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9

Short answer Short answer is no, it can't prove or disprove the existence of Planet 9. The reason is because, even if there is a significant difference between the barycenter of the solar system with and without Planet 9 we wouldn't be able to tell without hundreds if not thousands of years of precise data. If we don't have Planet 9 as a reference and are ...


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The motion of bodies orbiting anything, is described by orbital elements - their distance from the barycenter (central mass), direction, velocity - and so two bodies of mass significantly lower than the central mass, with identical orbital elements, will move along the same orbit; with very similar elements their orbits will be very similar. For example, an ...


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Planetary magnitudes vary not only according to the Sun’s luminosity, their own average albedo, and their distance from the Earth, but also from: Variations in their albedo across their surface. Their phase angle, for planets that we sometimes see as a crescent. Their inclination, for planets like Saturn and Uranus that have a different albedo at their ...


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This configuration should be possible if the planet is located sufficiently far from the barycentre (i.e. sufficiently far outside the outermost orbit of the triple): close-in orbits will end up being destabilised. So far there are no confirmed exoplanets orbiting the common barycentre of three (or more) stars: while planets are known in triple star systems, ...


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To quote from Batygin and Brown (2016) Trujillo & Sheppard (2014) point out that the Kozai mechanism allows libration about both ω = 0 as well as ω = 180, and the lack of ω ~ 180 objects suggests that some additional process originally caused the objects to obtain ω ~ 0. To this end, they invoke a strong stellar encounter to generate the desired ...


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Mars (the planet with an axial tilt closest to that of the Earth) has an axial tilt that varies greatly over time. For instance, 4 million years ago, the mean obliquity was $∼35 ± 10°$. From the same article: Earth on the other hand has had its rotational features stabilized due to the effect of our large Moon. So it is only by chance if the axial tilt of ...


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The formula for orbital period is given on Wikipedia: $$T=2\pi \sqrt\frac{a^3}\mu$$ where: $T$ is the orbital period in seconds $a$ is the orbit's semi-major axis in meters $\mu = GM$ is the standard gravitational parameter $G$ is the gravitational constant $M$ is the mass of the more massive body in kilograms So $T = 2 \pi \sqrt \frac { (172 \cdot 10^9) ^...


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If some mystical thing, for example an interstellar giant from the https://worldbuilding.stackexchange.com, would push the Sun, and only the Sun, then the rest of the solar system would not follow. They would eject from their orbit, or they would change orbit, depending upon the strength of the push. However, no such thing exists. What exists: the Sun is ...


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If you want correct values, you have to take into account the effects mentioned in Brandon Rhodes' answer. Nevertheless, here's how to do a quick-and-dirty calculation. The absolute magnitude of a planet is defined as the apparent magnitude if the Sun-planet and planet-observer distances are 1 au, at opposition. Assuming a diffuse disc reflector model, the ...


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