kg/m3).

V. Designing Mars with Two-Component Models

Now we add two components together for planets that have a high density core and a lower density mantle. These would have formed farther out than the orbit of Mercury but with masses lower than than of Earth. The mathematical model consists of a spherical core with a radius of Rc, surrounded by a spherical shell with an inner radius of Rc and an outer radius of Rp, where Rp is the observed planetary radius. Mathematically the model looks like this:

M = Dc x 4/3p Rc^{3 }+ Dm x 4/3p ( Rp^{3} – Rc^{3})

Draw a diagram of the planet’s interior showing Rc and Rp and confirm that this is the correct formula for the total mass of the planet where Dc is the core density, and Dm is the mantle density.

Test Case: An exoplanet is discovered with a mass of 5.97×10^24 kg and a radius of 6,378 kilometers. If the radius of its core is estimated to be Rc = 3,000 km and its core density is 7000 kg/m3, what is the average density of the mantle material?

Dm = (5.97×10^{24} – 7000 x 1.1×10^{20})/9.8×10^{20} = 5300 kg/m^{3}

Check your answer with ChatGPT using this query. A planet consists of a core region with a radius of Rc and a mantle region extending to the planet’s surface at a radius of Rp. If the planet is a perfect sphere with a radius Rp = 6378 km and Rc = 3000 km, with a total mass of 5.97×10^24 kg, for a core density of 7000 kg/cubic meters, what is the average mantle density? Give the answer to two significant figures.

Now lets use ChatGPT to generate some models and then we can select the best one. We will select a mantle density from three values, 2000, 3000 and 4000 kg/m3. The core density Dc will be fixed at Dc = 9000 kg/m3. We will use the measured radius for Mars of Rp = 3.4×10^{6} meters, and its total mass of Mm = 6.4×10^{23} kg. We then vary the core radius Rc. We will plot three curves on a graph of Rc versus Mm one for each value of the assumed mantle density. Use this ChatGPT query to generate your data points.

ChatGPT Query: A planet is modeled as a sphere with a radius of Rp=3.4×10^6 meters. It consists of a spherical core region with a radius of Rc surrounded by a spherical shell with an inner radius of Rc and an outer radius of Rp. The core of the planet has a density of 9000 kg/cubic meters. The radius of the core Rc = 30% of the planet’s radius. If the density of the mantle is 2000 kg/cubic meter, what is the total mass of the planet in multiples of the mass of Earth, which is 5.97×10^24 kg? Give your answer to two significant figures?

Repeat this query by changing the mantle density and the core radius values and then plot enough points along each density curve to see the trend clearly. An example of an Excel spreadsheet version of this data is shown in this graph:

This graph shows solutions for a two-component mars model where the mantle has three different densities (2000, 3000 and 4000 kg/m3). The average density of mars is 3900 kg/m3. Which core radius and mantle density combinations seem to be a better match for Mar’s total mass of 0.11 Mearth for the given density of the mantle?

VI: Modeling Terrestrial Planets with a three-component interior.

The most general exoplanet model has three zones; a dense core, a mantle and a low-density crust. This is the expected case for Earth-like worlds. Using our Earth as an example, rocky exoplanets have interiors stratified into three layers: Core, mantle, crust.

Core material is typically iron-nickel with a density of 9000 kg/m3

Mantle material is basaltic rock at a density of 4500 kg/m3

Crust is low-density silicate rich material with a density of 3300 kg/m3

The basic idea in modeling a planet interior is that with the three assumed densities, you vary the volume that they occupy inside the exoplanet until you match the actual mass (Mexo) in kilograms and radius (Rexo) in meters of the exoplanet that is observed. The three zones occupy the radii Rc, Rm, Rp

We will adjust the core and mantle radii until we get a good match to the exoplanet observed total mass and radius. Let’s assume that the measured values for the Super-Earth exoplanet mass is Mp = 2.5xEarth = 1.5×10^{25} kg, and its radius is Rp = 1.5xEarth = 9.6×10^{6} meters.

Core Volume Vcore = 4/3p Rc^{3}

Mantle Volume Vm = 4/3 p (Rm^{3} – Rc^{3})

Crust Volume Vcrust = 4/3 p (Rp^{3} – Rm^{3})

So the total Mass = (9000 Vcore + 4500Vm + 3300Vcrust)/Mp

Rc ,Rm and Rp are the core, mantle and planet radii in meters, and the total mass of the model is given in multiples of the exoplanet’s mass Mp.

Let’s do a test case that we work by hand to make sure we understand what we are doing.

Choose Rc = 30% of Rp and Rm = 80% of Rp. What is the predicted total mass of the exoplanet?

Rc = 0.3 x 9.6×10^{6} meters = 2.9×10^{6} meters.

Rm = 0.8x 9.6×10^{6} meters = 7.7×10^{6} meters.

