# Proof for elliptical orbits

Its mentioned in several books that a satellite launched with a velocity less than the escape velocity and other than the critical velocity will follow an elliptical orbit. However I can't find a derivation of its equation of trajectory.

• en.wikipedia.org/wiki/… May 25, 2020 at 6:31
• I don't see how this is a homework question. Mar 18 at 6:29

This problem is most easily done in polar coordinates. In polar coordinates, the Newtonian gravity is given by $$\mathbf{g} = -\frac{GM}{r^2} \mathbf{\hat{r}}$$
Applying Newton's Second Law in polar coordinates gives two differential equations: one radial and one angular: $$\ddot{r} - r \dot{\theta}^2 = -\frac{GM}{r^2} \tag {1}$$ $$r \ddot{\theta} + 2 \dot r \dot \theta = 0 \tag{2}$$
Multiplying $$(2)$$ by $$r$$, we observe that $$r^2 \dot \theta = h$$ is a constant. This is really just a result of conservation of angular momentum, as there is no force in the angular direction. $$h$$ is nothing more than the specific angular momentum of the orbiting body.
By substituting $$u = 1/r$$ into $$(1)$$ and $$(2)$$ and solving, it can be shown that $$\frac{\text{d}^2u}{\text{d} \theta^2} + u= \frac{GM}{h^2}$$ This can then be solved to obtain$$r=\frac{h^2}{GM (1+e \cos (\theta - \theta_0))}$$ where $$e$$ is the eccentricity.
This is the equation we were looking for: a conic section. $$e =0$$ gives a circle, $$0 \lt e \lt 1$$ gives an ellipse, $$e=1$$ gives a parabola, and $$e \gt 1$$ gives a hyperbola.