# Newton cannonball threshold velocity

In the classical "Newton cannonball experiment", it is known that as the velocity of the "horizontally" fired cannonball increases, the orbit changes, after an initial threshold is passed, from a circular closed orbit to an elliptical closed one, then to an open parabolic, and on to an open hyperbolic.

But where can i find a detailed analysis of this procedure? I'm particularly puzzled by the threshold value at which the orbit changes from a closed ellipse to an open parabola. It seems amazing that by a continuous variation of the initial velocity, a closed ellipse suddenly becomes an open parabola!

• What's the difference, in practice, between a closed ellipse whose apogee is a million light-years away (and with an orbital period longer than the age of the universe) and an open parabola? – Emilio Pisanty Nov 13 '17 at 19:48
• P.S., there's a name for that threshold: It's escape velocity. – Solomon Slow Nov 13 '17 at 21:10

You are essentially talking about Kepler's problem. Given 'horizontal' initial velocity $v_{0}$ at radius $R$, you can calculate both the energy and the angular momentum of the orbit

$$L=mv_{0}R$$ $$E=-\frac{GMm}{R}+\frac{1}{2}mv_{0}^2$$

Here $m$ is the mass of the particle and $M$ is the mass of earth. It can be shown that this type of motion posses an effective potential

$$V_{\rm eff}(r)=\frac{L^2}{2mr^2}-\frac{GMm}{r}$$

such that the total energy satisfies

$$E=\frac{1}{2}m\dot{r}^2+V_{\rm eff}(r)$$

When looking at this potential, which describes effective 1D motion in the radial direction, one can observe the different regimes you've described.

In particular, the case of parabolic orbit corresponds to $E=0$, which after solving gives you the threshold velocity

$$v_{\rm th}=\sqrt{\frac{2GM}{R}}$$