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I am working with a story. I wish to make my story somewhat logical by adding some science and thereby not making it to appear fantasy.
This story is on a fictional planet far away from Earth. Let's just consider that that fictional habitable planet is like Earth. This habitable planet gets fully eclipsed by another planet which revolves its parent star, thereby bringing darkness throughout the world once in 1.25 (earth) years. How long will it take for the ocean surface to freeze?
In another question, I calculate that the time it takes for the Earth to radiate off enough of its heat to drop the uppermost 100m to freezing would be 281 days.
However, that isn't your question. The land and the sea will cool at different rates, and the ocean surface will freeze later because vertical circulation will move any warmer ocean water from the deep upwards. The Arctic ocean is no more than 4 C when the upper layer has frozen, so lets calculate what it will take to drop the oceans to 5 C.
For a spherical cow model where the entire Earth's ocean is at 15 C, or 60 F, we can redo the calculation to see how long it will take all the Earth's oceans to radiate their heat into space. The imporant numbers here are the surface area of the Earth's oceans; $3.6\times10^{14}$ m$^2$, the volume of the oceans; $1.3\times10^{18}$ m$^3$; and the emissivity of the oceans: 0.95. From this we calculate the volumetric heat capacity of the oceans as $5.1\times10^{24}$ J/K
Plugging in to that other questions equations we get
$$ \begin{align} \frac{dT}{dt}\ &= \frac{0.95\cdot5.67\times10^{-8}\text{ J m}^{-2}\text{s}^{-1}\text{K}^{-4}\cdot T^4 \text{ K}^4\cdot3.6\times10^{14}\text{ m}^2}{5.1\times10^{24}\text{ J/K}}\\ &=3.8\times10^{-18}\cdot T^4 \text{ K/s} \end{align}$$
Integrating this over time, as in the other question, we see that in order for the Oceans to drop by 10 C it takes about 13 years.
$$\begin{align} \frac{dT}{dt} &= 1.0\times10^{-16}T^4\\ \int_{278\text{ K}}^{288\text{ K}}\frac{dT}{T^4}&=3.8\times10^{-18}\int dt\\ \frac{-1}{3}\cdot\left(\frac{1}{288^3}-\frac{1}{278^3}\right) &= 3.8\times10^{-18}t\\ t&=4.1\times10^{8}\text{ s} \end{align}$$
Obviously, these are very rough order of magnitude estimates, but you can see that even if the atmospheric temperatures approach freezing within a year, the oceans will take much longer to freeze because of stored heat in their vast depths.
I'll defer the answer to how long for an ocean to freeze over to Kingledion. There's no flaw in his answer. The next question becomes, how do you get a 10-year eclipse?
Well... you can't. First of all, please read this excellent answer at astronomy.SE. It talks about this, too, and comes to the same conclusion.
Problem #1
Assume the inner planet is 4X the diameter of the earth. It could have an orbit just a hair faster than Earth's that would let it cause an eclipse of 10 years (0.5 earth diameters/year faster than Earth). When you work out how much faster that it must orbit (convert orbital distance of Earth to units of "earth's diameter", call it "a", add 0.5 and call it "b", b/a kinda equals new period in terms of one year) and then play with this unbelievably cool orbital time calculator what you'll discover is that the center of a planet being 4X the diameter of the Earth must pass within (I kid you not) 418 miles (read that again) of the center of the Earth. Yup, they collided.
You could try to play games to make that work, but what you'll quickly discover is that the smaller you make the inner planet the closer it must be to the centerline of Earth's orbit to create the 10-year effect. You can't get any smaller than the diameter of the Earth or you won't have 100% occlusion. Ugh.
I'd like to point out that even if this did work, it would only happen once every bazillion years. It would be a bit impractical.
Problem #2
Let's try this from another angle. From here we learn the estimated umbral diameter (the shadow on the Earth) is...
$$\begin{align} Ud = \frac{2(mS - Ms)}{S - M} \end{align}$$
Where
We want Ud = Earth's diameter = 12,742 Km. Let's assume a better 0.8au (Venus is 0.72au) and find out how big it must be and how long the eclipse can be.
$$\begin{align} 6,371 = \frac{m(149,600,000) - (29,920,000)(695,700)}{149,600,000 - 29,920,000} \end{align}$$
m = 144,237 Km or more than 18X the diameter of the Earth. To give you an idea, Jupiter is only about 11X the diameter of the Earth. At 0.8au we have an orbital period of 0.72 year and, if my assumptions are correct (they might not be), an eclipse of about 12-16 hours with a "total blackout" of about 8 hours.
And that's the problem.
It's equivalent to the Transit of Venus if Venus were on a level plane with Earth's orbit and twice the diameter of Jupiter.
A nice side note
It appears you could have a super gas giant inside the orbit of Earth. Astronomers found a gas giant orbiting very close to a small star.
The planet in question (HATS-6b) weighs in at approximately the mass of Saturn, or 100 times the mass of Earth. But because of its close orbital distance, the star’s heat has caused HATS-6b’s gas to billow out, inflating the planet like a hot air balloon to the size of Jupiter. With an orbital period of just 3.3 days, astronomers say HATS-6b is significantly closer to its star than the much smaller Mercury is to our Sun.
Did you see that orbital period? 3.3 DAYS. Granted, it's closer than Mercury, but it means you could have a gas giant, or even a super gas giant, inside the orbit of earth. One large enough to complely occlude the sun from the entire planet. It wouldn't last 10 years...
But you would see it about 5 times every 4 years and it would cause wild-and-crazy weather during the "day of the long night." It would slow the earth down as it approached the eclipse and then speed it back up after (assuming it didn't drag Earth into becoming one of its moons...).
It's not the kind of cool you're looking for, but how cool is that?
