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Elementary Newtonian mechanics tells us that if a projectile is shot at 45 degrees from the ground, and a downward gravitational field is acting on it, it will follow a parabolic path and achieve maximum range.

But the Earth's gravitational field rather points radially inward, meaning that the projectile instead follows an elliptical path. What angle from the ground would allow it to achieve maximum range?

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    $\begingroup$ In true projectile, effect of air resistance is so much more than variation of gravitational field. $\endgroup$ – lucas Jun 24 '16 at 11:25
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    $\begingroup$ Way too true for me. I'll change the title $\endgroup$ – giobrach Jun 24 '16 at 11:27
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    $\begingroup$ If you want to consider to variation of $g$, then you should consider to curvature of the earth surface too. Because the earth's surface isn't flat (horizontal). $\endgroup$ – lucas Jun 24 '16 at 11:34
  • $\begingroup$ Interesting question. What attempt(s) have you made to solve it yourself? $\endgroup$ – sammy gerbil Jun 24 '16 at 12:38
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    $\begingroup$ You might want to edit the question to limit the initial velocity to values less than the orbital velocity ... if that's your intent. And that there is no atmosphere, and that the earth is spherical, and that you are at the equator. There are a lot of pendants on this site. :) $\endgroup$ – garyp Jun 24 '16 at 14:06
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If we are at the equator on the surface of an atmosphere-free, non-rotating, perfectly spherical planet (no mountains or trees, etc) then the optimal angle should be zero (launch towards the horizon, tangent to the planet's surface).

This assumes the initial velocity is the orbital speed at the surface of the planet, which is approximately the escape velocity divided by the square root of two. The range is then infinite, assuming by "range" we mean the difference between the place of launch and the place where the projectile hits (it never hits the ground, in this idealized situation). This is based purely on classical mechanics.

If by "range" we mean getting as far as possible from the planet, then the initial velocity needs to be escape velocity so that the projectile escapes from the planet entirely. If the planet is not rotating then I believe the direction of launch does not matter in this case.

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  • $\begingroup$ Interesting, and arguably a correct analysis, but I think implicit in the question is the initial speed being less than orbital speed. $\endgroup$ – garyp Jun 24 '16 at 14:05
  • $\begingroup$ Very insightful answer! Just a minor suggestion. Wouldn't it be perfectly okay even if we are not at the equator - we are anyway assuming perfect sphericity, right? $\endgroup$ – Dvij D.C. Jun 24 '16 at 14:10
  • $\begingroup$ @Dvij Yes you are right - good spot, and thanks for your comment! I first started to answer in the general case where there is planetary rotation as well as orbital speed around a home star. But this took away from the clarity so I edited the answer down to this one - guess I forgot to take that bit out about launching from the equator. $\endgroup$ – RiskyScientist Jun 24 '16 at 14:30

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