Then

Vcore = 4/3p (2.9×10^{6})^{3} = 1.0×10^{20} m^{3}

Vm = 4/3p ( (7.7×10^{6})^{3} – (2.9×10^{6})^{3}) = 1.8×10^{21} m^{3}

Then Mass = (9000 Vcore + 4500 Vm + 3300Vcrust)/Mp

Mass = (9×1023 kg + 8.1×1024 kg + 5.9×1024 kg)/1.5×1025 kg = 1.0 Mp

Now lets use ChatGPT to generate some models from which we can make a choice.

Enter the following query into ChatGPT to check your answer to the above test problem.

ChatGPT Query: A spherical planet with a radius of Rp consists of three interior zones; a core with a radius of Rcore, a mantle with an inner radius of Rc and an outer radius of Rm, and a crust with an inner radius of Rm and an outer radius of Rp=9.6×10^6 meters. If the density of the core is 9000 kg/m^3, the mantle is 4500 kg/m^3 and the crust is 3300 kg/m^3, What is the total mass of the planet if Rc = 30% of Rp and Rm = 80% of Rp? Give your answer for the planet’s total mass in multiples of the planet’s known mass of 1.5×10^25 kg, and to two significant figures.

Re-run this ChatGPT query but change the values for the mantle radius Rm and core radius Rc each time. Plot your models on a graph of Rc versus the calculated mass Mp on curves for which Rm is constant. An example of this plot is shown in the excel spreadsheet plot below.

For example, along the black curve we are using Rm=0.8. At Rc = 0.5 we have a model where the core extends to 50% of the radius of the exoplanet .The mantle extends to 80% of the radius, and so the crust occupies the last 20% of the radius to the surface. With densities of 9000, 4500 and 3300 kg/m3 respectively, the Y-axis predicts a total mass of about 1.1 times the observed mass of the exoplanet (1.00 in these units). With a bit of fine-tuning we can get to the desired 1.00 of the mass. But what about the solution at (0.3, 1.00) ? In fact, all of the solutions along the horizontal line along y = 1.00 are mathematically valid.

Question 1: The exoplanet is located close to its star where iron and nickel can remain in solid phase but the lower density silicates remain in a gaseous phase. Which of the models favors this location at formation?

Answer: The exoplanet should have a large iron/nickel core and not much of a mantle or crust. This favors solutions on the y=1.00 line to the right of x=0.5.

Question 2: The exoplanet is located far from its star where it is cool enough that silicates can condense out of their gas phase as the exoplanet forms. Which of the models favor this location?

Answer: The exoplanet will have a small iron/nickel core and a large mantle and crust. This favors models to the left of x= 0.5.

So here you have some examples for how ChatGPT can be used as an intelligent calculator once the students understands how to use the equations and is able to explain why they are being used for a given modeling scenario.

I would be delighted to get your responses and suggestions to this approach . Just include your comment in the Linkedin page where I have posted this idea.

Many people, including me, will try to capture some images of the eclipse on Monday, April 8. This blog is aimed at people in the Greater Washington DC area who will experience a partial eclipse. If you travel 6 hours due-west of Washington DC you will be on the Path of Totality and your experiences will be dramatically different.

In the Washington DC area, the eclipse will start at 2:04 PM with the dark lunar disk taking its first little bite out of the solar disk, and end at 4:33 PM as the moon leaves the disk. The maximum partial eclipse will occur at 3:20 PM when the moon will block about 89% of the solar disk. Here’s what that looks like:

You will notice a rapid darkening of the daylight sunshine so that instead of the normal mid-afernoon sun it will look more like early twilight for about 5-10 minutes before the daylight finally starts to return.

Safety:

This is a partial eclipse. Only use approved ‘eclipse glasses’ and not sunglasses. You will not be able to see the fabulous corona unless you are on the path of Totality.

I know it is tempting, but good photography practice is NOT to point your camera at the sun with no filter…including your smartphone. Smartphones have faint light meters for twilight photography and you run the risk of damaging this meter so that you may not be able to take low-light-level photos anymore.

Manage your expectations. You will not see the corona that everyone talks about. With your Eclipse Glasses you will see a sequence of partial stages that look something like this. This was taken by NASA/Noah Moran at the Johnson Space Center during the August 21, 2017 eclipse which was only a partial eclipse over Houston, TX. Also, instead of seeing a super-huge image with the naked eye, you will only see a disk as large as the full moon in the night sky.

Smartphone Photography Tips.

On sunday at 3:30 pm go outside and check that your viewing location will give you a good view of the sun. Put on your Eclipse Glasses and check that your view of the solar disk is unobscurred. The higher the sun is above the clutter at your chosen location the better your experience will be.

