How to find a hyperbolic escape orbit with a starting point and a desired velocity after sphere exit How would you find an escape orbit which includes a given starting point (not assuming a circular starting orbit) and escapes the sphere of influence of the current body with a given remaining velocity?
A problem perhaps similar to Lambert's Problem, but instead of an arrival location and time, it is a final velocity.
I have a solution that someone suggested to me, that takes the final velocity as a hyperbolic excess velocity, uses that to calculate specific energy and semi major axis of the departure orbit. However they then assume that you're departure burn will occur at the periapsis of the new orbit and derive the eccentricity, and determine other elements of the departure orbit using that. The approach would work, though you have to use various guesses as the starting position until you find one that results in an orbit departing in the correct direction and the start is always at the periapsis of the departure orbit.
I am hoping that there is an answer to this problem that doesn't require starting at the periapsis of the escape orbit, as I think that would broaden the range of potential departure directions from a given point. However I don't know how I would begin to calculate that.
 A: I assume that with starting point you mean a radius $r$, since this problem is point symmetric.
Than this problem is quite simple to solve by using the conservation of specific orbital energy,
$$
\frac{v^2}{2}-\frac{\mu}{r}=\epsilon,
$$
with
$$
\epsilon=\frac{v_{\infty}^2}{2},
$$
or
$$
\epsilon=\frac{v_{SOI}^2}{2}-\frac{\mu}{r_{SOI}},
$$
I was not sure whether you meant the hyperbolic excess velocity (velocity at $r=\infty$) or the velocity at the SOI.
So your starting velocity at your desired starting radius $r$ can be determined with
$$
v=\sqrt{v_{\infty}^2+\frac{2\mu}{r}},
$$
or
$$
v=\sqrt{v_{SOI}^2+2\mu\left(\frac{1}{r}-\frac{1}{r_{SOI}}\right)}.
$$
However this will just be the magnitude of the velocity. The velocity also has a direction, but this has not influence on the the specific orbital energy, so you are allowed to choose any direction.
When you specify in which direction you would like to leave the SOI than you can find one unique solution by using the conservation of specific angular momentum. This will return one trajectory but two velocity vectors, since the radial component can be positive or negative (after or before periapses passage).
A: In the classical two-body approximation, a given hyperbolic excess velocity fixes only the semi-major axis of the transfer hyperbola. Given also an initial position, you are still free to choose 2 Keplerian elements. 
That means you have an underconstrained problem; whatever the initial orbit is, and wherever the spacecraft is in that orbit, there is an infinite amount of possible trajectories starting at that point that have the given excess velocity. There is however always just one set of elements which would minimize the $\Delta V$ for that particular starting point.
And, as you may have expected, the overall minimum $\Delta V$ requirement for all points in the initial orbit is at pericenter, and the thrust vector parallel to the initial orbit -- the Oberth effect. 
Now, just to peak your interest -- when not using the two-body simplification and going to a more realistic, multi-body context (like the Earth/Moon system), you can use weak stability boundary theory  to reduce the $\Delta V$ requirements beyond what is possible with classical transfers as the one you describe. The interplanetary superhighway is a popular example of what is possible with this. Given its potential, and proven effective on several real space missions, it is a very active field of research.
A: If you have a body in a circular orbit, at ANY height above the surface; presumably in a vacuum, and at the velocity appropriate for that circular orbit, an (instantaneous) increase in (tangential) speed by a factor sqrt(2)  will convert the orbit to a parabola, with that point as perigee, or perihelion, or peri-whatever.
So any velocity greater than sqrt(2) times the circular orbit velocity, will give a hyperbolic escape orbit.
