Editing Projectile motion (section)

== Projectile motion on a planetary scale ==
[[File:Ballistic-trajectories-planet2.svg|thumb|left|Projectile trajectory around a planet, compared to the motion in a uniform gravity field]]
When a projectile travels a range that is significant compared to the Earth's radius (above ≈100&nbsp;km), the [[curvature of the Earth]] and the non-uniform [[Earth's gravity]] have to be considered. This is, for example, the case with spacecrafts and intercontinental [[Ballistic missile|missiles]]. The trajectory then generalizes (without air resistance) from a parabola to a Kepler-[[ellipse]] with one focus at the center of the Earth (shown in fig. 3). The projectile motion then follows [[Kepler's laws of planetary motion]].

The trajectory's parameters have to be adapted from the values of a uniform gravity field stated above. The [[Earth radius]] is taken as <var>R</var>, and <var>g</var> as the standard surface gravity. Let <math>\tilde v:=v/\sqrt{Rg}</math> be the launch velocity relative to the first cosmic or [[escape velocity]].

Total range <var>d</var> between launch and impact:
: <math> d = \frac{v^2 \sin(2 \theta)}{g} \Big/ \sqrt{1-\left(2-\tilde v^2\right)\tilde v^2\cos^2\theta}</math>     (where launch angle  <math>\theta=\tfrac12\arccos\left(\tilde v^2/(2-\tilde v^2)\right)</math>)
Maximum range of a projectile for optimum launch angle θ=45<sup>o</sup>:
: <math> d_{\mathrm{max}} = \frac{v^2}{g} \big/ \left(1-\tfrac12\tilde v^2\right)</math> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; with <math>v<\sqrt{Rg}</math>, the [[first cosmic velocity]]
Maximum height of a projectile above the planetary surface:
:<math> h = \frac{v^2 \sin^2\theta}{g} \Big/ \left(1-\tilde v^2+\sqrt{1-\left(2-\tilde v^2\right)\tilde v^2\cos^2\theta}\right) </math>
Maximum height of a projectile for vertical launch (<math>\theta=90^\circ</math>):
:<math> h_{\mathrm{max}} = \frac{v^2}{2g} \big/ \left(1-\tfrac12\tilde v^2\right) </math> &nbsp;&nbsp;&nbsp;&nbsp;&nbsp; with <math>v<\sqrt{2Rg}</math>, the [[second cosmic velocity]], 
Time of flight:
:<math> t = \frac{2v\sin\theta}{g} \cdot \frac{1}{2-\tilde v^2} \left(1 + \frac{1}{\sqrt{2-\tilde v^2}\,\tilde v\sin\theta}\arcsin\frac{\sqrt{2-\tilde v^2}\,\tilde v\sin\theta}{\sqrt{1-\left(2-\tilde v^2\right)\tilde v^2\cos^2\theta}}\right) </math>