<p>In short, it is using small angle approximation.</p>
<p>Long answer:</p>
<p>From your equation:</p>
<p><span class="math-container">$ds = r^2d\Omega$</span>, (I believe <span class="math-container">$r$</span> is the distance between Earth and Moon)</p>
<p>You substitute:
<span class="math-container">$ds = \frac{\pi D_l^2}{4}$</span> into the equation and move <span class="math-container">$d\Omega$</span> to the other side to get:</p>
<p><span class="math-container">$\Omega = \frac{\pi D_l^2}{4r^2}$</span></p>
<p>Note the angular diameter of the moon can be given by <span class="math-container">$tan$</span> function:</p>
<p><span class="math-container">$tan \varphi = \frac{D_l}{r}$</span>.</p>
<p>Because the apparent diameter of the Moon is so small, you can use the small angle approximation <em>when you convert <span class="math-container">$\varphi$</span> into radian</em>:</p>
<p><span class="math-container">$tan \varphi \approx \varphi$</span></p>
<p>Thus you can get:</p>
<p><span class="math-container">$tan^2 \varphi = \frac{D_l^2}{r^2} \approx \varphi^2$</span></p>
<p>Substitute the above equation back to:</p>
<p><span class="math-container">$\Omega = \frac{\pi}{4}\varphi^2$</span></p>
<p>To convert <span class="math-container">$\varphi$</span> to radian, remember that <span class="math-container">$\pi$</span> radian <span class="math-container">$=$</span> <span class="math-container">$180$</span> degree.</p>

CC BY-SA 4.0
2024-06-12T17:39:47.257Z
I wasn't aware of the small angle approximation, that is what i was missing. It is clear now thank you!
Orion.55290
CC BY-SA 4.0
2024-06-10T22:52:36.383Z
2024-06-10T22:52:36.383Z
Polaris5744
<p>I'm reading an astronomy textbook for high schoolers, it contains an exercise to calculate the visible angle ratio of the moon in the sky (the textbook is not in english so my terminology is probably wrong). The textbook also provides the answer, but I don't understand it.</p>
<p>The input for the exercise is the angle under which the moon is visible. It's specified that:</p>
<p><span class="math-container">$$\varphi = 31'$$</span></p>
<p>The diameter of the moon is also named <span class="math-container">$D_l$</span>.</p>
<p>The formula</p>
<p><span class="math-container">$$dS = {r}^2 d\Omega$$</span></p>
<p>is explained; <span class="math-container">$r$</span> being the distance from the observer and <span class="math-container">$d\Omega$</span> being the solid angle.</p>
<p>The textbook provides the following solution, which I don't understand:</p>
<p><span class="math-container">$$
dS = r^2d\Omega \implies \\ d\Omega=dS/r^2=\pi D_l^2/4r^2 = (\pi/4)\varphi^2= \pi/4 \times (31'/60' \times \pi/180)^2 \text{ sr} =6.39 \times 10^-{5} \text{ sr}
$$</span></p>
<p>I <em>think</em> it implies that:</p>
<p><span class="math-container">$$
\varphi = D_l/r
$$</span></p>
<p>But I'm not sure. I'm also confused about the degree angles and the steradian solid angles. I struggled to find clear explanations elsewhere online, I would be interested in any relevant link assuming only trigonometry prior knowledge.</p>

CC BY-SA 4.0
2024-06-10T03:33:18.267Z
*Welcome to Stack Exchange!* I've made some formatting edits, but please feel free to adjust further. I am also confused by $$\varphi = 31°$$ can you add any further information or context about how that was introduced? *Thanks!*
uhoh
CC BY-SA 4.0
2024-06-10T07:09:31.117Z
It is using small angle approximation
Polaris5744
CC BY-SA 4.0
2024-06-10T08:39:16.730Z
Are you sure it’s not  = 31′ (minutes instead of degrees)?
Pierre Paquette
CC BY-SA 4.0
2024-06-10T13:31:01.990Z
Yeah I was gonna ask the same thing -- 31 degrees would be a VERY BIG MOON.
Darth Pseudonym
CC BY-SA 4.0
2024-06-10T16:51:24.100Z
Yes, @PierrePaquette i should have written minutes and i corrected the post. thank you for your comment
Orion.55290
CC BY-SA 4.0
2024-06-10T19:46:29.163Z
What is "S"  ?..
James K
<mathematics><space-geometry><angular-distance>
visible solid angle of the moon
Orion.55290
# visible solid angle of the moon

(Asked by: username_0 on 2024-06-09)

I'm reading an astronomy textbook for high schoolers, it contains an exercise to calculate the visible angle ratio of the moon in the sky (the textbook is not in english so my terminology is probably wrong). The textbook also provides the answer, but I don't understand it.

The input for the exercise is the angle under which the moon is visible. It's specified that:

$$\varphi = 31'$$

The diameter of the moon is also named $D\_l$.

The formula

$$dS = {r}^2 d\Omega$$

is explained; $r$ being the distance from the observer and $d\Omega$ being the solid angle.

The textbook provides the following solution, which I don't understand:

$$
dS = r^2d\Omega \implies \\ d\Omega=dS/r^2=\pi D\_l^2/4r^2 = (\pi/4)\varphi^2= \pi/4 \times (31'/60' \times \pi/180)^2 \text{ sr} =6.39 \times 10^-{5} \text{ sr}
$$

I *think* it implies that:

$$
\varphi = D\_l/r
$$

But I'm not sure. I'm also confused about the degree angles and the steradian solid angles. I struggled to find clear explanations elsewhere online, I would be interested in any relevant link assuming only trigonometry prior knowledge.

**Tags:** mathematics, space-geometry, angular-distance
> **Comments:**
> **username_1**: *Welcome to Stack Exchange!* I've made some formatting edits, but please feel free to adjust further. I am also confused by $$\varphi = 31°$$ can you add any further information or context about how that was introduced? *Thanks!*  
> **username_2**: It is using small angle approximation  
> **username_3**: Are you sure it’s not  = 31′ (minutes instead of degrees)?  
> **username_4**: Yeah I was gonna ask the same thing -- 31 degrees would be a VERY BIG MOON.  
> **username_0**: Yes, @PierrePaquette i should have written minutes and i corrected the post. thank you for your comment  
> **username_5**: What is "S"  ?..  

---

## 1 answers

### Answer by username_2 on 2024-06-10. Score: 3 | Accepted answer

In short, it is using small angle approximation.

Long answer:

From your equation:

$ds = r^2d\Omega$, (I believe $r$ is the distance between Earth and Moon)

You substitute:
$ds = \frac{\pi D\_l^2}{4}$ into the equation and move $d\Omega$ to the other side to get:

$\Omega = \frac{\pi D\_l^2}{4r^2}$

Note the angular diameter of the moon can be given by $tan$ function:

$tan \varphi = \frac{D\_l}{r}$.

Because the apparent diameter of the Moon is so small, you can use the small angle approximation *when you convert $\varphi$ into radian*:

$tan \varphi \approx \varphi$

Thus you can get:

$tan^2 \varphi = \frac{D\_l^2}{r^2} \approx \varphi^2$

Substitute the above equation back to:

$\Omega = \frac{\pi}{4}\varphi^2$

To convert $\varphi$ to radian, remember that $\pi$ radian $=$ $180$ degree.

> **Comments:**
> **username_0**: I wasn't aware of the small angle approximation, that is what i was missing. It is clear now thank you!  

---

