Article 6A003 Solid angle of a star

Solid angle of a star

by
John
from John D. Cook on (#6A003)

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 4. If you're in a ship far from land, the solid angle of the sky is 2 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

solidangle3.svg

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.

solidangle4.svg

and so

solidangle5.svg

with an error on the order of (r/d)4. To put it another way, the error in our approximation for is on the order of ^2. The largest object in the sky is the sun, and it has apparent size less than 10-4, so is always small when looking at astronomical objects, and ^2 is negligible.

So for practical purposes, the apparent size of a celestial object is 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.

Square degrees

Sometimes solid angles are measured in square degrees, given by /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 /4 of a square.)

Examples

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.

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