Solid angle of celestial bodies

Question

I found online that by using the definition of angular diameter, the formula for the solid angle of a celestial object can be defined in terms of the radius of the object, $R$, and the distance from the observer to the object, $D$: $$\Omega = 2 \pi \left( 1-\frac{\sqrt{D^2-R^2}}{D} \right) : D \geq R $$

I know the definition of angular diameter, d, is:

$$d\equiv 2D \tan{\left(\frac{\delta}{2}\right)} $$ where $\delta$ is the angular diameter, $d$ is the physical diameter and $D$ is the distance from the observer to the object.

I also know that the definition of a solid angle (steradian) is

$$\Omega = \frac{A}{r^2} $$ where $A$ is the area of the spherical cap.

I'm unsure of how to use the last two equations to get to the first one. I have been unable to find a derivation for the first expression.

Answer 1

The first equation assumes that the object is a sphere. The second equation assumes that $D\gg R$, so that the difference between the visible diameter and real diameter is negligible. The definition of steradians involve projecting an object onto a spherical cap. That's how the first equation is derived: take a sphere at a distance $D>R$ from the observation point. Then project the sphere outward (or inward, either works) onto an imaginary sphere (in astronomy it's often called the "celestial sphere") to calculate the solid angle. The projection process involves taking lines that intersect the observation point and are tangent to the sphere.

Also, you don't derive the first from the latter two, you derive the second equation from the first and third. Another equation you'll need is that the area of a circular cap on the surface of a sphere is given by $$A = 4\pi R^2 \sin^2\left(\frac{\theta}{2}\right),$$ where $\theta$ is the polar angle from the center of the cap to its edge. This is a fact you can derive using calculus and spherical coordinates. In this case, $\theta = \delta / 2$.

Answer 2

I leave you a sketch with the proof