I would need to compute the solid angle of the Earth from a spacecraft when close to the Earth, where no small angles approximation can be made. I thought it wouldn't be that hard but I feel actually quite stuck on that problem...

If anyone has a hint that'd be great!



Solid angle

$\Omega = 2\pi(1-\cos \theta)$

$\theta = APO $

enter image description here

$(OA)^2 + (AP)^2 = (OP)^2$

OA = r - radius of earth and OP = r + h (height of satellite)

$r^2 +(AP)^2 = (r+h)^2$

$AP = \sqrt{2rh + h^2}$

$\cos \theta = AP / OP =\frac{ \sqrt{2rh + h^2}}{r + h}$

$\Omega = 2\pi \left( 1-\frac{ \sqrt{2rh + h^2}}{r + h} \right)$

For r= 6378 km and h = 400 km the solid angle $\Omega$ = 4.157 sR, and the half-angle (nadir to edge of Earth) is 70.22°.

  • $\begingroup$ +1 It's great when someone stops by and adds a new and accurate answer to an old question! I've adjusted a bit of the MathJax formatting, hope you don't mind. $\endgroup$ – uhoh Aug 21 '19 at 0:38

If anyone has a hint that'd be great!

Hint as requested:

  1. Calculate the half-angle $\theta$ of the Earth as seen from the height of your orbit. Half angle is measured from the axis (line from spacecraft to center of Earth) to the visible edge of the Earth' as seen from the Spacecraft. Use all the great mathematical help and diagrams in all of the excellent answers to the question Earth angular size looking from the ISS, and to the question How far into space does one have to travel to see the entire sphere of earth?.
  2. Use the equation for the solid angle $\Omega$ of a spherical cap as shown in this section in Wikipedia. Note, the sphere here is a mathematical unit sphere centered on the satellite, not the Earth, and $\theta$ is the half-angle that you calculated in step 1.
  3. Check your calculation against the following. If Earth's radius is 6378 km and the altitude is 400 km above that, the half-angle $\theta$ will be about 70.218 degrees, and the solid angle will be 4.157 sR which is 33.08% of 4π sR.
  4. (optional) If you are successful, post a new answer here outlining exactly what you did. Up votes on answers give 10 reputation points each, it's the quickest way to get to 50 so you can start leaving comments on other posts. If you have trouble with using MathJax for equations, just write them out and leave a comment and someone will format them for you.
  5. Have fun!
  • $\begingroup$ @oz380 how did it go? $\endgroup$ – uhoh Apr 26 '18 at 7:58

Hint: $$ \textrm{solid angle} = 4\pi \sin^2\left(\frac{\theta}{2}\right) $$

Where $\theta$ is the half cone angle.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.