What can JWST see?

The other day I ran across this photo of Saturn’s moon Titan taken by the James Webb Space Telescope (JWST).

If JWST can see Titan with this kind of resolution, how well could it see Pluto or other planets? In this post I’ll do some back-of-the-envelope calculations, only considering the apparent size of objects, ignoring other factors such as how bright an object is.

The apparent size of an object of radius r at a distance d is proportional to (r/d)². [1]

Of course the JWST isn’t located on Earth, but JWST is close to Earth relative to the distance to Titan.

Titan is about 1.4 × 1012 meters from here, and has a radius of about 2.6 × 106 m. Pluto is about 6 × 1012 m away, and has a radius of 1.2 × 106 m. So the apparent size of Pluto would be around 90 times smaller than the apparent size of Titan. Based on only the factors considered here, JWST could take a decent photo of Pluto, but it would be lower resolution than the photo of Titan above, and far lower resolution than the photos taken by New Horizons.

Could JWST photograph an exoplanet? There are two confirmed exoplanets are orbiting Proxima Centauri 4 × 1016 m away. The larger of these has a mass slightly larger than Earth, and presumably has a radius about equal to that of Earth, 6 × 106 m. Its apparent size would be 150 million times smaller than Titan.

So no, it would not be possible for JWST to photograph an expolanet.


[1] In case the apparent size calculation feels like a rabbit out of a hat, here’s a quick derivation. Imagine looking out on sphere of radius d. This sphere has area 4πd². A distant planet takes up πr² area on this sphere, 0.25(r/d)² of your total field of vision.

Earth : Jupiter :: Jupiter : Sun

The size of Jupiter is approximately the geometric mean of the sizes of Sun and Earth.

In terms of radii,

\frac{R_\Sun}{R_{\text{\Jupiter}}} \approx \frac{R_\Jupiter}{R_\Earth}

The ratio on the left equals 9.95 and the ratio on the left equals 10.98.

The subscripts are the astronomical symbols for the Sun (☉, U+2609), Jupiter (♃, U+2643), and Earth (, U+1F728). I produced them in LaTeX using the mathabx package and the commands \Sun, Jupiter, and Earth.

The the mathabx symbol for Jupiter is a little unusual. It looks italicized, but that’s not because the symbol is being used in math mode. Notice that the vertical bar in the symbol for Earth is vertical, i.e. not italicized.


Gravity on Jupiter

NASA image of Jupiter

I was listening to the latest episode of the Space Rocket History podcast. The show includes some audio from a documentary on Pioneer 11 that mentioned that a man would weigh 500 pounds on Jupiter.

My immediate thought was “Is that all?! Is this ‘man’ a 100 pound boy?”

The documentary was correct and my intuition was wrong. And the implied mass of the man in the documentary is 190 pounds.

Jupiter has more than 300 times more mass than the earth. Why is its surface gravity only 2.6 times that of the earth?

Although Jupiter is very massive, it is also very large. Gravitational attraction is proportional to mass, but inversely proportional to the square of distance.

A satellite in orbit 100,000 km from the center of Jupiter would feel 300 times as much gravity as one in orbit the same distance from the center of Earth. But the surface of Jupiter is further from its center of mass than the surface of Earth is from its center of mass.

The mass of Jupiter is 318 times that of Earth, and the its mean radius is 11 times that of Earth. So the ratio of gravity on the surface of Jupiter to gravity on the Earth’s surface is

318 / 11² = 2.63

Now suppose a planet had the same density as Earth but a radius of r Earth radii. Then its mass would be r³ times greater, but its surface gravity would only be r times greater since gravity follows an inverse square law. So if Jupiter were made of the same stuff as Earth, its surface gravity would be 11 times greater. But Jupiter is a gas giant, so its surface gravity is only 2.6 times greater.

Related posts

How to Organize Technical Research?


64 million scientific papers have been published since 1996 [1].

Assuming you can actually find the information you want in the first place—how can you organize your findings to be able to recall and use them later?

It’s not a trifling question. Discoveries often come from uniting different obscure pieces of information in a new way, possibly from very disparate sources.

Many software tools are used today for notetaking and organizing information, including simple text files and folders, Evernote, GitHub, wikis, Miro, mymind, Synthical and Notion—to name a diverse few.

AI tools can help, though they can’t always recall correctly and get it right, and their ability to find connections between ideas is elementary. But they are getting better [2,3].

One perspective was presented by Jared O’Neal of Argonne National Laboratory, from the standpoint of laboratory notebooks used by teams of experimental scientists [4]. His experience was that as problems become more complex and larger, researchers must invent new tools and processes to cope with the complexity—thus “reinventing the lab notebook.”