On Friday, Saturday or Sunday before the eclipse, place one of the lenses of the Eclipse Glasses over the selfi-camera lens located at the top edge of your smartphone just below the center of the top edge.

Start-up your camera and place it in selfi mode.

With the sun’s disk over your left or right sholder, check that your camera display shows a bright orange disk of the sun and adjust your camera angle so that the disk is centered and unobstructed by your head.

Your camera should automatically be able to focus on the edge of the sun disk and set the camera’s exposure. This photo on a cloudy morning on Friday April 4 without editing was automatically taken by my iPhone 13 Pro camera at 1/15 of a second at an ISO of 1600.

With no clouds, the sun disk should be crisp with a good clean edge. You may need to experiment with the manual focus if your camera allows you to do this.

6. You might want to experiment with manipulating this test image of the filtered sun to get the best clarity and background. With stray clouds this is a challenge as the image below, adjusted with Photoshop, shows. I adjusted the brightness and contrast.

The crispness of the solar disk was compromized by the diffusing of sunlight in the foreground clouds. You will have a better experience if there are no clouds in the way. To get an idea of what this optimal picture would look like in selfie-mode, here is an image of the moon taken by my iPhone 13 Pro in selfie mode. You will see a similar-sized solar disk with the filter covering the selfi camera lens but you will only see the sun disk and not the foreground trees etc. This was the best focus my camera in this mode was able to provide with its smaller lens.

A second mode of solar photography is to take your camera out of ‘selfie’ mode and use the normal forward camera. It has better lenses and resolution than the selfie camera. Here is an example of a photo of the moon taken in this mode. Notice that the lunar disk is much clearer. Your eclipse picture in this mode will have the same clarity but of course the sky and foreground will be completely black through the filter.

The set-up for a higher-resolution direct image requires some preparation. You might want to create a sun shield out of foamboard that covers a 1-foot x 1-foot area.

Cut out a square hole in the center that your camera lens can peak through as shown in the left-hand image.

Cover the camera lens opening with the Eclipse Glasses filter and secure all pieces in place with tape as shown in the middle picture.

When you want to photograph the sun, start up your camera in its normal ‘forward’ mode and place it over the filter opening as shown in the right-hand image. Keep your eye close to the camera display so that your head is shadowed by the shield. If you want, you can secure your smartpone to the foamboard with tape, but be sure that you place a 3×5 index card over the display so that you don’t get glue on it. Otherwise, you can hold the camera to the filter opening manually.

As before, your camera should be able to automatically focus on the eclipsed solar disk to give you the best clarity your particular camera is able to provide.

Good luck…but make sure you take the time to enjoy the eclipse and not worry about getting a perfect photo with your smartphone!!!

One year ago, I posted a fun problem of predicting when we will have the very last total solar eclipse viewable from Earth. It was a fun calculation to do, and the answer seemed to be 700 million years from now, but I have decided to revisit it with an important new feature added: The slow but steady evolution of the sun’s diameter. For educators, you can visit the Desmos module that Luke Henke and I put together for his students.

The apparent lunar diameter during a total solar eclipse depends on whether the moon is at perigee or apogee, or at some intermediate distance from Earth. This is represented by the two red curved lines and the red area in between them. The upper red line is the angular diameter viewed from Earth when the moon is at perigee (closest to Earth) and will have the largest possible diameter. The lower red curve is the moon’s angular diameter at apogee (farthest from Earth) when its apparent diameter will be the smallest possible. As I mentioned in the previous posting, these two curves will slowly drift to smaller values because the Moon is moving away from Earth at about 3cm per year. Using the best current models for lunar orbit evolution, these curves will have the shapes shown in the above graph and can be approxmately modeled by the quadratic equations:

Perigee: Diameter = T^{2} – 27T +2010 arcseconds

Apogee: Diameter = T^{2} -23T +1765 arcseconds.

where T is the time since the present in multiples of 100 million years, so a time 300 million years ago is T=-3, and a time 500 million years in the future is T=+5.

The blue region in the graph shows the change in the diameter of the Sun and is bounded above by its apparent diameter at perihelion (Earth closest to Sun) and below by its farthest distance called aphelion. This is a rather narrow band of possible angular sizes, and the one of interest will depend on where Earth is in its orbit around the Sun AND the fact that the elliptical orbit of Earth is slowly rotating within the plane of its orbit so that at the equinoxes when eclipses can occur, the Sun will vary in distance between its perihelion and aphelion distances over the course of 100,000 years or so. We can’t really predict exactly where the Earth will be between these limits so our prediction will be uncertain by at least 100,000 years. With any luck, however, we can estimate the ‘date’ to within a few million years.

