Interesting article, robphy!
We don't need to bound two parameters in order to get a parabolic approximation, though.
It is sufficient that [tex]\gamma=\frac{V_{0}^{2}}{gR}<<1[/tex]
where [tex]V_{0},\sqrt{gR}[/tex] are the launch speed and escape velocity, respectively (R is the radius of the Earth, g the acceleration due to gravity).
We have:
1. 2-D Coordinate choice:
Let y measure the radial distance above the Earth's surface, whereas [tex]x=R\theta[/tex] measures the arclength along the Earth's surface, where [tex]\theta[/tex] is the angular displacement.
2. Projectile description:
a) Position vector: [tex]\vec{r}(t)=(R+y(t))\vec{i}_{r}[/tex]
where [tex]\vec{i}_{r}[/tex] is the radial unit vector
(with [tex]\vec{i}_{\theta}=\frac{d\vec{i}_{r}}{d\theta}[/tex] orthogonal to [tex]\vec{i}_{r}[/tex])
We also define [tex]\epsilon=\frac{y(t)}{R}[/tex]
(which clearly must be greater than 0)
b) Velocities:
bi) [tex]\dot{y}[/tex] radial velocity component of projectile
bii) [tex]\dot{x}=R\dot{\theta}[/tex] arc-length velocity
biii) [tex](1+\epsilon)\dot{x}[/tex] projectile's velocity component in direction [tex]\vec{i}_{\theta}[/tex]
c) Geometric trajectory description:
We introduce [tex]Y(x(t))=y(t)[/tex], so that Y(x) is the traversed curve as seen in the 2-D coordinate system.
d) Initial conditions:
[tex]y(0)=0,x(0)=0,V_{y,0}=\dot{y}(0)=\dot{y}_{0},\dot{x}(0)=\dot{x}_{0},V_{x,0}(0)=(1+\frac{y(0)}{R})\dot{x}_{0}=\dot{x}_{0}[/tex]
3) Conservation of angular momentum
Angular momentum is conserved with respect to the Earth's center, i.e we have (cancelling the mass):
[tex]V_{x}=\frac{V_{x,0}}{1+\epsilon}=\frac{\dot{x}_{0}}{1+\epsilon},\dot{x}=\frac{\dot{x}_{0}}{(1+\epsilon)^{2}}[/tex]
4) Expressions used in the radial equation:
[tex]\frac{d^{2}\vec{r}}{dt^{2}}\cdot\vec{i}_{r}=(\ddot{y}-\frac{(1+\epsilon)}{R}\dot{x}^{2})=(\ddot{y}-\frac{\dot{x}^{2}_{0}}{R(1+\epsilon)^{3}})[/tex]
Furthermore:
[tex]\dot{y}=\frac{dY}{dx}\dot{x}=\frac{dY}{dx}\frac{\dot{x}_{0}}{(1+\epsilon)^{2}}[/tex]
And:
[tex]\ddot{y}=\frac{d^{2}Y}{dx^{2}}\frac{\dot{x}_{0}^{2}}{(1+\epsilon)^{4}}-(\frac{dY}{dx})^{2}\frac{2\dot{x}_{0}^{2}}{R(1+\epsilon)^{5}}[/tex]
5) Radial component of Newton's 2.law pr. unit mass
Thus, we gain:
[tex]\frac{d^{2}Y}{dx^{2}}=-\frac{g}{\dot{x}_{0}^{2}}(1+\epsilon)^{2}(1-\frac{\dot{x}_{0}^{2}}{gR}(\frac{2(\frac{dY}{dx})^{2}}{(1+\epsilon)^{3}}+\frac{1}{(1+\epsilon)}))[/tex]
6) Speed and curve length s:
We have the relation for the curve-length s:
[tex]\frac{ds}{dx}=\sqrt{1+(\frac{dY}{dx})^{2}}[/tex]
and also:
[tex]\frac{ds}{dx}=\frac{\frac{ds}{dt}}{\frac{dx}{dt}}=\frac{V}{V_{x}}=\frac{V}{\dot{x}_{0}}(1+\epsilon)[/tex]
where V is the speed of the particle.
Thus, we may rewrite Newton's law as follows:
[tex]\frac{d^{2}Y}{dx^{2}}=-\frac{g}{\dot{x}_{0}^{2}}(1+\epsilon)^{2}(1-\frac{V^{2}}{gR}(\frac{2}{(1+\epsilon)}+(\frac{\dot{x}_{0}}{V})^{2}(\frac{1}{(1+\epsilon)}-\frac{2}{(1+\epsilon)^{3}}))[/tex]
7) Order of magnitude estimates
a) The bracket [tex](\frac{2}{(1+\epsilon)}+(\frac{\dot{x}_{0}}{V})^{2}(\frac{1}{(1+\epsilon)}-\frac{2}{(1+\epsilon)^{3}}))[/tex] is positive, and if the internal bracket [tex]}(\frac{1}{(1+\epsilon)}-\frac{2}{(1+\epsilon)^{3}})[/tex] is negative, then we have:
[tex]0\leq(\frac{2}{(1+\epsilon)}+(\frac{\dot{x}_{0}}{V})^{2}(\frac{1}{(1+\epsilon)}-\frac{2}{(1+\epsilon)^{3}}))\leq\frac{2}{1+\epsilon}\leq{2}[/tex]
whereas if the internal bracket is positive, then the whole bracket is maximized when V is minimized, i.e, at the apogee where [tex]\frac{dY}{dx}=0[/tex]
Thus, in order to neglect the bracket expression with respect to unity, we must either have
[tex]\frac{2V^{2}}{gR}\leq\frac{2V_{0}^{2}}{gR}=2\gamma<<1[/tex]
or:
[tex]\frac{\dot{x}_{0}^{2}}{gR(1+\epsilon)}\leq\frac{V_{0}^{2}}{gR}=\gamma<<1[/tex]
It remains to be shown that [tex]\gamma<<1\to\epsilon<<1[/tex]
This is easiest done by energy conservation, noting that
[tex]\gamma<<1\to\frac{V_{a}^{2}}{gR}<<1[/tex]
where [tex]V_{a}[/tex] is the speed at the apogee.
It follows, that the allowable value of [tex]\epsilon[/tex]:
[tex]\gamma<<1\to\epsilon\approx\frac{\dot{y}_{0}^{2}}{2gR}\frac{1}{1-\frac{\dot{y}_{0}^{2}}{2gR}}<<1[/tex]