Lets say Earth as a moon of a gas giant. Lets say it orbited the gas giant the size of Saturn every 24 hours. I understand that the part facing away from the gas giant would have a day/night cycle like the planet Earth but what about the side facing the gas giant? Would it get any sunlight before the gas giant eclipsed the sun? Im interested in what that would look like also?
Let's assume the planet, or moon in this case, is tidally locked so that its rotational period is equal to its orbital period. It would have a regular day/night cycle except for when the Sun is eclipsed by the gas giant. A 24 hours orbit is very very close and there would probably be a great many solar eclipses. In fact, there may be one every day. But as the gas giant would only block light for a portion of the orbit, there wouldn't be any point on the moon's surface that was in perpetual nighttime. Also, as the gas giant orbits around the Sun the time of the eclipse would gradually change over the course of the year with different parts of the moon being eclipsed at different times of the year.
It's relatively straightforward. But perhaps you'd like to calculate this yourself? First, let's find out the radius of the orbit. You want to take the force of gravity between two bodies of masses equal to those of Earth's and Saturn's respectively, and compare that force to the one needed to keep the "moon" in a nice circular orbit(for simplicity's sake, and a good approximation anyway) - that is: centripetal force. You end up with F_{g}=F_{c} Where F_{g}=GmM/r^{2} F_{c}=mrω^{2} and the angular velocity ω=2∏/T where T is one day(or 86400s) So: GM/r^{2}=r4∏^{2}/T^{2} GMT^{2}=r^{3}4∏^{2} finally r=(GMT^{2}/4∏^{2})^{1/3} I'll let you plug in all the numbers.(remember to use T in seconds) With that ready, let's see what size will the Saturn's disc appear on the sky: 2R/r=θ where 2R is Saturn's diameter, r is the distance we've just calculated, and θ is the angular size(in radians) it'll have. Now we know how big it is on the sky. How fast does it move across it then? It goes a full circle in 1day, so its apparent angular velocity on the sky is: ε=2∏/T with that velocity, it obscures any point in the sky for: t=θ/ε seconds. Now, at the distance of 10AU(Saturn's orbit) from the Sun, the Sun's disc is 1/100th of the half a degree in angular size we see on Earth. So let's just treat it as a point-like source of light that spends the above calculated t time obscured by Saturn's disc. And that's your eclipse duration. Of course, that'd only work for the "moon"'s orbit perfectly aligned with that of Saturn's around the Sun. Should it be inclined, the eclipses' periods will change. If the "moon"'s orbit inclination is higher than θ, then there will be times in the ~30 year long, well, year, when there are no eclipses, and times when the eclipses are shorter than the t we calculated, varying in a sinusoidal fashion from 0 to t. In such case the eclipses would last full t time precisely twice in a year. If the inclination is less than θ, the eclipse times will similarly vary between t and some value x, where t>x>0. Yeah, so I did go ahead and plugged some numbers in, just to check if the results are roughly sensible, and I got: r=200000km {~1/2th the distance from Earth to Moon and some 50000 km above the Satrun's Roche limit} θ=0.6 rad {which is roughly 33 degrees, or 1/6th of the sky} ε=7*10^{-5} rad/s t=~8500s {2.3 hours} An orbit inclined as much as, or more, than 33 degrees for a large moon seems unlikely, so the actual eclipse times would most likely never go below two hours for modestly inclined orbits.
Let's try again. Also I find that choice of different units can make the computations easier to follow and check for errors. 24 h orbit around Earth is about 42 200 km from Earth centre Saturn is about 95 Earth masses As noted in the previous derivation, the radius of a given period orbit is proportional to the cube root of the primary mass; cube root of 95 is about 4,55 so we are looking at orbital radius of about 190 000 km (as stated, about half Earth-Moon distance). The length of orbit is about 1 200 000 km and speed on orbit about 50 000 km/h. At equinoxes, the daily eclipse would be almost 2 and a half hours. This would still leave over 9 and a half hours of sunlight. Add the shine of the planet at night, and some twilight in planet´s atmosphere at beginnings and ends of eclipses. But I believe there was a gross error in estimating the orbital angle needed to evade eclipse. YES, the size, as in diametre, of Saturn would be about 33 degrees. But you will NOT need to get as far as diametre to evade the eclipse! It is sufficient to get as far as radius. Earth inclination 23,5 degrees has sine of about 0,4 so at solstices, the centre of Earth would pass about 76 000 km from the shadow centreline, and the winter polar circle about 6000 km closer. Still 70 000 km Since the polar radius of Saturn is only 55 000 km, over the 190 000 km the shadow would diverge by around 1800 km (relying on Earth at 1 au), to then 55 900 km (excessive precision here meant for illustration) of penumbra and 54 100 km of umbra. Both much less than the 70 000 km from polar circle to eclipse centreline.
Interesting question. A 24-hour orbit is very close, maybe too close to be realistic. I think Io is about one-and-one half days. Every point on the "moon" would get sunlight, though some points maybe half of what other points get By fiddling with the inclination of the "lunar" orbit you can get whatever frequency of eclipse you like, I would think. But normally there would be many eclipses, many more than here on Earth. Would the difference in temperature during the orbit be noticeable, as the planet went closer and further from the Sun? I'd think it would be. Jupiter has a huge EM field and that would make a big difference. Look up the "Io flux tube." There is a great deal of energy there. There would be so much light reflected from the giant that the phase would make a big difference. When "full" then night would be like a day with overcast, I would expect.
Of smaller bodies, Thebe is 16 hours; Mimas is just under 23 hours. Obviously depending on inclination! In case of low inclination, you have the demonstration that the subplanetary point gets about 80 % the sunlight of far side by time, and slightly less by amount (can someone compute how much?), whereas the poles obviously get 0% by amount. You have a hard maximum on inclination at 90 degrees. Hardly. Earth goes 2,5 million km closer and further, and we cannot notice. Yes, and Saturn has far weaker field in comparison. Perhaps, yes. We are speaking of a disc roughly 5000 times bigger in area than Moon. How will the albedo compare, for a gas giant in habitable zone? At subplanetary point, the planet would never be less than half at night. Crescent planet can only be seen at night on east and west edges of near side. Precisely how is east, west, north and south defined outside Earth?