For the TL;DR, see the bottom of this answer.

Okay, so first of all, the orbital period of the gas giant around its star is $256\times 24$ hours, and I'd like to establish the distance from the planet to its star. Since you haven't specified anything about the star itself, I'll go with our Sun for simplicity's sake. Also for simplicity's sake (or to retain everybody's sanity, including my own) I will approach this as two two-body problems rather than a three-body problem. This reduces the attainable precision, but greatly simplifies the math. For a representative gas giant, I'll use Jupiter.

As an approximation for the planet's orbit around its star, we can use the formula for a small body orbiting a central body:

$$r=\sqrt[3]{\frac{\mu T^2}{4{\pi }^2}}$$

where:

• $r$ is the orbit's semi-major axis in meters (note: this is not the same thing as the orbital altitude, but can be approximated as the orbital radius)
• $\mu$ is the standard gravitational parameter, $\mu =GM$
  • $G$ is the gravitational constant, in units relevant here $6.67408\times {10}^{-11}{\text{m}}^3{\text{kg}}^{-1}{\text{s}}^{-2}$
  • $M$ is the mass of the central body (in this case, the star) in kg
  • ${\mu }_{\text{Sun}}\approx 1.327\times {10}^{20}{\text{m}}^3{\text{s}}^{-2}$
• $T$ is the orbital period in seconds

We know that the desired $T=256\times 24\times 60\times 60=22118400$ seconds. Let's plug all those values in and see what comes out:

$$r=\sqrt[3]{\frac{1.327\times {10}^{20}\times 22118{400}^2}{4{\pi }^2}}\approx \sqrt[3]{1.644442\times {10}^{33}}\approx 1.1803375\times {10}^{11}$$

So your planet orbits at a distance of about $1.2\times {10}^8$ km, or 120 million km, to its star. This is comparable to Venus' orbit around the Sun (Venus' semi-major axis is about $1.08\times {10}^8$ km, with an orbital period of $224.7\times 24$ hours). That's awfully close for a gas giant in anything resembling our solar system, but it's the only way to get the planet orbital period you ask for while keeping the star Sun-like. You could twiddle the knob for star mass ($M=M_{\text{Sun}}$, influencing ${\mu }_{\text{Sun}}$ above) until you are happy with the outcome; for inspiration, look no further than to Wikipedia's list of main sequence star example parameters which gives the stars' mass in terms of solar masses, from which you can calculate the corresponding value for $\mu$.

The arc length of a circle sector is given by $L=\theta \times r$ where $\theta$ is the angle subtended. We know the approximate arc length (the diameter of the Sun: twice its radius of $695700$ km) and distance ($1.2\times {10}^8$ km) and want angle subtended, so we get $$2\times 695700\text{km}=\theta \times 1.2\times {10}^8\text{km}\Rightarrow \theta =\frac{2\times 695700}{1.2\times {10}^8}=0.011595$$

Because $\theta$ comes out in radians, we multiply by 57.296° to get the angle subtended in degrees, which turns out to be 39.86 arcminutes or 0.664 degrees. A quick check against Wikipedia gives the Sun's angle subtended from Earth (at an orbital radius of $1.5\times {10}^8$ km) as 31.6-32.7 arcminutes, so while possibly not perfect, this is well within the ballpark. The same calculation for an orbital radius of $1.5\times {10}^8$ km gives 31.9 arcminutes, squarely in the range given.

You specified the orbital period of the moon around the gas giant to be 192 hours, or $192\times 3600=691200$ seconds. We can use the vis-viva equation to calculate the corresponding orbital radius. We have $$v^2=\mu (\frac{2}{r}-\frac{1}{a})$$

For a circular orbit, $r=a$ (orbital radius is equal to the semi-major axis of the orbit) and thus $${\left(\frac{2\pi r}{T}\right)}^2=\mu (\frac{2}{r}-\frac{1}{r})$$

We have ${\mu }_{\text{Jupiter}}\approx 1.267\times {10}^{17}{\text{m}}^3{\text{s}}^{-2}$ and $T=691200\text{s}$. By rearranging, we get $$r=\sqrt[3]{\mu {\left(\frac{T}{2\pi }\right)}^2}\approx 1153080\text{km}$$

So the Earth-like moon orbits the gas giant at an orbital radius of about 1.15 million km, because that's the orbital radius (for a perfectly circular orbit, one with eccentricity $e=0$ or semi-major axis equal to semi-minor axis) that corresponds with the desired orbital period. This happens to be very similar to the radius of the orbit of Ganymede (which is 1.07M km, and an eccentricity of about 0.0013 in case you wondered), providing a nice sanity check for the result; Ganymede orbits Jupiter in 171 hours, only slightly less than your desired 192 hours, so at least to a first order approximation this checks out.

