The apparent size of a distant object can be measured by projecting the object onto a unit sphere around the observer and calculating the area of the projected image.
A unit sphere has area 4π. If you’re in a ship far from land, the solid angle of the sky is 2π steradians because it takes up half a sphere.
If the object you’re looking at is a sphere of radius r whose center is a distance d away, then its apparent size is
steradians. This formula assumes d > r. Otherwise you’re not looking out at the sphere; you’re inside the sphere.
If you’re looking at a star, then d is much larger than r, and we can simplify the equation above. The math is very similar to the math in an earlier post on measuring tapes. If you want to measure the size of a room, and something is blocking you from measuring straight from wall to wall, it doesn’t make much difference if the object is small relative to the room. It all has to do with Taylor series and the Pythagorean theorem.
Think of the expression above as a function of r and expand it in a Taylor series around r = 0.
with an error on the order of (r/d)4. To put it another way, the error in our approximation for Ω is on the order of Ω². The largest object in the sky is the sun, and it has apparent size less than 10-4, so Ω is always small when looking at astronomical objects, and Ω² is negligible.
So for practical purposes, the apparent size of a celestial object is π times the square of the ratio of its radius to its distance. This works fine for star gazing. The approximation wouldn’t be as accurate for watching a hot air balloon launch up close.
Sometimes solid angles are measured in square degrees, given by π/4 times the square of the apparent diameter in degrees. This implicitly uses the approximation above since the apparent radius is r/d.
(The area of a square is diameter squared, and a circle takes up π/4 of a square.)
When I typed
3.1416 (radius of sun / distance to sun)^2
into Wolfram Alpha I got 6.85 × 10-5. (When I used “pi” rather than 3.1416 it interpreted this as the radius of a pion particle.)
When I typed
3.1416 (radius of moon / distance to moon)^2
I got 7.184 × 10-5, confirming that the sun and moon are approximately the same apparent size, which makes a solar eclipse possible.
The brightest star in the night sky is Sirius. Asking Wolfram Alpha
3.1416 (radius of Sirius / distance to Sirius)^2
we get 6.73 × 10-16.
2 thoughts on “Solid angle of a star”
> (When I used “pi” rather than 3.1416 it interpreted this as the radius of a pion particle.)
Wait, what? Would it always do that, or is this new, or did being next to the word “radius” influence its interpretation and this is some ChatGPT cuckoo-bananas stuff? Because that sounds like pretty bad behavior for a computer program to me.
“When I used “pi” rather than 3.1416 it interpreted this as the radius of a pion particle.”
This is an aliasing artifact from the underlying Mathematica language. Using “pi” or “Pi” will give some goofy cuckoo-bananas results. But you can get the constant pi by using Evaluate[Pi] to disambiguate on Mathematica’s terms.
This was a fun one to debug; I just discovered the WolframAlpha function in Mathematica! Slick.