The question is formulated as follows:
A satellite is some distance, r, from the centre of the earth, traveling at a speed $2v_e$, where $v_e$ is the escape velocity. The angle between the velocity and the satellite-earth line is 45 degrees. Express the distance of closest approach between the satellite and the earth centre.
My thoughts.
First I assume only Earth's gravitational acceleration is acting on the satellite. Then, I conclude the trajectory is a hyperbolic trajectory because the speed is larger than the escape velocity.
I can calculate the semi-major axis $a$ of the Kepler orbit using the vis-viva equation (as we know the velocity at a distance $r$ for body earth with gravitational parameter $GM$)
The closest distance to Earth occurs at periapsis $r_p = -a(e-1)$ where $e$ denotes the eccentricity.
Now my question is how to determine the eccentricity?
Is the following perhaps a good method? As the angle $\theta = 45$ degrees between the velocity vector and the radius vector is known, we assume that the angle between the hyperbola asymptotes is equal to $2\theta = 90 ^{\circ}$. In that case we can use $e = 1/cos(\theta)$ to calculate the eccentricity.
Combining all this together with the fact that $V_{esc} = \sqrt{2GM/r}$ we can find that $r_p = r/6$.
I am unsure about the eccentricity calculations. Could anyone advise me on this one?
An alternative to calculate the eccentricity would be using the eccentricity vector, but then I am do not obtain the elegant solution of $r_p = r/6$.