Now in previous calculations it was assumed that the physical diameter of the Sun remained constant and only the Earth-Sun distance affected the angular diameter of the Sun. In fact, our Sun is an evolving star whose physical diameter is slowly increasing due to its evolution ‘off the Main Sequence’. Stellar evolution models can determine how the Sun’s radius changes. The figure below comes from the Yonsei-Yale theoretical models by Kim et al. 2002; (Astrophysical Journal Supplement, v.143, p.499) and Yi et al. 2003 (Astrophysical Journal Supplement, v.144, p.259).

The blue line shows that between 1 billion years ago and today, the solar radius has increased by about 5%. We can approximate this angular diameter change using the two linear equations:

Perihelion: Diameter = 18T + 1973 arcseconds.

Aphelion: Diameter = 17T + 1908 arrcseconds.

where T is the time since the present in multiples of 100 million years, so a time 300 million years ago is T=-3, and a time 500 million years in the future is T=+5. When we plot these four equations we get

There are four intersection points of interest. They can be found by setting the lunar and solar equations equal to each other and using the Quadratic Formula to solve for T in each of the four possible cases.:

Case A : T= 456 million years ago. The angular diameter of the Sun and Moon are 1890 arcseconds. At apogee, this is the smallest angular diameter the Moon can have at the time when the Sun has its largest diameter at perihelion. Before this time, you could have total solar eclipses when the Moon is at apogee. After this time the Moon’s diameter is too small for it to block out the large perihelion Sun disk and from this time forward you only have annular eclipses at apogee.

Case B : T = 330 million years ago and the angular diameters are 1852 arcseconds. At this time, the apogee disk of the Moon when the Sun disk is smallest at aphelion just covers the solar disk. Before this time, you could have total solar eclipses even when the Moon was at apogee and the Sun was between its aphelion and perihelion distance. After this time, the lunar disk at apogee was too small to cover even the small aphelion solar disk and you only get annular eclipses from this time forward.

Case C : T = 86 million years from now and the angular diameters are both 1988 arcseconds. At this time the large disk of the perigee Moon covers the large disk of the perihelion Sun and we get a total solar eclipse. However before this time, the perigee lunar disk is much larger than the Sun and although this allows a total solar eclipese to occur, more and more of the corona is covered by the lunar disk until the brightest portions can no longer be seen. After this time, the lunar disk at perigee is smaller than the solar disk between perihelion and aphelion and we get a mixture of total solar eclipses and annular eclipses.

Case D : T = 246 million years from now and the angular diameters are 1950 arcseconds. The largest lunar disk size at perigee is now as big as the solar disk at aphelion, but after this time, the maximum perigee lunar disk becomes smaller than the solar disk and we only get annular eclipses. This is approximately the last epoc when we can get total solar eclipses regardless of whether the Sun is at aphelion or perihelion, or the Moon is at apogee or perigee. The sun has evolved so that its disk is always too large for the moon to ever cover it again even when the Sun is at its farthest distance from Earth.

The answer to our initial question is that the last total solar eclipse is likely to occur about 246 million years from now when we include the slow increase in the solar diameter due to its evolution as a star.

Once again, if you want to use the Desmos interactive math module to exolore this problem, just visit the Solar Eclipses – The Last Total Eclipse? The graphical answers in Desmos will differ from the four above cases due to rounding errors in the Desmos lab, but the results are in close accord with the above analysis solved using quadratic roots.

During a solar eclipse, the lansdcape will slowly dim until it is nearly complete darkness along the path of totality. other observers wil see te landscape dim a bit but then brighten to normal intensity. If you didn;t know that an eclipse was going on you might not even notice the dimming, mistaking it for a cloud passing across the sun. The geometric condition for this dimming have to do with the area of exposed solar surface and how this changes as the disk of the moon passes across it. Below is a simple mathematical model for ambient light dimming that you can put to the test the next time a solar eclipse passes over your geographic location.

I have reanalyzed the geometry and defined it in terms of the center-to-center distance, L, between the sun and moon, and their respective radii Rs and Rm as the figure of the upper half-plane of the intersection shows, with the yellow area on the left representing the disk of the sun and the white area on the right the disk of the moon. This problem was previously considered in 2000 by British astronomer David Hughes who used the distance defined by the segment FE, which he called alpha, but L = 1+M-a. The figure shows the moon overlapping the disk of the sun in a lens-shaped zone whose upper half is represented by the area AFDE.

The basic idea is that we want to compute the area of the lunar arc cap AFD by computing the area of the sector BAF and subtracting the triangle BAE from the sector area. That leaves the area of the cap as the left-over area. We perform the same calculation for the solar sector CAE and subtract the triangle CAD from this. The resulting area of the full lens-shaped overlap region is then

Occulting Area = 2x(AreaAFD + AreaAED).

Because of the geometry, the resulting area should only depend on the center-to-center separation and the radii of the sun and moon. You should not have to specify any angles as part of the final calculation. In the following we will use degree measure for all angles.

