<p>I am not sure of any direct measurement, but it can be figured out rather easily and accurately from theory with a little bit of maths. We have the following parameters for Metis and Jupiter (taken off Wikipedia):</p>

<ul>
<li>Diameter of Jupiter: about 140 Mm (megameters, 1 Mm = 1000 km), so radius ~70 Mm. Compare Earth at ~6.5 Mm radius, 13 Mm diameter.</li>
<li>Orbital distance for Metis from Jupiter's center: 128 Mm</li>
<li>Eccentricity: effectively zero, so we can treat it <em>very</em> accurately as a circular orbit.</li>
<li>Orbital period: 25.4690 ks (kiloseconds, 1 ks = 1000 s). Compare against Earth's day, 86.4 ks, and orbital period of its Moon, about 2500 ks (more accurate: 2360 ks).</li>
<li>Also, one more relevant point is the distance to the Sun and its diameter, to find the apparent size: average distance to the Sun is 779 Gm (<span class="math-container">$7.79 \times 10^5\ \mathrm{Mm}$</span>), and Sun's diameter roughly 1 Gm (<span class="math-container">$10^3\ \mathrm{Mm}$</span>).</li>
</ul>

<p>The way we will reason now is as follows. Jupiter "moves" in one circuit across the sky dome of an observer on Metis in one orbit. Hence, the time of eclipse is the time required for Jupiter to "move", from arrival of its leading edge to departure of its trailing edge, across the image of the Sun on the same sky dome. Equivalently, then, it is the time required for it to move through an <em>angular</em> distance equal to its own angular diameter plus <em>two</em> angular diameters of the Sun (one on first contact, the other before last contact):</p>

<p><span class="math-container">$$\mbox{Eclipse time} := \frac{2 \cdot (\mbox{Angular diameter of Sun}) + \mbox{Angular diameter of Jupiter}}{\mbox{Angular speed of Jupiter on sky dome}}$$</span></p>

<p>Now the denominator is easy: a full revolution is <span class="math-container">$2\pi\ \mathrm{rad}$</span> or about <span class="math-container">$6283\ \mathrm{mrad}$</span>, so we just divide that by 25.4690 ks, 4o get</p>

<p><span class="math-container">$$\mbox{Angular speed of Jupiter on sky dome} \approx 246.7\ \frac{\mbox{mrad}}{\mbox{ks}}$$</span></p>

<p>Hence, now all we need is the angular diameters of the Sun and Jupiter in milliradians and we'll be done.</p>

<p>This is easily doable with a little bit of trigonometry: while I do not have a diagram on hand, you can easily find one. The angular size of an object of girth <span class="math-container">$D$</span> emplaced at a distance <span class="math-container">$d$</span> from an observer is</p>

<p><span class="math-container">$$\mbox{Angular size} := 2 \cdot \tan^{-1}\left(\frac{D}{2d}\right)$$</span></p>

<p>Note that you'll often see the approximation</p>

<p><span class="math-container">$$\mbox{Angular size} \approx \frac{D}{d}$$</span></p>

<p>but this is only good when the object is very far, i.e. <span class="math-container">$d \gg D$</span>. In this case, though, that is <em>not</em> the case at least for Jupiter, so we have to use the full formula with the inverse tangent. Using the figures just given, we have</p>

<p><span class="math-container">$$\mbox{Angular size of Jupiter} = 2 \cdot \tan^{-1}\left(\frac{140\ \mathrm{Mm}}{2 \cdot 128\ \mathrm{Mm}}\right) \approx 1.000\ \mathrm{rad}$$</span></p>

<p>or <em>accurately</em>(!) <span class="math-container">$1000\ \mathrm{mrad}$</span>. The Sun is far enough that we <em>can</em> just use the approximation and get 1/779 rad or about <span class="math-container">$1.28\ \mathrm{mrad}$</span>. Hence the total angular distance to cover is 1.28 + 500 + 1.28 ~ 503 mrad, and the eclipse time equals</p>

<p><span class="math-container">$$\mbox{Eclipse time} = \frac{1000\ \mathrm{mrad}}{246.7\ \mathrm{mrad/ks}} \approx 4.08\ \mathrm{ks}$$</span></p>

<p>or for so many still not fully accustomed to the SI, a bit longer than an hour (3.6 ks), and in local measures, a bit less than 1/5 of a Metian day-month.</p>

<p>NOTE: I should point out a caveat of this is is that that this doesn't take into account the inclination of orbits - both Jupiter's and Metis's which may make the Sun in effect "encounter" Jupiter at a different point than simply cutting along its equator. But it should still give you an idea - <strong>4 ks</strong> or so, plus-minus some amount that'd be too much work to get.</p>

CC BY-SA 4.0
2019-12-21T03:33:56.760Z
Wow, so it spends about 16% of its time fully in shadow.  Thanks!
cowlinator
CC BY-SA 4.0
2019-12-21T03:42:30.907Z
@cowlinator : Yep (4/25 ~ 16%). Unless I made some error somewhere in this so I'd like others to check it.
The_Sympathizer
CC BY-SA 4.0
2019-12-21T02:38:18.123Z
2019-12-21T02:43:31.073Z
2019-12-21T02:43:31.073Z
The_Sympathizer
<p>How long does Jupiter obscure the sun, from the perspective of Metis, the innermost moon of Jupiter?</p>

<p>In other words, how long does Metis remain in the shadow of Jupiter?</p>

<p>If it varies by time of year, what are the minimum/maximum values?</p>

<jupiter><natural-satellites><eclipse-non-earth-related>
How long do solar eclipses last on Metis?
cowlinator
# How long do solar eclipses last on Metis?

