The "pole" you are referring to is also known as a space elevator. Creating a cable of sufficient tensile strength is currently unfeasible, but carbon nano- and macrotubes are promising.
Your question can be generalized, ignoring Earth's non-spherical shape, to apply to any position in space around earth by replacing "height above the equator" with "distance from Earth's rotation axis" (henceforth r).
All questions except for the one for the minimum r to achieve orbit can be answered using elemental math and energy conservation.
The velocity at any r can be easily calculated via
$$ \omega = \frac{2 \cdot \pi}{\textit{1 sidereal day}} $$
Note the difference: One year has 366(.2425) sidereal days, but only 365(.2425) days
$$ v(r) = \omega r \approx 7.29212 \cdot 10^{-5} \frac{r}{s} $$
At ground level ($r=6378137 m$), that is $465 \frac{m}{s}$. At geostationary altitude, it's about $3\frac{km}{s}$. The speed of light is reached at a distance of about 27 AU (read: far, far away).
The specific (per unit of mass) energy of your orbit, $ E_{spec} $, is the sum of the specific kinetic and potential energies, and negative for a bound (circular, elliptical and/or sub-orbital) orbit.
$$ E_{spec,kin}(r) = \frac{1}{2} v(r)^2 = \frac{1}{2} \omega^2 r^2 $$ $$ E_{spec,pot}(r) = \int_{r}^{\inf}{\frac{GM}{r^2}dr} = -\frac{GM}{r} $$
The parabolic escape orbit has a specific energy of zero:
$$ E_{spec}(r) = 0 $$ $$ \frac{GM}{r} = \frac{1}{2} \omega^2 r^2 $$ $$ 2 GM = \omega^2 r^3 $$ $$ r = \sqrt[3]{\frac{2 GM}{\omega^2}} $$ $$ r = 53123487m $$
That is 46,745 km above the surface of earth, or 10,959 km above geostationary altitude. 921
Note that slightly lower altitudes will also ultimately lead to an escape due to the three-body nature of the Earth/Sun/Object system (further reading).
All trajectories above this altitude are hyperbolic, with increasing eccentricities. Between the geostationary and escape altitude, you have various elliptical orbits with eccentricities ranging from zero (at geostationary altitude) to one (at escape altitude). The Periapsis (closest-to-earth point) is at release altitude. The calculation of the Apoapsis (furthest-from-earth) altitude and further orbital parameters requires <PERSON> equations; this site provides an excellent listing of all relevant equations.
We already know the Periapsis altitude, plus the orbit's specific energy $E_{spec}$. 246