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 $r_p = r/6$.
Best Answer
While calculating the orbital parameters works in this case, a simpler method is just to use energy and angular momentum conservation (as mentioned by fibonatic.) First, from energy conservation we have $$ -\frac{GMm}{r} + \frac{1}{2} m v^2 = -\frac{GMm}{r_p} + \frac{1}{2} m v_p^2 $$ Second, we have angular momentum conservation. At perigee, the satellite will have $dr/dt = 0$; this means that its velocity and radial vector will necessarily be at right angles, and therefore $L_p = m v_p r_p$. Thus, $$ m v r \sin (45^\circ) = m v_p r_p $$ This is a system of two equations in two unknowns ($v_p$ and $r_p$), and therefore can be easily solved. That said, I don't get $r_p = r/6$ when I solve these equations, so there's an error somewhere (either in my derivation or in yours.)