Editing Orbit equation (section)

==Low-energy trajectories==
If the central body is the Earth, and the energy is only slightly larger than the potential energy at the surface of the Earth, then the orbit is elliptic with eccentricity close to 1 and one end of the ellipse just beyond the center of the Earth, and the other end just above the surface. Only a small part of the ellipse is applicable.

If the horizontal speed is <math>v\,\!</math>, then the [[periapsis distance]] is <math>\frac{v^2}{2g}</math>. The energy at the surface of the Earth corresponds to that of an elliptic orbit with <math>a=R/2\,\!</math> (with <math>R\,\!</math> the radius of the Earth), which can not actually exist because it is an ellipse fully below the surface. The [[Specific orbital energy#Rate of change|energy increase with increase]] of <math>a</math> is at a rate <math>2g\,\!</math>. The maximum height above the surface of the orbit is the length of the ellipse, minus <math>R\,\!</math>, minus the part "below" the center of the Earth, hence twice the increase of <math>a\,\!</math> minus the periapsis distance. At the top{{of what|date=November 2020}} the potential energy is <math>g</math> times this height, and the kinetic energy is <math>\frac{v^2}{2}</math>. This adds up to the energy increase just mentioned. The width of the ellipse is 19 minutes{{why|date=November 2020}} times <math>v\,\!</math>.

The part of the ellipse above the surface can be  approximated by a part of a parabola, which is obtained in a model where gravity is assumed constant. This should be distinguished from the parabolic orbit in the sense of astrodynamics, where the velocity is the [[escape velocity]].

See also [[trajectory]].