While acknowledging the value of paper notebooks, he found electronic methods essential because of distributed teammates. In his view many streams of notes are probably necessary, using tools such as GitLab and Jupyter notebooks. Crucial is the actual discipline and methodology of notetaking, for example a hierarchical organization of notes (separating high-level overview and low-level details) that are carefully written to be understandable to others.

A totally different case is the research methodology of 19th century scientist Michael Faraday. He is not to be taken lightly, being called by some “the best experimentalist in the history of science” (and so, perhaps, even compared to today) [5].

A fascinating paper [6] documents Faraday’s development of “a highly structured set of retrieval strategies as dynamic aids during his scientific research.” He recorded a staggering 30,000 experiments over his lifetime. He used 12 different kinds of record-keeping media, including lab notebooks proper, idea books, loose slips, retrieval sheets and work sheets. Often he would combine ideas from different slips of paper to organize his discoveries. Notably, his process to some degree varied over his lifetime.

Certain motifs emerge from these examples: the value of well-organized notes as memory aids; the need to thoughtfully innovate one’s notetaking methods to find what works best; the freedom to use multiple media, not restricted to a single notetaking tool or format.

Do you have a favorite method for organizing your research? If so, please share in the comments below.


[1] How Many Journal Articles Have Been Published? https://publishingstate.com/how-many-journal-articles-have-been-published/2023/

[2] “Multimodal prompting with a 44-minute movie | Gemini 1.5 Pro Demo,” https://www.youtube.com/watch?v=wa0MT8OwHuk

[3] Geoffrey Hinton, “CBMM10 Panel: Research on Intelligence in the Age of AI,” https://www.youtube.com/watch?v=Gg-w_n9NJIE&t=4706s

[4] Jared O’Neal, “Lab Notebooks For Computational Mathematics, Sciences, Engineering: One Ex-experimentalist’s Perspective,” Dec. 14, 2022, https://www.exascaleproject.org/event/labnotebooks/

[5] “Michael Faraday,” https://dlab.epfl.ch/wikispeedia/wpcd/wp/m/Michael_Faraday.htm

[6] Tweney, R.D. and Ayala, C.D., 2015. Memory and the construction of scientific meaning: Michael Faraday’s use of notebooks and records. Memory Studies8(4), pp.422-439. https://www.researchgate.net/profile/Ryan-Tweney/publication/279216243_Memory_and_the_construction_of_scientific_meaning_Michael_Faraday’s_use_of_notebooks_and_records/links/5783aac708ae3f355b4a1ca5/Memory-and-the-construction-of-scientific-meaning-Michael-Faradays-use-of-notebooks-and-records.pdf

Constellations in Mathematica

Mathematica has data on stars and constellations. Here is Mathematica code to create a list of constellations, sorted by the declination (essentially latitude on the celestial sphere) of the brightest star in the constellation.

constellations = EntityList["Constellation"]
sorted = SortBy[constellations, -#["BrightStars"][[1]]["Declination"] &]

We can print the name of each constellation with

Map[#["Name"] &, sorted]

This yields

{"Ursa Minor", "Cepheus", "Cassiopeia", "Camelopardalis", 
…, "Hydrus", "Octans", "Apus"}

We can print the name of the constellation along with its brightest star as follows.

Scan[Print[#["Name"], ", " #["BrightStars"][[1]]["Name"]] &, sorted]

This prints

Ursa Minor, Polaris
Cepheus, Alderamin
Cassiopeia, Tsih
Camelopardalis, β Camelopardalis
Hydrus, β Hydri
Octans, ν Octantis
Apus, α Apodis

Mathematica can draw star charts for constellations, but when I tried

Entity["Constellation", "Orion"]["ConstellationGraphic"]

it produced extraneous text on top of the graphic.

Related posts

How to memorize the periodic table

Periodic table image


Memorizing the periodic table has some practical value, especially if you’re a chemist, but in any case it’s an interesting exercise, easier to do than it may sound. And it’s a case study for how you might memorize other things of more practical value to you personally.

Major system pegs

The Major system is a way to associate consonant sounds to numbers. You can fill in vowels and semivowels as you please to turn the sequence of consonant sounds into words, preferably words that create a vivid image in your mind.

You can pick a canonical encoding of each number to create a set of pegs and use these to memorize numbered lists. Although numbers can be encoded many ways, a set of pegs is a one-to-one mapping to numbers. To pull up the nth item in the list, recall what image you’ve associated with the peg image for n.

