# Tag Info

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: $$... 6 The definition of Hill sphere is the region where the given object's gravity is dominant. In this area the object's gravity pulls more strongly than anything else; and everything else combined. The primary competition for a planet is the sun. The further you get from the sun then the weaker its gravity is. This means it's easier for Neptune's gravity to ... 3 The sidereal year is 6 hours and 9 minutes longer than a 365 day calendar year. To simplify your question, assuming you an your friend were born on the same date and time in 1983 and 1984 you would be 365 days, 6 hours, and 9 minutes older than your friend. Since there is no leap day between the two times you are considering, it doesn't affect the ... 2 I take it there isn't exactly a coincidence between you finding my question regarding geostationary orbits and you asking this question about Hill Spheres? :-) The chart you found does seem counter-intuitive at first. But consider this chart: https://upload.wikimedia.org/wikipedia/en/timeline/5fb1322f537f8a55d85170976c150191.png (I wish I could add it ... 2 I'll use YYYY-MM-DD notation. You were born 1983-02-06 07:00. Your friend was born 1984-02-07 01:00. There was no leap day between your birthdays, which makes the calculation a little simpler. The most recent leap day before your birthday was 1980-02-29; the earliest leap day after your friend's birthday was 1984-02-29. (No, 1982 was not a leap year, and ... 2 Comets don't cross Earth's orbit really. Orbits are one-dimensional objects and their chance of crossing in 3D space is 0. Henceforth, I consider a comet at distance 1AU from the Sun. What's the maximum speed of a returning comet at 1AU from the Sun? This can be easily worked out from the orbial energy$$ E = \frac{1}{2}v^2 - ...