3 ms·

Technically its suborbital

by preisschild 3y ago

thelittleone 3y ago
Isn't orbit above 125 miles (200km)?
- madaxe_again 3y ago
  Orbit is when you drop an apple off a dining room table - it’s just a very crappy orbit. “On orbit” typically means in a stable orbit around a body - but in the case of starship, it could have been on orbit, the delta v is more than sufficient, but that’s an unsafe configuration if you don’t know your engines will relight.
- apendleton 3y ago
  Orbit is mostly about speed, not about height. Going straight up and down doesn't count, even if you pass the height that some orbital vehicles attain. This vehicle is actually going orbital speed, but not quite orbital height (or rather, it's in an "orbit" that has a very eccentric elliptical shape that would cause it to hit the atmosphere on its way back around; it'd be well above a typical orbital height at apogee, though).
  - ta1243 3y ago
    Technically you can orbit the earth at a pretty low speed. Geostationary satellites orbit about 5,500kph, which is far slower than the 26,000kph starship reached today. You just need to get high enough, and be pointing in the right direction.
- KineticLensman 3y ago
  Orbit is when you turn the engines off and stay up there.
- hoorayimhelping 3y ago
  Space is generally defined to be 100km above earth's surface, an arbitrary point called the Karman Line [1]. Orbit is when an object is traveling so fast that it reaches the horizon of a body before the body's gravity can pull it down to the surface, but perpetually. It's basically perpetually falling around the body. Imagine one of those guys in a wingsuit skimming along the surface of a mountain, never actually touching the surface. It's similar to that, but at a much higher scale. 1) https://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_line https://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_line
- usrusr 3y ago
  Orbit is a speed, not a distance. You could do a suborbital hop that has its highest point beyond the orbit of the moon. (if you aim very, very well)
  - WithinReason 3y ago
    At the same time, if your velocity vector is pointing towards the ground you're not achieving orbit no matter your speed
    - the8472 3y ago
      It works if you're a black hole or a chunk of degenerate matter. For transatmospheric orbits being a large ball of iridium will work too.
- malfist 3y ago
  It's in space (above the Kármán line), but not in orbit. Orbit implies it has the velocity for staying in space, which isn't the intention here. If anything goes wrong they want to be able to not leave junk in orbit
ghufran_syed 3y ago
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...
dotnet00 3y ago
Technically it's a transatmospheric orbit. That is, an orbit such that it'd stay up there if the atmosphere were not present. The difference between this and full orbit is just a few seconds longer burn, so it's a difference with little meaning in terms of proving out the ability to reach orbit.
- perihelions 3y ago
  I don't believe that's the case—see the math in my sibling comment.
  - dotnet00 3y ago
    The number going around from people who typically do this kind of thing is a ~55x235km orbit: https://twitter.com/planet4589/status/1768270310199935299 https://twitter.com/planet4589/status/1768270310199935299 I'm not really in a place to judge your math right now to really add anything on that.
    - perihelions 3y ago
      Yeah, that guy's absolutely a domain expert! But note that he writes -55 km, not +55 km—that is a suborbital trajectory. Its perigee is below the earth's surface; -55 km is a negative altitude. (He's also clearly using a different set of data than I have access to. I can't read the context of the Twitter thread so I don't know what numbers he's looking at).
      - dotnet00 3y ago
        Ooh that's a good catch, I subconsciously substituted the -55 for ~55!

Technically its suborbital

thelittleone 3y ago

Isn't orbit above 125 miles (200km)?

madaxe_again 3y ago
Orbit is when you drop an apple off a dining room table - it’s just a very crappy orbit. “On orbit” typically means in a stable orbit around a body - but in the case of starship, it could have been on orbit, the delta v is more than sufficient, but that’s an unsafe configuration if you don’t know your engines will relight.
apendleton 3y ago
Orbit is mostly about speed, not about height. Going straight up and down doesn't count, even if you pass the height that some orbital vehicles attain. This vehicle is actually going orbital speed, but not quite orbital height (or rather, it's in an "orbit" that has a very eccentric elliptical shape that would cause it to hit the atmosphere on its way back around; it'd be well above a typical orbital height at apogee, though).
- ta1243 3y ago
  Technically you can orbit the earth at a pretty low speed. Geostationary satellites orbit about 5,500kph, which is far slower than the 26,000kph starship reached today. You just need to get high enough, and be pointing in the right direction.
KineticLensman 3y ago
Orbit is when you turn the engines off and stay up there.
hoorayimhelping 3y ago
Space is generally defined to be 100km above earth's surface, an arbitrary point called the Karman Line [1]. Orbit is when an object is traveling so fast that it reaches the horizon of a body before the body's gravity can pull it down to the surface, but perpetually. It's basically perpetually falling around the body. Imagine one of those guys in a wingsuit skimming along the surface of a mountain, never actually touching the surface. It's similar to that, but at a much higher scale. 1) https://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_line https://en.wikipedia.org/wiki/K%C3%A1rm%C3%A1n_line
usrusr 3y ago
Orbit is a speed, not a distance. You could do a suborbital hop that has its highest point beyond the orbit of the moon. (if you aim very, very well)
- WithinReason 3y ago
  At the same time, if your velocity vector is pointing towards the ground you're not achieving orbit no matter your speed
  - the8472 3y ago
    It works if you're a black hole or a chunk of degenerate matter. For transatmospheric orbits being a large ball of iridium will work too.
malfist 3y ago
It's in space (above the Kármán line), but not in orbit. Orbit implies it has the velocity for staying in space, which isn't the intention here. If anything goes wrong they want to be able to not leave junk in orbit

ghufran_syed 3y ago

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...

dotnet00 3y ago

Technically it's a transatmospheric orbit. That is, an orbit such that it'd stay up there if the atmosphere were not present. The difference between this and full orbit is just a few seconds longer burn, so it's a difference with little meaning in terms of proving out the ability to reach orbit.

perihelions 3y ago
I don't believe that's the case—see the math in my sibling comment.
- dotnet00 3y ago
  The number going around from people who typically do this kind of thing is a ~55x235km orbit: https://twitter.com/planet4589/status/1768270310199935299 https://twitter.com/planet4589/status/1768270310199935299 I'm not really in a place to judge your math right now to really add anything on that.
  - perihelions 3y ago
    Yeah, that guy's absolutely a domain expert! But note that he writes -55 km, not +55 km—that is a suborbital trajectory. Its perigee is below the earth's surface; -55 km is a negative altitude. (He's also clearly using a different set of data than I have access to. I can't read the context of the Twitter thread so I don't know what numbers he's looking at).
    - dotnet00 3y ago
      Ooh that's a good catch, I subconsciously substituted the -55 for ~55!