On a moon of a gas giant, there would be a substantial amount of time where the giant blocks sunlight from reaching the moon. How do I determine how long this total eclipse would be?
| Notes: | |
|---|---|
| Orbiting distance: | 932,438 km |
| Orbital period: | 33.11 days |
| Rotation period: | 36 hours |
| Gas giant radius: | 60,613 km |
| Gas giant rings radius: | 141,286 km |
| Gas giant mass: | 50 earths |
| Gas giant density: | 322 kg/m3 |
| Gas giant apparent size from moon: | 7.4 degrees or 14.74x Luna |
Let's assume that the gas giant is sufficiently far away from the star that the diameter of the shadow at the distance of the satellite is approximately equal to the diameter of the gas giant.
$D_\text{shadow} = 2 \times 60{,}613\,\text{km} = 121{,}226\,\text{km}$
Let's also assume that everything rotates and revolves prograde, which is the most usual situation.
Moreover, let's assume that the orbital plane of the satellite lies in the orbital plane of the gas giant. For example, the orbits of the Galilean satellites of Jupiter come quite close to this assumption. Moreover, let's assume that the satellite has negligible obliquity.
For an observer on the surface of the satellite, the eclipse begins when the observation point enters the shadow of the gas giant, and ends when the observation point leaves the shadow of the gas giant, provided that both the entering and the leaving of the shadow happen in daytime.
The center of the satellite moves on its orbit at about
$v_\text{center} = 2 \times \pi \times 932{,}438\,\text{km} / 33.11 / 24 = 7732.75\,\text{km/h}$,
so that the center of the satellite will cross the shadow in
$t_\text{center} = D_\text{shadow} / v_\text{center} = 121{,}226 / 7732.75 = 16.44\,\text{hours}$
Now of course the observer is moving in the oppposite direction, so that its effective speed when crossing the shadow of the planet is lower. On the average, an observer sitting on the equator of the satellite moves against the progress of the satellite at a speed equal to the diameter of the satellite divided by 18 hours, but we don't know the diameter of the satellite.
Let's assume that the satellite is about the size of Ganymede, with a diameter of some 5300 km. Then the effective speed of an observer on the equator of the satellite with respect to the shadow of the gas giant is
$v_\text{observer} = v_\text{center} - 5300 / 18 = 7078.3\,\text{km/h}$
So that the time for the observer to cross the shadow of the planet is
$t_\text{observer} = D_\text{shadow} / v_\text{observer} = 121{,}226 / 7078.3 = 17.12\,\text{hours}$
Overall, the maximum duration of the eclipse will be about 17.12 hours for an observer on the equator of the satellite and about 16.44 hourse for an observer at one of the poles of the satellite.
But this is of course the maximum duration of the eclipse, which is achievable only if the eclipse begins not more than about 0.88 hours after sunrise for the observer on the equator and no more than 1.56 hours after sunrise for the observer at one of the poles. If the eclipse begins later, than it will end after sunset; and if it begins before sunrise, of course its duration will be decreased by the corresponding amount of time.
In comments, the radius of the star was clarified to be about 669,766 km, and the orbital radius of the gas giant was given as 114,000,000 km. This decreases the diameter of the shadow at the distance of the satellite by about
$2 \times \displaystyle \frac{669{,}766 - 60{,}613}{114{,}000{,}000} \times 932{,}438 = 9{,}965\,\text{km}$,
so that the diameter of the shadow is only
$D_\text{shadow} = 2 \times 60{,}613\,\text{km} - 9{,}965\,\text{km} = 111{,}261\,\text{km}$
This will decrease the maximum duration of the total eclipse in proportion, so that at the equator it will be 15.71 hours, and at the poles 15.09 hours. But on the other hand, the partial eclipse will be increased by the same amount...