The area of the sector of a circle is just A = (Theta/360)piR^{2} so that gives us the first two relationships:

To simplify the problem, we are only interested in the fraction of the full sun disk that is illuminated. The full sun has an area of pi Rs^{2}, so we divide Am and As by pi Rs^{2} , and if we define Rs=1.0 and M = Rm/Rs we get:

Although M is fixed by the solar-lunar ratio, we seem to have two angular variables alpha and theta that we also have to specify. We can reduce the number of variables because the geometry gives a relationship between these two angles because they share a common segment length given by h.

so that the EQ-1 for A can be written entirely in terms of the center-to-center distance, L, and moon-to-sun disk ratio M = Rm/Rs. This is different than the equation used by Hughes, which uses the width of the lens (the distance between the lunar and solar limbs) segment FDE=a as the parameter, which is defined as L = 1+M-a.

During a typical total solar eclipse lasting 4 minutes, we can define L as

L = 1900 – 900*(T/240) arcseconds where T is the elapsed time from First Contact in seconds. Since L is in units of the current solar diameter (1900 asec) we have

EQ 3) L = 1 – T/480.

If we program EQ 1, 2, and 3 into an excel spreadsheet we get the following plot for the April 8, 2024 eclipse.

First Contact occurs at 16:40 UT and Fourth Contact occurs at 19:57 UT so the full duration is 197 minutes. During this time L varies from -(1+M) to +(1+M). For the April 8, 2024 eclipse we have the magnitude M = 1.0566, so L varies from -2.0566 (t=0) to +2.0566 (t=197m). As the moon approaches the full 4-minute overlap of the solar disk between L=-0.05 and L=+0.05 (t =97m to t=102m), we reach full eclipse.

We can re-express this in terms of the landscape lighting. The human eye is sensitive to a logarithmic variation in brightness, which astronomers have developed into a ‘scale of magnitudes’. Each magnitude represents the minimum change in brightness that the human eye can discern and is equivalent to a factor change by 2.51-times. The full-disk solar brightness is equal to -26.5m, full moon illumination is -18.0m on this scale. The disk brightness, S, is proportional to the exposed solar disk area, where E is the solar surface emission in watts/m^{2} due to the Planck distribution for the solar temperature of T=5770 k. This results in the formula:

m = -26.5 – 2.5log10(F)

where F is the fraction of the full disk exposed and is equal to Equation 1. For a sun disk where 90% has been eclipsed, f=0.10 and the dimming is only 2.5log(1/10) = 2.5m. How this translates into how humans perceive ambient lighting is complicated.

The concept of a Just Noticeable Difference is an active research area in psychophysics. In assessing heaviness, for example, the difference between two stimuli of 10 and 11 grams could be detected, but we would not be able to detect the difference between 100 and 101 grams. As the magnitude of the stimuli grow, we need a larger actual difference for detection. The percentage of change remains constant in general. To detect the difference in heaviness, one stimulus would have to be approximately 2 percent heavier than the other; otherwise, we will not be able to spot the difference. Psychologists refer to the percentages that describe the JND as Weber fractions, named after Ernst Weber (1795-1878), a German physiologist whose pioneering research on sensation had a great impact on psychological studies. For example, humans require a 4.8% change in loudness to detect a change; a 7.9% change in brightness is necessary. These values will differ from one person to the next, and from one occasion to the next. However, they do represent generally accurate values.

The minimum perceivable light intensity change is sometimes stated to be 1%, corresponding to +5.0^{m}, but for the Weber Fraction a 7.8% change is required in brightness corresponding to only -2.5log(0.078) = +2.7m. This is compounded by whether the observer is told beforehand that a change is about to happen. If they are not informed, this threshold magnitude dimming could be several magnitudes higher and perhaps closer to the +5.0^{m} value.

Heliophysics is an area of space science, named by NASA, which focuses on the matter and energy of our Sun and its effects on the solar system. It also studies how the Sun varies and how those changes pose a hazard to humans on Earth and in space.

The Heliophysics Big Year is a global celebration of solar science and the Sun’s influence on Earth and the entire solar system.During the Heliophysics Big Year, you will have the opportunity to participate in many solar science events such as watching solar eclipses, experiencing an aurora, participating in citizen science projects, and other fun Sun-related activities. For details, have a look at the NASA video that describes it in more detail [HERE].

This 14-month series for science and math educators focuses on heliophysics topics with related math problems at three levels: elementary, middle, and high school. It is sponsored by NASA’s Heliophysics Education Activation Team.

Each month, I will be hosting a webinar on the theme-of-the-month and also providing some math-oriented activities that go along with he theme. Here is a list of the Webinar viewing dates. When each webinar is completed, you can view a recorded version of it at the link provided. Visit this blog page a day before the scheduled program to register.

