4 ms·

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,

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?

perihelions 3y ago
I did the math. So, the minimal orbit at 180 km, the one that grazes the earth's surface at its perigee, is about 7.75 km/s or 27,900 kph. The other commenters are right: it was never technically orbital. 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
- ghufran_syed 3y ago
  Thanks! On one of the feeds, it sounds like they didn't want to leave a bunch of debris in orbit if they had an anomaly
- BenoitP 3y ago
  > 26000 km/h > 27,900 kph Seems like they want to test the limit. Same speeds as LEO, but guaranteed to come down.
- grecy 3y ago
  That is really cool. For those of us that are space nerds but don't have the depth of understanding, do you mind walking us through the calculation above?
  - perihelions 3y ago
    I apologize I don't have a good explanation of it at hand! It's a form of the vis-viva equation [0] that's basically a restatement of conservation of energy. It derives the (scalar) speed of an object in a 2-body orbit, at any position within that orbit, as a simple function of their separation distance. 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 https://en.wikipedia.org/wiki/Vis-viva_equation#Equation [1] https://en.wikipedia.org/wiki/Semi-major_and_semi-minor_axes#Ellipse https://en.wikipedia.org/wiki/Semi-major_and_semi-minor_axes... [2] https://en.wikipedia.org/wiki/Standard_gravitational_parameter https://en.wikipedia.org/wiki/Standard_gravitational_paramet...
    - grecy 3y ago
      No apology needed, thankyou very much!
- eagerpace 3y ago
  But it would have been if they executed the burn in at a slightly different angle.

perihelions 3y ago

I did the math. So, the minimal orbit at 180 km, the one that grazes the earth's surface at its perigee, is about 7.75 km/s or 27,900 kph. The other commenters are right: it was never technically orbital. 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

ghufran_syed 3y ago
Thanks! On one of the feeds, it sounds like they didn't want to leave a bunch of debris in orbit if they had an anomaly
BenoitP 3y ago
> 26000 km/h > 27,900 kph Seems like they want to test the limit. Same speeds as LEO, but guaranteed to come down.
grecy 3y ago
That is really cool. For those of us that are space nerds but don't have the depth of understanding, do you mind walking us through the calculation above?
- perihelions 3y ago
  I apologize I don't have a good explanation of it at hand! It's a form of the vis-viva equation [0] that's basically a restatement of conservation of energy. It derives the (scalar) speed of an object in a 2-body orbit, at any position within that orbit, as a simple function of their separation distance. 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 https://en.wikipedia.org/wiki/Vis-viva_equation#Equation [1] https://en.wikipedia.org/wiki/Semi-major_and_semi-minor_axes#Ellipse https://en.wikipedia.org/wiki/Semi-major_and_semi-minor_axes... [2] https://en.wikipedia.org/wiki/Standard_gravitational_parameter https://en.wikipedia.org/wiki/Standard_gravitational_paramet...
  - grecy 3y ago
    No apology needed, thankyou very much!
eagerpace 3y ago
But it would have been if they executed the burn in at a slightly different angle.