It's at 26000 km/h which sufficient for orbital velocity. It looks like it's in an elliptical orbit. I guess we'll see soon if they need to do a de-orbit burn, or if the orbit just intersects the atmosphere and they use atmospheric braking?
edit:
(let* ((μ 398600.0) ;; km^3/s^2
(r 6371.0) ;; km
(peri (+ r 0.0))
(apo (+ r 180.0))
(a (* 0.5 (+ peri apo))))
(sqrt (* μ
(- (/ 2.0 apo)
(/ 1.0 a)))))
;; 7.745844595118488> 27,900 kph
Seems like they want to test the limit. Same speeds as LEO, but guaranteed to come down.
In the form I'm using, I'm using standard parameters of an elliptical orbit: the periapsis (the closest approach to the center of mass of the massive body (which is a focal point of the ellipse which the orbit traces)), apoapsis (farthest distance), and semimajor axis (their arithmetic mean [1]). I'm evaluating the orbital velocity at the highest point, the apoapsis. μ is a short form for the product G*M, the standard gravitational parameter [2] of Earth (which is known to much higher precision than either the universal gravitational constant G, or the mass of the earth M, individually).
The particular orbit I'm applying it to is one whose periapsis is equal to the Earth's radius—an orbit that touches the surface of the Earth. This is the dividing line for orbital / suborbital: a suborbital trajectory is one that (mathematically) goes beneath the Earth's surface.
[0] https://en.wikipedia.org/wiki/Vis-viva_equation#Equation
[1] https://en.wikipedia.org/wiki/Semi-major_and_semi-minor_axes...
[2] https://en.wikipedia.org/wiki/Standard_gravitational_paramet...