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A black hole forms if the gravitational force overcomes all other forces resisting the collapse. In stars, gas pressure balances gravity and they are in hydrostatic equilibrium. A star constantly looses energy by radiation, but that is compensated by fusion in its core. Once the fusion energy dries up, the (core of the) star contract. This releases ...


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You are correct in saying that a star loses a lot of its mass in a supernova. However, there is a reason why the star still becomes a black hole. Actually, I suppose the question here is "Why doesn't a star become a black hole before it even undergoes a supernova?" There is a reason for a supernova (I'll assume you're talking about type II supernovae, which ...


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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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There is no gravitational phenomenon associated with exact alignment. (moreover, exact alignment never happens, there is always a small deviation). The directions of tidal forces are aligned and hence sum-up for alignment, but this maximum is quite broad and nothing spectacular happens near exact alignment. You need close to exact alignment (depending on ...


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Referring to your #gravity tag this would only create a peak in the planets tidal forces. But this would be a visible effect in case the planets would be big and close enough... Last night happened that with Sun-Earth-Moon although they were not aligned as you say. To sum up: a proper alignment would increase the effect of tidal forces.


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You may want to use perturbation theory. This only gives you an approximate answer, but allows for analytic treatment. Your force is considered a small perturbation to the Keplerian elliptic orbit and the resulting equations of motion are expanded in powers of $K$. For linear perturbation theory, only terms linear in $K$ are retained. This simply leads to ...



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