HBY Webinar programs:

December 19, 2023 – Citizen Science Projects – Do Auroras Ever Touch Ground?

January 16, 2024 – The Sun Touches Everything – Solar Panel Math and Sunlight Energy.

February 20, 2024 – Fashion – How do color filters work?

March 19, 2024 – Experiencing the Sun – Predicting Solar Storms.

April 16, 2024 – Total Solar Eclipse – The Last Total Solar Eclipse!

May 21, 2024 – Visual Art – Is the Sun Really Yellow?

June 18, 2024 – Performance Art – Do Songs About the Sun Follow the Sunspot Cycle?

July 16, 2024 – Physical and Mental Health – How Old is Sunlight?

August 20, 2024 – Back to School – Can You Accurately Draw the Solar Corona in Under 5 Minutes?

September 17, 2024 – Environment and Sustainability – Interplanetary solar electricity for spacecraft.

October 15, 2024 – Solar Cycle and Solar Max – Predicting the Next Sunspot Cycles and Travel to Mars.

November 19, 2024 – Bonus Science – TBD

December 17, 2024 – Parker’s Perihelion – TBD

To participate in a webinar, held at 7:00pm Eastern Time, you first need to register for it by clicking on the appropriate link below and filling out the registration form. You will then receive via email the link to the live program.

This is my new book for the general public about our sun and its many influences across the solar system. I have already written several books about space weather but not that specifically deal with the sun itself, so this book fills that gap.

We start at the mysterious core of the sun, follow its energies to the surface, then explore how its magnetism creates the beautiful corona, the solar wind and of course all the details of space weather and their nasty effects on humans and our technology.

I have sections that highlight the biggest storms that have upset our technology, and a discussion of the formation and evolution of our sun based on Hubble and Webb images of stars as they are forming. I go into detail about the interior of our sun and how it creates its magnetic fields on the surface. This is the year of the April 2024 total solar eclipse so I cover the shape and origin of the beautful solar corona, too. You will be an expert among your friends when the 2024 eclipse happens.

Unlike all other books, I also have a chapter about how teachers can use this information as part of their standards-based curriculum using the NASA Framework for Heliospheric Education. I even have a section about why our textbooks are typically 10 – 50 years out of date when discussinbg the sun.

For the amateur scientists and hobbyists among you, there is an entire chapter on how to build your own magnetometers for under $50 that will let you monitor how our planet is responsing to solar storms, which will become very common during the next few years.

There hasn’t been a book like this in over a decade, so it is crammed with many new discoveries about our sun during the 21st century. Most books for the general public about the sun have actually been written in a style appropriate to college or even graduate students.

My book is designed to be understandable by my grandmother!

Generally, books on science do not sell very well, so this book is definitely written without much expectation for financial return on the effort. Most authors of popular science books make less than $500 in royalties. For those of you that do decide to get a copy, I think it will be a pleasurable experience in learning some remarkable things about our very own star! Please do remember to give a review of the book on the Amazon page. That would be a big help.

Yep…I want to get the e-book version ($5): Link to Amazon.

Yep…I want to get the paperback version ($15): Link to Amazon.

Oh…by the way…. I am a professional astronomer who has been working at NASA doing research, but also education and public outreach for over 20 years. Although I have published a number of books through brick-and-morter publishing houses, I love the immediacy of self-publishing on topics I am excited about, and seeing the result presented to the public within a month or two from the time I get the topic idea. I don’t have to go through the lengthy (month-year) tedium of pitching an idea to several publishers who are generally looking for self-help and murder mysteries. Popular science is NOT a category that publishers want to support, so that leaves me with the self-publishing option.

Other books you might like:

Exploring Space Weather with DIY Magnetometers. ($7). Link to Amazon.

History of Space Weather: From Babylon to the 21st Century. (paperback, $30) (ebook, $5). Link to Amazon.

Solar Storms and their Human Impacts (e-book; $2) Link to Amazon.

The 23rd CycleL Learning to live with a stormy star. – Out of print.

This page is a supplement to my new book ‘Exploring Space Weather with DIY Magnetometers‘, which is avalable at Amazon by clicking [HERE]. This $8.00 B/W book contains 146 pages and has 116 illustrations and figures that describe six different magnetometers that you can build for under $60.00. I will be posting updates to my magnetometer designs on this page along with new storm data examples as the current sunspot cycle progresses. For convenience, the sections below are tied to updates to the corresponding chapters in the book. In each instance, I have compared my design data for the magnetic D-component (angular deviation from True Geographic North) to the corresponding data from the Fredericksburg Magnetic Observatory (FRD) in Virginia.

What is Space Weather?

