The Earthly Ellipse!

The first thing to note is that the Earth will be at the focus of the elliptical path. Now, let the distance of the particle from the earth at either apogee or perigee be $d$ . Now, $d$ will take two values: $a(1+e)$ or $a(1-e)$ where $a, e$ are the semi major axis and eccentricity of the elliptical orbit respectively.

Now, by conservation of angular momentum, we have:

$u\cos\theta R = v d$ where $v$ is the velocity of the particle at the distance $d$ . Now here's the benefit of choosing apogee or perigee. The velocity vector and the position vector (w.r.t centre of earth) of the particle are perpendicular, so the angular momentum is simply $mvd$ - magnitude wise that is.

Also, energy is conserved. So,

$\dfrac{u^2}2 - \dfrac{ GM } {R} = \dfrac{v^2}2 - \dfrac{ GM } {d}$ .

Substituting for $v$ in the above equation, we get a quadratic in $d$ , which after some algebra, can be written as:

$d^2\left( u^2 - \dfrac{2GM}{R} \right) + (2GM)d - R^2u^2\cos^2\theta = 0$

Now, here comes the trick! We're not interested in $d$ . We're only interested in $a$ . Even if we solve for $d$ , we'll be stuck with two values, and those two will give us two equations in $a$ and $e$ . So what we do is, find the sum of the values of $d$ . That is $d_1 + d_2 = \dfrac{2GM} { \dfrac{2GM}{R} - u^2 }$ . But the values of $d$ are $a(1+e)$ and $a(1-e)$ . The sum of which is $2a$ . So, we have:

$\dfrac{2GM} { \dfrac{2GM}{R} - u^2 } = 2a$ .

Plugging the values, we get: $a = 3000 km$ .

Because the "launch speed" is so low (about that of throwing a rock), we already know that the maximum altitude of the trajectory is going to be very small compared to the radius of the Earth (the "uniform gravity" approximation that we learn when we start off in mechanics tells us this is about 5 meters). As I mention in a comment below, the launch angle does not affect the total mechanical energy of the orbit, which determines the semi-major axis of the (theoretical) elliptical path. So the apogee for the ellipse (the greatest distance from the "gravitational focus" *) is very close to the radius of the Earth.

$^*$ We typically assume that the density distribution of the Earth is radially symmetric (no dependence on direction) in these problems, in which case the gravitational focus of the trajectory can be taken as the center of the Earth.

Were it free to do so, the thrown object would follow an ellipse with its perigee very close to that focus, since the total mechanical energy of the object is essentially just the gravitational potential energy in the field of Earth. In the absence of any sort of resistance, the object would "fall" along an extremely narrow ellipse, make a very tight turn around the center of the Earth, and "climb" back along the ellipse until it reaches about 5 meters above the surface again.

So the major axis of the ellipse would be pretty nearly just the radius of the Earth. Since we are taking that to be 6000 km., the semi-major axis is half of that, or 3000 km.

(Incidentally, the eccentricity of this ellipse is extremely close to 1 , which is why we describe the "sub-orbital" trajectory of a ballistic trajectory in introductory mechanics, in the absence of air resistance, as a parabola, the conic section with eccentricity of exactly 1 . )