The formula for computing the length of the umbra (central shadow) of an eclipse is $$L=\frac{r\times R_o}{R_s-R_o}$$ where $r$ is the distance from the star to the occulting object (in our case, the gas giant), $R_o$ is the radius of the occulting object, and $R_s$ is the radius of the star. We thus have a shadow cone of length (where all distance and size values are in kilometers) $$L=\frac{1.2\times {10}^8\times 71492}{695700-71492}\approx \frac{8.58\times {10}^{12}}{624208}\approx 13.74\times {10}^6$$

Because $1.15\times {10}^6<13.74\times {10}^6$, the moon passes through the shadow cone cast by the planet, so we have a full eclipse (the moon passes through the umbra cast by the planet). Now, for how long does the eclipse last?

By considering the shadow cone to be a triangle with the base length of the diameter of the planet and the height calculated above, we can use Pythagoras' theorem to calculate the length of the resulting hypotenuse. (This turns out to be almost identical to the height, approximately $1.37400\times {10}^7$ versus the height approximately $1.37439\times {10}^7$ km.) We can then apply the intercept theorem which states that when dividing a triangle by a line parallell to the base of the triangle, the length of the new base line is to the original base line as the hypotenuse of the part of the triangle is to the total hypotenuse length. By approximating the required inner height as the orbital radius of the moon around the gas giant, we end up with $$\frac{DE}{2\times 71492}=\frac{1150000}{13740000}\approx \frac{0.083697}{1}$$ where $DE$ is the length of the line connecting the edges of the triangle at the orbital radius of the moon. Hence the occluded path for the moon is approximately $\frac{2\times 71492}{0.083697}\approx 1.708\times {10}^6$ km. (This is really the base of a circle segment where the moon traces the circle segment, but the difference is small enough to be negligible at these levels of precision.)

The circumference of a circle of radius $1.15\times {10}^6$ km is $$2\pi r=2\pi \times 1.15\times {10}^6\approx 7.226\times {10}^6\text{km}$$

Thus, passing through the umbra cast by the planet takes $\frac{1.708\times {10}^6}{7.226\times {10}^6}\approx 0.2364$ of the orbital period of the moon. Multiplying by the orbital period of 192 hours gives us a duration of 45.4 hours within the total eclipse zone (the umbra).

Note that there are three things I am actually ignoring in the calculations above. First, I'm positing that all bodies are orbiting within your solar system's ecliptic; if their orbits are inclined relative to one another, you need to take the angle at which they are orbiting (the inclination) into account. Doing so complicates the math a fair bit for no significant gain, as bodies that form naturally within a solar system are likely to orbit close to the ecliptic. I'm leaving that entirely as an exercise for the reader.

Second, I'm ignoring the planet's orbital motion around the star. When the (Earth-like) moon is orbiting around the (gas giant) planet, and the (gas giant) planet is orbiting around the star, this is going to have the effect of either making the apparent eclipse slightly shorter or slightly longer. (Which it happens to become depends on the relative direction of the orbital movement.) I'm too lazy to account for this, so I just don't, but it shouldn't be more than some trigonometry if you really care to do that part of the math yourself.

Third, I'm ignoring the fact that the Earth-sized moon is going to tug a little at the planet. The barycenter of the system won't actually be at the planet's center, but a little outside of the planet's center, which will cause the two to be joined in somewhat of an orbital dance. This is very similar to how, in our solar system, Jupiter perturbs the Sun, despite being only $\frac{1}{1047}$ the mass, or how Earth's moon perturbs Earth.

TL;DR:

One set of values that match your criteria are:

• Star mass $1.99\times {10}^{30}$ kg (1 solar mass) (by choice)
• Star diameter $1391400$ km (1 solar diameter) (by choice)
• Planet mass $1.8986\times {10}^{27}$ kg (1 Jupiter mass) (by choice)
• Planet diameter $142984$ km (1 Jupiter diameter) (by choice)
• Planet orbital period around star $256\times 86400$ seconds (by decree)
• Planet orbital radius around star $1.2\times {10}^8$ km
• Moon orbital period around planet $192\times 3600$ seconds (by decree)
• Moon orbital radius around planet $1.15\times {10}^6$ km
• Eclipse length $45.4\times 3600$ seconds

There are many other sets of values that can match your criteria. If you aren't happy with the above, just pick different values for the masses and radii involved, and recompute; there is nothing magical about the sizes of Sun or Jupiter. Just don't forget to change the value of $\mu$ accordingly!

posted about 8 years ago

Canina‭

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