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As Kerbal Space Program has taught me, it’s all very well reaching ballistic apogee but you need to apply thrust again at apogee to circularize orbit (raise you

As Kerbal Space Program has taught me, it’s all very well reaching ballistic apogee but you need to apply thrust again at apogee to circularize orbit (raise your perigee to the same altitude as you are right now.) Otherwise you’ll just come straight back down again. How can this device fling a functioning rocket motor into space?

dmead 5y ago
staged rockets also fling rocket motors into space. /shrug
- boardwaalk 5y ago
  …at maybe 10g, not 10,000g.
jeffwass 5y ago
I think launch angle can be adjusted, it's just straight up for the photo. Eg, see the rendering further down the page on the grassy hill which shoots at an oblique angle.
- gorgoiler 5y ago
  Your periapsis — the lowest part of your orbit — is always the place you return to. Throwing harder or at a different angle just determines how high you get when at your highest point, before coming all the way back down again. http://www.scielo.org.mx/img/revistas/rmaa/v52n2/0185-1101-rmaa-52-02-283-gf5.jpg http://www.scielo.org.mx/img/revistas/rmaa/v52n2/0185-1101-r... If you apply thrust at A then you have the solid line: an elliptical orbit. If you just throw something hard all you have is the dashed line: go up then come back down again. (In this diagram, substitute “initial circular orbit” for “surface of Earth” :)
  - Symmetry 5y ago
    That's why spinlaunch is flinging a whole rocket high into the air rather than just the payload.
  - sidewndr46 5y ago
    This isn't the case at all. The trap your reasoning falls into is assuming that for some reason an object must be in an orbit after being thrown. Obviously, an orbit comes back to it's original position at some point. In many cases atmospheric drag converts what would be an orbit into burning up in the atmospheric or smashing into the ocean somewhere. But if you throw something hard enough from the Earth's surface, it absolutely does not have to return to that position. You would just need to throw it hard enough that it was at escape velocity. Due to air friction, the actual speed you would need to throw the object would be flat out absurd if on Earth at sea level. But on a body like the moon with no atmosphere it isn't that bad at all. The bonus to this is the direction doesn't matter at all really. Anything other than straight down is fine. Now, where this does become problematic is when the velocity you would need to achieve is higher than the speed of light. At that point you're basically on a neutron star or some supermassive planet.
    - bigyikes 5y ago
      > Now, where this does become problematic is when the velocity you would need to achieve is higher than the speed of light. Isn’t this sort of a definition of a black hole? My non-physicist intuition tells me a neutron star can’t have an escape velocity higher than the speed of light, otherwise the light wouldn’t escape
    - gorgoiler 5y ago
      No one here is talking about Spinlaunch achieving Earth escape velocity, only orbital velocity.
    - mlindner 5y ago
      We're not talking about escape velocity though. (And if we're going to nitpick, even on an escape trajectory, the launch point is still on the hyperbola.)
colechristensen 5y ago
Rockets already have to withstand some pretty extreme structural challenges. The spinning would certainly involve a bit of a different challenge but not necessarily out of scale for things they already have to survive.

dmead 5y ago

staged rockets also fling rocket motors into space. /shrug

boardwaalk 5y ago
…at maybe 10g, not 10,000g.

jeffwass 5y ago

I think launch angle can be adjusted, it's just straight up for the photo. Eg, see the rendering further down the page on the grassy hill which shoots at an oblique angle.

gorgoiler 5y ago
Your periapsis — the lowest part of your orbit — is always the place you return to. Throwing harder or at a different angle just determines how high you get when at your highest point, before coming all the way back down again. http://www.scielo.org.mx/img/revistas/rmaa/v52n2/0185-1101-rmaa-52-02-283-gf5.jpg http://www.scielo.org.mx/img/revistas/rmaa/v52n2/0185-1101-r... If you apply thrust at A then you have the solid line: an elliptical orbit. If you just throw something hard all you have is the dashed line: go up then come back down again. (In this diagram, substitute “initial circular orbit” for “surface of Earth” :)
- Symmetry 5y ago
  That's why spinlaunch is flinging a whole rocket high into the air rather than just the payload.
- sidewndr46 5y ago
  This isn't the case at all. The trap your reasoning falls into is assuming that for some reason an object must be in an orbit after being thrown. Obviously, an orbit comes back to it's original position at some point. In many cases atmospheric drag converts what would be an orbit into burning up in the atmospheric or smashing into the ocean somewhere. But if you throw something hard enough from the Earth's surface, it absolutely does not have to return to that position. You would just need to throw it hard enough that it was at escape velocity. Due to air friction, the actual speed you would need to throw the object would be flat out absurd if on Earth at sea level. But on a body like the moon with no atmosphere it isn't that bad at all. The bonus to this is the direction doesn't matter at all really. Anything other than straight down is fine. Now, where this does become problematic is when the velocity you would need to achieve is higher than the speed of light. At that point you're basically on a neutron star or some supermassive planet.
  - bigyikes 5y ago
    > Now, where this does become problematic is when the velocity you would need to achieve is higher than the speed of light. Isn’t this sort of a definition of a black hole? My non-physicist intuition tells me a neutron star can’t have an escape velocity higher than the speed of light, otherwise the light wouldn’t escape
  - gorgoiler 5y ago
    No one here is talking about Spinlaunch achieving Earth escape velocity, only orbital velocity.
  - mlindner 5y ago
    We're not talking about escape velocity though. (And if we're going to nitpick, even on an escape trajectory, the launch point is still on the hyperbola.)

colechristensen 5y ago

Rockets already have to withstand some pretty extreme structural challenges. The spinning would certainly involve a bit of a different challenge but not necessarily out of scale for things they already have to survive.