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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.

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    $\begingroup$ 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! $\endgroup$
    – uhoh
    Commented Jun 10 at 3:33
  • $\begingroup$ It is using small angle approximation $\endgroup$ Commented Jun 10 at 7:09
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    $\begingroup$ Are you sure it’s not 𝜑 = 31′ (minutes instead of degrees)? $\endgroup$ Commented Jun 10 at 8:39
  • $\begingroup$ Yeah I was gonna ask the same thing -- 31 degrees would be a VERY BIG MOON. $\endgroup$ Commented Jun 10 at 13:31
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    $\begingroup$ Yes, @PierrePaquette i should have written minutes and i corrected the post. thank you for your comment $\endgroup$ Commented Jun 10 at 16:51

1 Answer 1

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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.

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  • $\begingroup$ I wasn't aware of the small angle approximation, that is what i was missing. It is clear now thank you! $\endgroup$ Commented Jun 12 at 17:39

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