I was trying to solve this problem myself but i dont know how to work out certain variables like gravity, centripetal force and the changing impact point due to the curve of a planet (any planet).

So i wondered if you guys could try and help me giving out some of the intuition to get the result and i also wondered if its possible to describe the motion of a projectile over a planet using polar coordinates.

So yeah, we have a initial launching point from the surface of a circle, and then a initial velocity and initial angle of launch, how do i know what will the distance over the surface up until the impact point be for that projectile accounting for the planetary body circular shape and the changing gravitational force?

  • $\begingroup$ This has effectively been answered here. $\endgroup$ Jan 1, 2022 at 17:32

2 Answers 2


If you are happy to ignore atmospheric drag and the rotation of the planet then the projectile is in orbit as soon as it is launched. As long as it’s launch speed is lower than the planet’s escape velocity then it will follow an elliptical path until the ellipse intersects the surface of the planet again. The orbital parameters are determined by the speed and direction of launch.

If your planet has an atmosphere and/or is rotating then the solution will be more complex.

  • 2
    $\begingroup$ So in effect you will need to try and find two intersection points between a circle (the Earth) and an ellipse (the trajectory of the projectile) which is a geometry problem. $\endgroup$
    – Farcher
    Jan 1, 2022 at 17:16
  • 1
    $\begingroup$ @Farcher Yes. However, you must first compute the elliptical orbit, which is a physical problem... $\endgroup$
    – Alfred
    Jan 1, 2022 at 17:31
  • $\begingroup$ thanks for the input! i think you guys gave me some confidence on the direction i should go for to solve it, do you also have any sort of reading material on that theme? (by that i mean, orbital mechanics for beginners (?)) $\endgroup$
    – jfinizolas
    Jan 12, 2022 at 22:48

I tried to figure this too. I had this hypothesis: if rocket is launched to some angle instead of straight up, Apoapsis should be more high because planet has curvature.

I tried to calculate how high my Apoapsis would be with this hypothesis. Tried this at Kerbal Space Program and whenever i launched any other angle than UP, Apoapsis was lower.

I figured out that while you move horizontal, gravity pull changes coordination of trajectory, so curvature doesn't have any affect on calculations and therefore distance traveled is same as calculating it with trajectory formula.

Distance traveled horizontally formula is: 2*cos(angle)Velocitysin(angle)*velocity/gravity.


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