1
$\begingroup$

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: $$\arctan\left(\frac{3476}{384402}\right)=0.51^\text{o}$$

$\endgroup$
2
  • 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, 2020 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, 2020 at 21:50

1 Answer 1

3
$\begingroup$

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!

$\endgroup$
5
  • $\begingroup$ 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, 2020 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, 2020 at 5:57
  • $\begingroup$ Indeed! Sorry, thanks, etc. $\endgroup$
    – uhoh
    Aug 7, 2020 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, 2020 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. $\endgroup$ Aug 7, 2020 at 8:25

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .