The Titan with a vision of the future

Let the point of projection be $P$ and earth be $E$ with mass M.

for the whole problem, think 2D. we will rotate our envelope perimeter along PE and get the ellipsoid

let the mass of a fireball be m.now take any fireball orbit. let the length of semi major axis be $a$ .

we know $a=-\frac{GMm}{2E}$ where E is the total mechanical energy of the fireball.

escape velocity is $V_e=\sqrt{\frac{2GM}{d}}$ . let velocity of throw be $V = xV_e$ where $x=0.6$

so $E = -\frac{GMm(1-x^2)}{d}$

so $a = \frac{d}{2(1-x^2)}$

now for the orbit,let the second focus be $S$ . now $SP+PE=2a , PE = d => SP=2a-d$ ,i.e locus of S is a circle or radius $2a-d$ and center P.

for the envelope: it should be the locus of a point which can never be inside any orbit, but must touch atleast one orbit. so X be a point on the envelope.

$SX+EX \ge 2a$ for all S and equality holds for atlest one S.

minimum of SX is $\left|PX - (2a-d)\right|$ .

so equality must hold at $SX = \left|PX - (2a-d)\right|$

so $\left|PX - (2a-d)\right|+EX = 2a$

obviously $PX > 2a-d$ because every point between two foci of an an ellipse lies inside the ellipse.

so $PX+EX = 4a-d$ and $EP=d$ . so envelope is an ellipse with foci E&P

so $e = \frac{EP}{PX+EX} = \frac{1-x^2}{1+x^2} = \frac{8}{17}$

$\frac{1}{e} = 2.125$

A Computed Picture depicting the formation of ellipsoid: