Solid Angle of the Sun Derivation

  • #1

Main Question or Discussion Point

I read in a paper the following passage:

"We take the sun to subtend a linear angle of 32 arc-minutes. The solid angle is derived as [tex]\Omega=\pi sin^{2}16'=6.8x10^{-5} sr[/tex]"

I don't understand how the formula to go from linear angle to solid angle is just found by taking the area treating sin(16') as a radius. Can someone explain?
 

Answers and Replies

  • #2
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Use diameter of 900,000 miles and distance of 94 million miles. So linear subtended angle is d = 9.57 milliradians = 32.9 arc-minutes. So radius is 16' and area is pi R2, so solid angle is pi x sin2(16') = 7.2 x 10-5 sterads.
 
  • #3
662
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Use diameter of 900,000 miles and distance of 94 million miles. So linear subtended angle is d = 9.57 milliradians = 32.9 arc-minutes. So radius is 16' and area is pi R2, so solid angle is pi x sin2(16') = 7.2 x 10-5 sterads.
This derivation is not clear to me. You say that "radius is 16' ", so you're equating a length to an angle, then you insert sine of that angle without explanation.

Here's what I would say instead: The solid angle is defined as the ratio of the subtended area (pi*r^2 for the "disk" of the Sun) to the square of the radial distance to that area (R, the distance to the Sun). This would give us pi*r^2/R^2.

In this case, we are given the subtended angle rather than either radius value, so for convenience we use the small angle approximation:

r/R = tan(theta) ~ sin(theta), where theta is half the angle subtended by the disk. Plug that in the expression above and you'll get the desired expression.
 
  • #4
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This derivation is not clear to me. You say that "radius is 16' ", so you're equating a length to an angle, then you insert sine of that angle without explanation..
You are correct. Actually, to get the correct value, we have to go back to my post:
Use diameter of 900,000 miles and distance of 94 million miles. So linear subtended angle is d = 9.57 milliradians.
Solid angle = pi d2/4 = pi (.00957)2/4 = 7.193 x 10-5 sr
 
  • #5
haruspex
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The previous posts give a perfectly good approximation for the case in question, but I don't want anyone to go away thinking it's exactly right - it would break down at larger angles.
The solid angle is not based on the area of the disc (or other 2-d shape) at a given distance. It is based on a section of the surface of a sphere.
E.g. for the sun, imagine a transparent sphere of unit radius centred on the observer. Draw the outline of the sun on the sphere, as perceived by the observer. The solid angle is the area of the spherical cap inside that drawn circle. This is slightly greater than the area of the flat 2-d circle.
In the extreme, the solid angle subtended by the entire enclosing sphere is 4 pi.
 

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