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    $\begingroup$ Did you have a look at the derivation of the escape velocity? That will answer your question. $\endgroup$
    – AtmosphericPrisonEscape
    Commented Nov 10, 2023 at 0:59
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    $\begingroup$ Note (to everyone) that it is the escape speed, since you are taking the square root of an equation for $v_{esc}^2$. The direction in which the object is launched does not matter (in Newtonian physics, and so long as it is directed away from the surface), so it is not an "escape evelocity". $\endgroup$
    – ProfRob
    Commented Nov 10, 2023 at 12:16
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    $\begingroup$ @AtmosphericPrisonEscape - Not all AstronomySE users (nor all AstronomySE readers) have mastered integral calculus to the point where the answer of this question is obvious or the derivation of escape velocity is easy to follow and understand. I wouldn't close the question because it's useful for a subset of readers. $\endgroup$
    – Pere
    Commented Nov 10, 2023 at 14:50
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    $\begingroup$ Something that always felt funny to me was that clearly the direction matters in practical situations. You won't escape if you plow into the planet. But the formulation of escape velocity is based around a point-source gravitational object (or, more completely, that the radius of the object is much smaller than the dimensions of the orbit). Remembering that the surface of the planet doesn't exist was helpful for me in accepting that direction doens't matter. $\endgroup$
    – Cort Ammon
    Commented Nov 10, 2023 at 16:03
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    $\begingroup$ @Pere This was in relation to the rudeness of OP's comments, which were abusive of anyone prospectively helping them. They, by now, put their tail between their legs and deleted those comments. $\endgroup$
    – AtmosphericPrisonEscape
    Commented Nov 10, 2023 at 19:40