For example, you could encode 16 as dish, tissue, touché, Hitachi, etc. If you want to remember that sulfur has atomic number 16 you could use any of those images. But if you wanted to remember that the 16th element is sulfur, you need to have a unique peg associated with 16.

Learning pegs is more work than hanging things on pegs. But once you have a set of pegs, you can reuse them for memorizing multiple lists. For example, you could use the same pegs to memorize the periodic table and the ASCII table.

Atomic numbers

Allan Krill has written up a way to associate each element with a peg. You could use his suggestions, but you’ll almost certainly need to customize some of them. It’s generally hard to use anyone else’s mnemonics. What works for one person may not for another.

To memorize the periodic table, you first come up with pegs for the numbers 1 through 118. Practice those and get comfortable with them. This could take a while, but it’s reusable effort. Then associate an image of each element with its corresponding peg. For example, polonium is element 84. If your peg for 84 is fire, you might imagine someone playing polo on a field that’s on fire.

Element symbols

Every element has a one- or two-letter symbol, and most of these are easy: Ti for titanium, U for uranium, etc. Some seem completely arbitrary, such as Hg for mercury, but these you may already know. These names seem strange because they are mnemonic in Latin. But the elements with Latin names are also the ones that were discovered first and are the most common. You probably know by osmosis, for example, that the symbol for iron is Fe.

The hard part is the second letter, if there is a second letter. For example, is does Ar stand for argon or arsenic? Is the symbol for thulium Th or Tl or Tm?

When you associate an element image with a peg image, you could add a third image for the second letter of the element symbol, using the NATO phonetic alphabet if you know that. For example, the NATO word for S is Sierra. If your peg for 33 is mummy, you might imagine a mummy drinking a bottle of Sierra Springs® water laced with arsenic.

Related posts

Image from OpenStax Biology 2e. CC BY Attribution license.

Homework problems are rigged

This post is a follow-on to a discussion that started on Twitter yesterday. This tweet must have resonated with a lot of people because it’s had over 250,000 views so far.

You almost have to study advanced math to solve basic math problems. Sometimes a high school student can solve a real world problem that only requires high school math, but usually not.

There are many reasons for this. For one thing, formulating problems is a higher-level skill than solving them. Homework problems have been formulated for you. They have also been rigged to avoid complications. This is true at all levels, from elementary school to graduate school.

A college school student tutoring a high school student might notice that homework problems have been crafted to always have whole number solutions. The college student might not realize how his own homework problems have been rigged analogously. Calculus homework problems won’t avoid fractions, but they still avoid problems that don’t have tidy solutions [1].

When I taught calculus, I looked around for homework problems that were realistic applications, had closed-form solutions, and could be worked in a reasonable amount of time. There aren’t many. And the few problems that approximately satisfy these three criteria will be duplicated across many textbooks. I remember, for example, finding a problem involving calculating the mass of a star that I thought was good exercise. Then as I looked through a stack of calculus texts I saw that the same homework problem was in most if not all the textbooks.

But it doesn’t stop there. In graduate school, homework problems are still crafted to avoid difficulties. When you see a problem like this one it’s not obvious that the problem has been rigged because the solution is complicated. It may seem that you’re able to solve the problem because of the power of the techniques used, but that’s not the whole story. Tweak any of the coefficients and things may go from complicated to impossible.

It takes advanced math to solve basic math problems that haven’t been rigged, or to know how to do your own rigging. By doing your own rigging, I mean looking for justifiable ways to change the problem you need to solve, i.e. to make good approximations.

For example, a freshman physics class will derive the equation of a pendulum as

y″ + sin(y) = 0

but then approximate sin(y) as y, changing the equation to

y″ + y = 0.

That makes things much easier, but is it justifiable? Why is that OK? When is that OK, because it’s not always.

The approximations made in a freshman physics class cannot be critiqued using freshman physics. Working with the un-rigged problem, i.e. keeping the sin(y) term, and understanding when you don’t have to, are both beyond the scope of a freshman course.

Why can we ignore friction in problem 5 but not in problem 12? Why can we ignore the mass of the pulley in problem 14 but not in problem 21? These are questions that come up in a freshman class, but they’re not freshman-level questions.


[1] This can be misleading. Students often say “My answer is complicated; I must have made a mistake.” This is a false statement about mathematics, but it’s a true statement about pedagogy. Problems that haven’t been rigged to have simple solutions often have complicated solutions. But since homework problems are usually rigged, it is true that a complicated result is reason to suspect an error.