(Asked by: username_0 on 2019-12-21)

How long does Jupiter obscure the sun, from the perspective of Metis, the innermost moon of Jupiter?

In other words, how long does Metis remain in the shadow of Jupiter?

If it varies by time of year, what are the minimum/maximum values?

**Tags:** jupiter, natural-satellites, eclipse-non-earth-related


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## 1 answers

### Answer by username_1 on 2019-12-21. Score: 1 | Accepted answer

I am not sure of any direct measurement, but it can be figured out rather easily and accurately from theory with a little bit of maths. We have the following parameters for Metis and Jupiter (taken off Wikipedia):

* Diameter of Jupiter: about 140 Mm (megameters, 1 Mm = 1000 km), so radius ~70 Mm. Compare Earth at ~6.5 Mm radius, 13 Mm diameter.
* Orbital distance for Metis from Jupiter's center: 128 Mm
* Eccentricity: effectively zero, so we can treat it *very* accurately as a circular orbit.
* Orbital period: 25.4690 ks (kiloseconds, 1 ks = 1000 s). Compare against Earth's day, 86.4 ks, and orbital period of its Moon, about 2500 ks (more accurate: 2360 ks).
* Also, one more relevant point is the distance to the Sun and its diameter, to find the apparent size: average distance to the Sun is 779 Gm ($7.79 \times 10^5\ \mathrm{Mm}$), and Sun's diameter roughly 1 Gm ($10^3\ \mathrm{Mm}$).

The way we will reason now is as follows. Jupiter "moves" in one circuit across the sky dome of an observer on Metis in one orbit. Hence, the time of eclipse is the time required for Jupiter to "move", from arrival of its leading edge to departure of its trailing edge, across the image of the Sun on the same sky dome. Equivalently, then, it is the time required for it to move through an *angular* distance equal to its own angular diameter plus *two* angular diameters of the Sun (one on first contact, the other before last contact):

$$\mbox{Eclipse time} := \frac{2 \cdot (\mbox{Angular diameter of Sun}) + \mbox{Angular diameter of Jupiter}}{\mbox{Angular speed of Jupiter on sky dome}}$$

Now the denominator is easy: a full revolution is $2\pi\ \mathrm{rad}$ or about $6283\ \mathrm{mrad}$, so we just divide that by 25.4690 ks, 4o get

$$\mbox{Angular speed of Jupiter on sky dome} \approx 246.7\ \frac{\mbox{mrad}}{\mbox{ks}}$$

Hence, now all we need is the angular diameters of the Sun and Jupiter in milliradians and we'll be done.

This is easily doable with a little bit of trigonometry: while I do not have a diagram on hand, you can easily find one. The angular size of an object of girth $D$ emplaced at a distance $d$ from an observer is

$$\mbox{Angular size} := 2 \cdot \tan^{-1}\left(\frac{D}{2d}\right)$$

Note that you'll often see the approximation

$$\mbox{Angular size} \approx \frac{D}{d}$$

but this is only good when the object is very far, i.e. $d \gg D$. In this case, though, that is *not* the case at least for Jupiter, so we have to use the full formula with the inverse tangent. Using the figures just given, we have

$$\mbox{Angular size of Jupiter} = 2 \cdot \tan^{-1}\left(\frac{140\ \mathrm{Mm}}{2 \cdot 128\ \mathrm{Mm}}\right) \approx 1.000\ \mathrm{rad}$$

or *accurately*(!) $1000\ \mathrm{mrad}$. The Sun is far enough that we *can* just use the approximation and get 1/779 rad or about $1.28\ \mathrm{mrad}$. Hence the total angular distance to cover is 1.28 + 500 + 1.28 ~ 503 mrad, and the eclipse time equals

$$\mbox{Eclipse time} = \frac{1000\ \mathrm{mrad}}{246.7\ \mathrm{mrad/ks}} \approx 4.08\ \mathrm{ks}$$

or for so many still not fully accustomed to the SI, a bit longer than an hour (3.6 ks), and in local measures, a bit less than 1/5 of a Metian day-month.

NOTE: I should point out a caveat of this is is that that this doesn't take into account the inclination of orbits - both Jupiter's and Metis's which may make the Sun in effect "encounter" Jupiter at a different point than simply cutting along its equator. But it should still give you an idea - **4 ks** or so, plus-minus some amount that'd be too much work to get.

> **Comments:**
> **username_0**: Wow, so it spends about 16% of its time fully in shadow.  Thanks!  
> **username_1**: @username_0 : Yep (4/25 ~ 16%). Unless I made some error somewhere in this so I'd like others to check it.  

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