Chapter 2: Earth’s Magnetic Field

Chapter 3: Basic Soda Bottle Designs (under $10.00)

Chapter 5: A Dual Hall Sensor Design ($20.00)

Chapter 6: The Smartphone Magnetometer

Chapter 7: The Photocell Comparator ($40.00)

Chapter 8: The Arduino Magnetometer with the RM3100 sensor ($60.00)

This zoomed-in image of Uranus, captured by Webb’s Near-Infrared Camera (NIRCam) Feb. 6, 2023, reveals stunning views of the planet’s rings. The planet displays a blue hue in this representative-color image, made by combining data from two filters (F140M, F300M) at 1.4 and 3.0 microns, which are shown here as blue and orange, respectively.

What a typical Press Release might tell you.

Uranus has eleven known rings that contain dark, boulder-sized particles. Although the rings are nearly-perfect circles, they look like ellipses in some pictures because the planet is tilted as we are viewing it and so the rings appear distorted. They appear compressed in the side-to-side direction but are their normal undistorted distance from Uranus in the top-down direction in this image.

The outermost ‘epsilon’ ring is made up of ice boulders several meters across. The other rings are made up mainly of icy chunks darkened by rocks. The rings are thin ( less than 200 meters), narrow (less than 100 km), and dark compared to the rings of other planets. They are actually so dark they reflect about as much light as charcoal. The rings have a temperature of about 77 Kelvin according to U.C. Berkeley astronomers.

Press Releases and textbook entries depend on scientists interpreting the data they gather to create a better picture of what they are seeing. I am going to show you in this Blog just how some of those numbers and properties are derived from the data astronomers gather. I have created three levels of interpretation and information extraction that pretty nearly match three different education levels and the knowledge about math that goes along with them.

I. Space Cadet (Grades 3-6)

This is the easiest step, and the first one that astronomers use to get a perspective on the sizes of the things they are seeing. In education, it relies on working with simple proportions. The scale bar in the picture tells you how the size of the picture relates to actual physical sizes.

1) Use a millimeter ruler to calculate the diameter of Uranus in kilometers.

2) Place your ruler vertically across the disk of Uranus. Measure the vertical (undistorted) distance from the center of Uranus to the center of the outer ‘Epsilon’ ring. How many kilometers is this ring from the center of Uranus?

3) Compare the size of the Uranus ring system to the orbit of our moon of 385,000 km. Would the moon’s orbit fit inside the ring system or is it bigger?

II. Space Commander (Grades 7-9)

This level of interpretation is a bit more advanced. If you are in Middle School, it works with concepts such as volume, mass, speed and time to extract more information from the data in the image, now that you have completed the previous-level extraction.

1) What is the circumference of the Epsilon ring in kilometers?

2) The rings of Uranus are only about 200 meters thick and no more then 100 km wide, which means their cross-section is that of a skinny rectangle. What is the cross sectional of the area of the Epsilon ring in square-kilometers assuming it is a rectangle with these dimensions?

3) What is the volume of the ring in cubic-kilometers if Volume = Area x Circumference?

4) If the volume of Earth is 1 trillion cubic kilometers, how much volume does the Epsilon ring occupy in units where Earth = 1?

5) With sizes of about7-meters in diameter, what is the average mass of a single boulder if it is made from ‘dirty ice’ with a density of 2000 kg/m^3?

6) Scientists estimate that there could be as many as 25 billion boulders in the Epsilon ring. From your answer to the previous question, what is the total mass of this ring?

7) [Speed and time] The particles in the Epsilon ring travel at a speed of 10.7 km/s in their orbit around Uranus. How many hours does it take for a boulder to complete one orbit?

III. Junior Scientist (College, Graduate school)

This is a very hard level that requires that you understand how to work with algebraic equations, evaluate them for specific values of their constants and variables, and in physics understand Planck’s Black Body formula. You also need to understand how radiation works as you calculate various properties of a body radiating electromagnetic radiation. Undergraduate physics majors work with this formula as Juniors, and by graduate school it is simply assumed that you know how to apply it in the many different guises in which bodies emit radiation.

The Planck black body formula is given in the frequency domain by

where E is the spectral energy density of the surface in units of Watts/meter^2/Steradians/Hertz and the constants are h = 6.63×10^43 Joules Sec; c = 3.0×10^8 meters/sec and k = 1.38 x 10^-23 Joules/k.

If the temperature of the Epsilon ring material is T=77 k, what is E in the 1.4-micron band of the Webb (F140M) and in the 1.3- millimeter band of the Atacama Large Millimeter Array (ALMA) for a single ring particle?

What does your answer to Question 1 tell you about the visibility of the rings seen by the Webb at 1.4-microns?

If the 1.3-millimeter total flux in the Epsilon ring was measured by ALMA to be 112milliJanskies (Table 3), about how many particles are contained in the Epsilon ring if it consists of dirty ice boulders with a 7-meter diameter? Note: 1 Jansky = 10^-26 watts/m^2/Hz. Assume a distance to Uranus of 1 trillion meters.

