Not being a math-minded person at all, however basic it might appear to be to all of you in this particular question, I would like to get the following calculation confirmed.

I'm trying to calculate the angular diameter of two natural satellites orbiting a planet, as seen from the planet.

  • Satellite A is located 568350 km from the planet and has a diameter of 4400 km.
  • Satellite B is located 357094 km from the planet and has a diameter of 1900 km.

For Satellite A, we would therefore get: $$\arctan\left(\frac{4400}{568350}\right)=0.44^\text{o}$$

For Satellite B, we would therefore get : $$\arctan\left(\frac{1900}{357094}\right)=0.3^\text{o}$$

Similarly, for our moon, the same calculation goes as follows: $$\arctan\left(\frac{3476}{384402}\right)=0.51^\text{o}$$

  • 2
    $\begingroup$ Hello! In order to calculate the angular diameter, you have to insert the satellite's actual diameter in the calculation - it appears to me that you calculated the angular diameter of the planet as seen from the satellites. $\endgroup$ – Jonas Aug 6 at 20:15
  • 1
    $\begingroup$ Looks like your first result is in radians instead of degrees, and I agree with Jonas. $\endgroup$ – Mike G Aug 6 at 21:50

To find the angular diameter of a satellite you need to find $$\arctan\left(\frac{\text{diameter of satellite}}{\text{distance to satellite}}\right)$$

As you have been using the diameter of the planet, your formulae are wrong.

You should also make use of the small-angle approximations $\arctan(x)\approx x$ for small x in radians. So to get the value in degrees you can do $$\frac{180^\circ\times\text{diameter of satellite}}{\pi\times\text{distance to satellite}}$$

And you should round your values. You don't need 15 significant figures of accuracy!

| improve this answer | |
  • $\begingroup$ Technically speaking a better formula is $2 \arctan(r/d)$ since the base of the right triangle is from the observer to the center of the object; at a 10° diameter it's only a 0.7% effect but the error scales as the fourth power so it's 3% off at a diameter of 20°, but of course this is still for a flat disk rather than a sphere. $\endgroup$ – uhoh Aug 6 at 23:41
  • 1
    $\begingroup$ No I haven't "As you have been using the diameter of the planet, your formulae are wrong." $\endgroup$ – James K Aug 7 at 5:57
  • $\begingroup$ Indeed! Sorry, thanks, etc. $\endgroup$ – uhoh Aug 7 at 7:42
  • 1
    $\begingroup$ Also it is the bit about "for a disk not a sphere" is why I say go with small angle approximations. If the angle is big enough to use "real" trigonometry then it is big enough for you do deal with the actual shape $\endgroup$ – James K Aug 7 at 7:52
  • $\begingroup$ Thank you for this. I was so wrong it's funny, or not. Well, I've updated my question, adding the diameters of the two satellites. $\endgroup$ – CyanideBaby Aug 7 at 8:25

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.