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Comets don't cross Earth's orbit really. Orbits are one-dimensional objects and their chance of crossing in 3D space is 0. Henceforth, I consider a comet at distance 1AU from the Sun. What's the maximum speed of a returning comet at 1AU from the Sun? This can be easily worked out from the orbial energy $$E = \frac{1}{2}v^2 - ... 1 There are relatively big varieties, but most of them is between 10 and 70 km/s. If a comet is a periodic comet, that means it needs to have an elliptic orbit around the Sun. That gives an upper limit to its speed of the escape speed from the solar system on the orbit of the Earth. That is around 40 km/s. But this 40 km/s is in the reference frame of the ... 6 Hill sphere is the region of space around a satellite where the satellite wins the gravitational tug-of-war with its primary. If the mass of the primary object is M, mass of the satellite is m, semi-major axis of satellite is a, and eccentricity of the orbit of the satellite is e, then the radius r of the Hill sphere for satellite is given by:$$ ...

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The definition of Hill sphere is the region where the given object's gravity is dominant. In this area the object's gravity pulls more strongly than anything else; and everything else combined. The primary competition for a planet is the sun. The further you get from the sun then the weaker its gravity is. This means it's easier for Neptune's gravity to ...

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I take it there isn't exactly a coincidence between you finding my question regarding geostationary orbits and you asking this question about Hill Spheres? :-) The chart you found does seem counter-intuitive at first. But consider this chart: https://upload.wikimedia.org/wikipedia/en/timeline/5fb1322f537f8a55d85170976c150191.png (I wish I could add it ...

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I'll use YYYY-MM-DD notation. You were born 1983-02-06 07:00. Your friend was born 1984-02-07 01:00. There was no leap day between your birthdays, which makes the calculation a little simpler. The most recent leap day before your birthday was 1980-02-29; the earliest leap day after your friend's birthday was 1984-02-29. (No, 1982 was not a leap year, and ...

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The sidereal year is 6 hours and 9 minutes longer than a 365 day calendar year. To simplify your question, assuming you an your friend were born on the same date and time in 1983 and 1984 you would be 365 days, 6 hours, and 9 minutes older than your friend. Since there is no leap day between the two times you are considering, it doesn't affect the ...

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charon and pluto are tidally locked, as is the moon to the earth. so you could say that earth is in a geostationary orbit with the moon. in fact, the earth rotation has been and will continue to be slowed by the moon until the moon is in a geostationary orbit with earth.

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Yes. Charon is in Pluto synchronous orbit. Pluto and Charon are mutually tide locked.

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Of course, a natural satellite (moon) could have an orbital period equal to the spin period of its host (provided such an orbit would be accessible). However, the tidal friction that may generate such a locking is quite weak, so this would have to be a rare chance. Moreover, perturbations to the orbit from other moons or their host star may put the moon out ...

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It would require a very precise trajectory for an asteroid to end up in geostationary orbit. It doesn't happen by chance. Space flight providing companies have to make a real effort to put their customers' communication satellites there. And geostationary isn't a very stable kind of orbit. The varying gravity of the Moon pulls satellites out of their ...

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