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