Assuming a star with similar properties to our sun, could a gas giant orbit it with an orbital period similar to that of Saturn, but, at a much closer orbital distance, more similar to 1 AU?
1 Answer
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Gas giants can be found very close to their suns. The closest ones are then called Hot Jupiters.
But not with a orbital period of 30 years. Everything that orbits a star with one solar mass at 1 AU will also have an orbital period of 1 year, according to Keplers 3rd law:
$$ T^{2}=\frac{4 \pi^{2}}{G M} a^{3}. $$
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$\begingroup$ OK. That fits with Ethereal Mechanics as well. The model where the aether moves around stars (and galaxies, planets), and planets move with the aether. Explaining Michelson–Morley experiment giving null result. In EM, the orbital period is equivalent to the speed of the aether, and like in a hurricane, decreases with distance. $\endgroup$– orbusDec 13, 2019 at 11:18
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2$\begingroup$ @orbus For a 30 year period at 1 AU, the combined mass of the planet & star equals 1/900 solar masses. But for a planet with the mass of Saturn, the "star" mass is just under 2.9 times the mass of Saturn, which is far too small for a star: it's actually lighter than Jupiter. $\endgroup$– PM 2RingDec 13, 2019 at 11:56
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2$\begingroup$ @orbus The exact Kepler's law is $T^2 = 4\pi^2a^3/G(M+m)$, where $m$ is usually neglected if it's much smaller than $M$. You can play around with values of $T$, $a$, $M$, and $m$ to see if any values fulfil your scenario. For Sirius, I think you're out of luck, though… $\endgroup$– pelaDec 13, 2019 at 12:08
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3$\begingroup$ @orbus You are using $a$ in AU but the rest of the quantities in SI units. Using multiple systems of units in a calculation often makes the answer difficult to interpret. If you want a form that's easier to use, $$T^{2}=\frac{a^{3}}{M+m}$$ works for $T$ in years, $a$ in AU, and $M,m$ in solar masses. $\endgroup$ Dec 13, 2019 at 12:45
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1$\begingroup$ @orbus Do you know Python? With the module
astropyyou can do calculations with units, so that you don't have to think about that. I love it! $\endgroup$– pelaDec 13, 2019 at 14:28