<p>To find the angular diameter of a satellite you need to find
<span class="math-container">$$\arctan\left(\frac{\text{diameter of satellite}}{\text{distance to satellite}}\right)$$</span></p>
<p>As you have been using the diameter of the planet, your formulae are wrong.</p>
<p>You should also make use of the small-angle approximations <span class="math-container">$\arctan(x)\approx x$</span> for small x in radians. So to get the value in degrees you can do
<span class="math-container">$$\frac{180^\circ\times\text{diameter of satellite}}{\pi\times\text{distance to satellite}}$$</span></p>
<p>And you should round your values. You don't need 15 significant figures of accuracy!</p>

CC BY-SA 4.0
2020-08-06T23:41:52.813Z
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.
uhoh
CC BY-SA 4.0
2020-08-07T05:57:16.457Z
No I haven't "As you have been using the diameter of the planet, your formulae are wrong."
James K
CC BY-SA 4.0
2020-08-07T07:42:03.760Z
Indeed! Sorry, thanks, etc.
uhoh
CC BY-SA 4.0
2020-08-07T07:52:44.053Z
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
James K
CC BY-SA 4.0
2020-08-07T08:25:17.320Z
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.
CyanideBaby
CC BY-SA 4.0
2020-08-06T22:21:42.963Z
2020-08-06T22:21:42.963Z
James K
<p>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.</p>
<p>I'm trying to calculate the angular diameter of two natural satellites orbiting a planet, as seen from the planet.</p>
<ul>
<li><strong>Satellite A</strong> is located 568350 km from the planet and has a diameter of 4400 km.</li>
<li><strong>Satellite B</strong> is located 357094 km from the planet and has a diameter of 1900 km.</li>
</ul>
<p>For <strong>Satellite A</strong>, we would therefore get:
<span class="math-container">$$\arctan\left(\frac{4400}{568350}\right)=0.44^\text{o}$$</span></p>
<p>For <strong>Satellite B</strong>, we would therefore get :
<span class="math-container">$$\arctan\left(\frac{1900}{357094}\right)=0.3^\text{o}$$</span></p>
<p>Similarly, for our <strong>moon</strong>, the same calculation goes as follows:
<span class="math-container">$$\arctan\left(\frac{3476}{384402}\right)=0.51^\text{o}$$</span></p>

CC BY-SA 4.0
2020-08-06T20:15:53.113Z
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.
jng224
CC BY-SA 4.0
2020-08-06T21:50:50.643Z
Looks like your first result is in radians instead of degrees, and I agree with Jonas.
Mike G
<the-moon><natural-satellites><angular-diameter>
Angular diameter result confirmation
CyanideBaby
# Angular diameter result confirmation

(Asked by: username_0 on 2020-08-06)

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

**Tags:** the-moon, natural-satellites, angular-diameter

> **Comments:**
> **username_1**: 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.  
> **username_2**: Looks like your first result is in radians instead of degrees, and I agree with Jonas.  

---

## 1 answers

### Answer by username_3 on 2020-08-06. Score: 3 | Accepted answer

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!

> **Comments:**
> **username_4**: 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.  
> **username_3**: No I haven't "As you have been using the diameter of the planet, your formulae are wrong."  
> **username_4**: Indeed! Sorry, thanks, etc.  
> **username_3**: 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  
> **username_0**: 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.  

---