Alien astronomers and Benford’s law

alien astronomer image created by DALL-E

In 1881, astronomer Simon Newcomb noticed something curious. The first pages in books of logarithms were dirty on the edge, while the pages became progressively cleaner in later pages. He inferred from this that people more often looked up the logarithms of numbers with small leading digits than with large leading digits.

Why might this be? One might reasonably expect the numbers that came up in work to be uniformly distributed. But as often the case, it helps to ask “Uniform on what scale?”

Newcomb might have imagined his counterpart on another planet. This alien astronomer might have 12 fingers [1] and count in base 12. Base 10 is not inevitable, even for creatures with 10 fingers: the ancient Sumerians used a base-60 number system.

If Newcomb’s twelve-fingered counterpart had developed logarithms but not digital computers, he might have tables of duodecimal logarithms bound into books, and he too might noticed that pages with small leading (duo)digits are more frequently referenced. Both astronomers would naturally look up the logarithms of physical constants, physical distances, and so fort, numbers that vary over a practically unlimited range. The unlimited range is important.

On what scale could both astronomers see the leading digits uniformly distributed?

If Newcomb needed to look up the logarithms of numbers over a limited range, say from 1 to 106, each with equal probability, then the leading digits would be uniformly distributed. But our alien astronomer would have no special interest in the number 106. He might want to look at numbers between 1 and 126. The leading digits of numbers over this range would be uniformly distributed when represented in base 12, but not when represented in base 10. The choice of upper limit introduces a bias in one base or another.

Now suppose the numbers that both astronomers used in their work were uniformly distributed on a logarithmic scale. Newcomb conjectured that the numbers that came up in practice were uniformly distributed in their logarithms base 10. Our alien astronomer might conjecture the same thing for logarithms base 12. And both could be right. So would a third astronomer working in base 42. All logarithms are proportional, and so numbers uniformly distributed on a log scale using one base are uniformly distributed on a log scale using any other base.

Benford’s law says that the leading digits of numbers that come up in practice are uniformly distributed on a log scale. This applies to base 10, but also any other base, such as base 100. If you looked at the first two digits and thought of them as single base-100 digits, Benford’s law still applies.

But who is Benford? True to Stigler’s law of eponymy, Newcomb’s observation is named after physicist Frank Benford who independently made the same observation in 1938 and who tested it more extensively.

Let’s look at a set of physical constants and see how well Benford’s law applies. I took at list of physical constants from NIST and made a histogram of the leading digits to compare with what one would expect from Benford’s law.

If one were to write the NIST constants in base 12 and repeat the exercise, the result would look similar.

Related posts

[1] The image at the top of the post was created by DALL-E. There is a slight hint of an extra finger. DALL-E usually has a hard problem with hands, adding or removing fingers. But my attempts to force it to draw a hand with an extra finger were not successful.

Oval orbits?

Johannes Kepler thought that planetary orbits were ellipses. Giovanni Cassini thought they were ovals. Kepler was right, but Cassini wasn’t far off.

In everyday speech, people use the words ellipse and oval interchangeably. But in mathematics these terms are distinct. There is one definition of an ellipse, and several definitions of an oval. To be precise, you have to say what kind of oval you have in mind, and in the context of this post by oval I will always mean a Cassini oval.

Ellipses and ovals each have two foci, f1 and f2. Let d1(p) and d2(p) be the distances from a point p to each of the foci. For an ellipse, the sum d1(p) + d2(p) is constant. For an oval, the product d1(p) d2(p) is constant.

In [1] the authors argue that just as planetary orbits are nearly circles, they’re also nearly ovals. This post will look at how far the earth’s orbit is from a circle and from an oval.

We need a way to specify which oval we want to compare to the ellipse of earth’s orbit. We’ll do this by equating the major and minor semi-axes of the two curves. These are usually denoted a and b, but the same variables have a different meaning in the context of ovals, so I’ll denote them by M for major and m for minor.

The equation of an ellipse is

(x/M)² + (y/m)² = 1

and the equation of an oval is

((x + a)² + y²) ((xa)² + y²) = b².

Setting x = 0 in the equation of an oval tells us

m² = ba²

and setting y = 0 tells us

M² = b + a².


b = (M² + m²)/2


a² = (M² – m²)/2.

For the earth’s orbit, M = 1.00000011 and m = 0.99986048 measured in AU, astronomical units. So or oval has parameters

a = 0.011816102


b = 0.99986060.

If you plot Kepler’s ellipse and Cassini’s oval for earth’s orbit at the same time, you can’t see the difference.

Planet orbits are nearly circular. If we compare a circle of radius 1 AU with Kepler’s ellipse we get a maximum error of about 1 part in 10,000.

But if we compare Cassini’s oval with Kepler’s ellipse we get a maximum error of about 1 part in 100,000,000.


In short, a circle is a good approximation to earth’s orbit, but a Cassini oval is four orders of magnitude better.

It would be difficult to empirically distinguish an ellipse from an oval as the shape of earth’s orbit, but theory is clearly on Kepler’s side since his ellipses fall out of Newton’s laws. Cassini’s error was more qualitative than quantitative.

More orbital mechanics posts

[1] Kepler’s ellipse, Cassini’s oval and the trajectory of planets. B Morgado1 and V Soares. 2014 Eur. J. Phys. 35 025009 DOI 10.1088/0143-0807/35/2/025009

Sphere of influence

Apollo 11 orbit diagram via NASA

Suppose a spaceship is headed from the earth to the moon. At some point we say that the ship has left the earth’s sphere of influence is now in the moon’s sphere of influence (SOI). What does that mean exactly?

Wrong explanation #1

One way you’ll hear it described is that the moon’s sphere of influence is the point at which the earth is no longer pulling on the spaceship, but that’s nonsense. Everything has some pull on everything else, so how do you objectively say the earth’s pull is small enough that we’re now going to call it zero? And as we’ll see below, the earth’s pull is still significant even when the spaceship leaves earth’s SOI.

Wrong explanation #2

Another explanation you’ll hear is the moon’s sphere of influence is the point at which the moon is pulling on the spaceship harder than the earth is. That’s a better explanation, but still not right.

The distance from the earth to the moon is about 240,000 miles, and the radius of the moon’s SOI is about 40,000 miles. So when a spaceship first enters the moon’s SOI, it is five times closer to the moon than to the earth.

Newton’s law of gravity says gravitational force between two bodies is proportional to the product of their masses and inversely proportional to the square of the distance. The mass of the earth is about 80 times that of the moon. So at the moon’s SOI boundary, the pull of the earth is 80/25 times as great as that of the moon, about three times greater.

Correct exlanation

So what does sphere of influence mean? The details are a little complicated, but essentially the moon’s sphere of influence is the point at which it’s more accurate to say the ship is orbiting the moon than to say it is orbiting the earth.

How can we say it’s better to think of the ship orbiting the moon than the earth when the earth is pulling on the ship three times as hard as the moon is? What matters is not so much the force of earth’s gravity as the effect of that force on the equations of motion.

The motion of an object between the earth and the moon could be viewed as an orbit around earth, with the moon exerting a perturbing influence, or as an orbit around the moon, with the earth exerting a perturbing influence.

At the boundary of the moon’s SOI the effect of the earth perturbing the ship’s orbit around the moon is equal to the effect of the moon perturbing its orbit around the earth. It’s a point at which it is convenient to switch perspectives. It’s not a physical boundary [1]. Also, the “sphere” of influence is not exactly a sphere but an approximately spherical region.

The moon has an effect on the ship’s motion when it’s on our side of the moon’s SOI, and the earth still has an effect on its motion after it has crossed into the moon’s SOI.

Calculating the SOI radius

As a rough approximation, the SOI boundary is where the ratio of the distances to the two bodies, e.g. moon and earth, equals the ratio of their masses to the exponent 2/5:

r/R = (m/M)2/5.

This approximation is better when the mass M is much larger than the mass m. For the earth and the moon, the equation is good enough for back-of-the-envelope equations but not accurate enough for planning a mission to the moon. Using the round numbers in this post, the left side of the equation is 1/5 = 0.2 and the right side is (1/80)0.4 = 0.17.


Everything above has been in the context of the earth-moon system. Sphere of influence is defined relative to two bodies. When we spoke of a spaceship leaving the earth’s sphere of influence, we implicitly meant that it was leaving the earth’s sphere of influence relative to the moon.

Relative to the sun, the earth’s sphere of influence reaches roughly 600,000 miles. You could calculate this distance using the equation above. A spaceship like Artemis leaves the earth’s sphere of influence relative to the moon at some point, but never leaves the earth’s sphere of influence relative to the sun.

Related posts

[1] The sphere of influence sounds analogous to a continental divide, where rain falling on one side of the line ends up in one ocean and rain falling on the other side ends up in another ocean. But it’s not that way. I suppose you could devise an experiment to determine which side of the SOI you’re on, but it would not be a simple experiment. An object placed between the earth and the moon at the SOI boundary would fall to the earth unless it had sufficient momentum toward the moon.