Imagine if Jupiter orbited the Sun at the Earth's orbital distance of 1AU.
Would a planet of this size orbiting the Sun take 1 Earth year (365 days) to complete an orbit, or would the size of Jupiter affect its rotation and orbit?
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If Jupiter was at 1 AU from the Sun, its orbit about the Sun would be about 4 hours and 10 minutes shorter than a sidereal year. That's an effect, but it's not very much of one, due to Jupiter's mass being about 1/1000th of the Sun's mass. For a more significant effect, if our Moon was much smaller than it is, it would orbit the Earth about 4 hours and 1 minute longer than a sidereal month. The Moon's mass is about 0.0123 Earth masses.
The period of the Keplerian orbit of an object of negligible mass about some massive object is $$P = 2\pi\sqrt{\frac{a^3}{GM}}$$ In the above, $a$ is the semimajor axis length, $G$ is the Newtonian gravitational constant, and $M$ is the mass of the central object. When the orbiting object has non-negligible mass $m$, the above expression needs to be modified to $$P = 2\pi\sqrt{\frac{a^3}{G(M+m)}}$$
answered Aug 17, 2016 at 13:32
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$\begingroup$ Just to put in some numbers, if I use the full equation for the period, I get the Earth's orbital time to be $P_{Earth} = 365.268\:\mathrm{days}$ whereas for Jupiter I get $P_{Jupiter} = 365.095\:\mathrm{days}$. This means Jupiter's mass does have a measurable difference in the length of the year if it orbited at $1\:AU$, albeit not a very practical difference. At worst, the system of leap days would change under such an orbit. $\endgroup$– zephyrAug 17, 2016 at 14:00