If the volume of the Epsilon ring is about 6.4×10^6 cubic kilometers, about how far apart are the boulders in the ring?

From your answer to Question 4, if the relative speeds of the ring boulders are about 1 millimeter/sec (see Henrik Latter), how often will these boulders collide?

So, how did you do?

You may not have gotten all of the answers correct, but the important thing is to appreciate just how astronomers extract valuable information about an object from its image at different wavelengths. Other than visiting the object in person, there is no other way to learn about distant objects than to gather data and follow many different series of steps to extract information about them. Some of these series of steps are pretty simple as you experienced at the Space Cadet-level. Others can be very difficult as you saw at the Junior Scientist-level. What is interesting is that, as our theoretical knowledge improves, we can create new series of steps to extract even more information from the data. Comparing the ring brightness at two or more wavelengths, we can deduce if the boulders are made of pure ice or dark rocks. By studying how the thickness of the ring changes along its circumference relative to the locations of nearby moons, we can study gravity and tidal waves in the ring bodies that tell us about how the density of the ring particles change. Astronomy is just another name for Detective Work but where the ‘crime scene’ is out of reach!

Answers

I.1 Close to 50,700 km.

I.2 Close to 51,000 km radius 102,000 km diameter.

I.3 The moons orbit radius is 385,000 km. The outer Epsilon ring could easily fit inside our Moon’s orbit.

II.1 Circumference is about 2(pi)R, so with R = 51,000 km, the circumference is 320,000 km.

II.2 The cross-section is 100 km long and 200 meters tall so its area is just A = 0.2 km x 100 km = 20 square km.

II.3 Volume = Area x Circumference = (20 km^2)x (3.2×10^5 km) = 6.4 x 10^6 km^3.

II.5 Mass = volume x density. A single boulder with a radius of 3.5 meters has a volume of 4/3 pi (3.5)^3 = 179 m^3. Then its mass is 179 m^3 x 2000 kg/m^3 = 359000 kg or 359 tons.

II.6 There are 25 billion of these boulders so their total mass is 25 billion x 359 tons = 9.0×10^15 kg or 9 trillion tons. According to calculations in 1989, the mass of the Epsilon ring is estimated to be closer to 10^{16} kg or 10 trillion tons [Nature].

II.7 51,000 km/(10.7 km/s) = 8.3 hours.

III.1 Working with the Planck equation is always a bit tricky. The units for B will be Watts/m^2/Str/Hertz. First lets evaluate the equation for its constants h, c and k.

For the Webb infrared band of 1.4-microns, its frequency is 3×10^8/1.4×10^-6 or 7.1×10^13 Hertz. For ALMA the frequency is 231 GHz or 2.31×10^11 Hertz. For a temperature of 77 k, we get B(Webb) = (5.26×10^-9)(5.8×10^-20) = 3.0×10^-28 watts/m^2/Hz/str. For ALMA it will be B(ALMA) = (1.8×10^-16)(6.46) = 1.2×10^-15 Watts/m^2/Hz/str.

III.2 The visibility of the ring by its thermal emission cannot be the cause of its brightness in the Webb F140M band because its black body emission at 1.4 microns is vanishingly small due to the huge exponential factor. It must be caused by reflected visible light from the sun.

III.3 In the ALMA 1.3-millimeter band, the flux density from a black body at a temperature of 77 k is 1.2×10^-15 watts/m^2/Hertz/str. One Jansky unit equals 10^-26 watts/m^2/Hz, so the black body emission in the ALMA band can be re-written as 1.2×10^11 Janskies/str. The angular radius of the boulder at the distance of Uranus ( 1 trillion meters) is given in radians by Q = 3.5 meter/1 trillion meters = 3.5×10^-12 radians. The angular area is then AQ = pi Q^2 = 3.9×10^-23 steradians. Then the black body flux density from this boulder is about S=1.2×10^11 Jy/str x 3.9×10^-23 str so S=4.5×10^-12 Janskies. But ALMA detects a total emission from this ring of about 0.112 Janskies, so the number of emitters is just 0.112 Jy/4.5×10^-12 Jy/boulder = 2.5×10^10 boulders or 25 billion boulders.

III.4 The ring volume is 6.4 x 10^6 cubic kilometers, so each boulder occupies a volume in the ring of about 6.4×10^6 km^3/25×10^9 boulders= 256,000 cubic meters/boulder. The average distance between them is L = V^0.333 = 63 meters.

III.5 Time = distance/speed so 63m/0.001 m/sec = 17.5- hours. The Epsilon particle orbit speed is about 8.3 hours (see II.7), so the particles collide about twice every orbit.

An astronomer's point-of-view on matters of space, space travel, general science and consciousness