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.

**Read More!**

NASA Press Release: Webb Scores Another Ringed World

Mashable.com. Surprise! Uranus has rings!

Research paper. Thermal emission from uranian ring system

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 about** **7-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.4 Volume = 6.4 x 10^6 km^3 / 1 trillion km^3 = 0.0000064 Earths.

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- hour**s. The Epsilon particle orbit speed is about 8.3 hours (see II.7), so the particles collide about twice every orbit.