4 ms·

Eh, not really. It's not getting to altitude that's expensive, it's speed. Orbit is blisteringly fast.

AdamTReineke 10y ago
It's expensive to build up speed while fighting through the lower atmosphere. Density drops pretty fast[1], so you don't need to fight it with the rocket if the launch vehicle has already paid the cost. [1] https://upload.wikimedia.org/wikipedia/commons/d/de/Atmosphere_model.png https://upload.wikimedia.org/wikipedia/commons/d/de/Atmosphe...
- patall 10y ago
  Why aren't space ports then at higher altitudes but mostly at sea-level like cape canaveral? At the beginning the rocket flies comparably slow, so the lower atmosphere density is not that important, its the speed that is hard.
  - msandford 10y ago
    Because there aren't any places in the US where you've got 20kft mountains right next to the ocean, on the east coast. The idea is that you launch over the ocean, near the equator so that you're getting the first ~1000mph for "free" since you're launching with the earth's rotation instead of against it. Next to the ocean is for range safety. If it's over the ocean and it blows up you don't rain down parts on populated areas. If you tried launching from the top of a mountain in the Rockies and something went wrong it could theoretically rain down on DC which would definitely get your funding cut.
    - Retric 10y ago
      No need for the east coast. Hawaii's highest point is 13,800 ft above sea level and much closer to the equator both significant advantages. The real advantage to Cape Canaveral is it was in the middle of nowhere with lot's of open land.
  - deleted 10y ago
    [deleted]
astrodust 10y ago
It's a lot easier to get up to speed if you don't have to deal with cutting through a heavy chunk of atmosphere on the way up. Plus, rocket designs could vary, be less concerned about aerodynamics, if they were starting from a launch platform already in very thin air. A wider, shorter rocket might work better in thin air, for example, but that would be impractical launching from sea-level as is tradition.
- msandford 10y ago
  I have this dream that someday we'll build a SR-710 that's 10x the size and carry a BIG payload. And then putting stuff into orbit will basically be strapping your thing to a "big" upper stage (but small relative to a whole rocket stack), loading it into the cargo bay and then some fancy flying. I know it'll never happen for a variety of reasons. But it's such a cool plane and when you think about cruising at mach 3 (2000mph+, 15% of escape velocity (http://www.sr-71.org/blackbird/sr-71/) http://www.sr-71.org/blackbird/sr-71/)) it seems like it could make the rocket equation for a plane-launched rocket much, much more forgiving.
  - Scramblejams 10y ago
    One big reason it'll never happen: The square-cube law. https://en.wikipedia.org/wiki/Square-cube_law https://en.wikipedia.org/wiki/Square-cube_law
    - astrodust 10y ago
      Physics is such a killjoy.
    - msandford 10y ago
      Well 10x doesn't necessarily mean 10x longer, it might mean something with 10x the total wing area and thus 10x the lifting capacity. Which might only be say 3x the length and 3x the width.
- Symmetry 10y ago
  With a wider, shorter rocket and lower pressure you could also use wider engine bells, for more efficient thrust.
TallGuyShort 10y ago
When you're dealing with LEO velocities, the difference between the speed of the aircraft they launched with (significantly <M1) and a drifting balloon is relatively negligible considering the benefit of start at the edge of the atmosphere inexpensively, I would think. Definitely a new technique that would have some hard problems, but not fundamentally infeasible, I would think.

It's expensive to build up speed while fighting through the lower atmosphere. Density drops pretty fast[1], so you don't need to fight it with the rocket if the launch vehicle has already paid the cost. [1] https://upload.wikimedia.org/wikipedia/commons/d/de/Atmosphere_model.png https://upload.wikimedia.org/wikipedia/commons/d/de/Atmosphe...

patall 10y ago
Why aren't space ports then at higher altitudes but mostly at sea-level like cape canaveral? At the beginning the rocket flies comparably slow, so the lower atmosphere density is not that important, its the speed that is hard.
- msandford 10y ago
  Because there aren't any places in the US where you've got 20kft mountains right next to the ocean, on the east coast. The idea is that you launch over the ocean, near the equator so that you're getting the first ~1000mph for "free" since you're launching with the earth's rotation instead of against it. Next to the ocean is for range safety. If it's over the ocean and it blows up you don't rain down parts on populated areas. If you tried launching from the top of a mountain in the Rockies and something went wrong it could theoretically rain down on DC which would definitely get your funding cut.
  - Retric 10y ago
    No need for the east coast. Hawaii's highest point is 13,800 ft above sea level and much closer to the equator both significant advantages. The real advantage to Cape Canaveral is it was in the middle of nowhere with lot's of open land.
- deleted 10y ago
  [deleted]

astrodust 10y ago

It's a lot easier to get up to speed if you don't have to deal with cutting through a heavy chunk of atmosphere on the way up. Plus, rocket designs could vary, be less concerned about aerodynamics, if they were starting from a launch platform already in very thin air. A wider, shorter rocket might work better in thin air, for example, but that would be impractical launching from sea-level as is tradition.

msandford 10y ago
I have this dream that someday we'll build a SR-710 that's 10x the size and carry a BIG payload. And then putting stuff into orbit will basically be strapping your thing to a "big" upper stage (but small relative to a whole rocket stack), loading it into the cargo bay and then some fancy flying. I know it'll never happen for a variety of reasons. But it's such a cool plane and when you think about cruising at mach 3 (2000mph+, 15% of escape velocity (http://www.sr-71.org/blackbird/sr-71/) http://www.sr-71.org/blackbird/sr-71/)) it seems like it could make the rocket equation for a plane-launched rocket much, much more forgiving.
- Scramblejams 10y ago
  One big reason it'll never happen: The square-cube law. https://en.wikipedia.org/wiki/Square-cube_law https://en.wikipedia.org/wiki/Square-cube_law
  - astrodust 10y ago
    Physics is such a killjoy.
  - msandford 10y ago
    Well 10x doesn't necessarily mean 10x longer, it might mean something with 10x the total wing area and thus 10x the lifting capacity. Which might only be say 3x the length and 3x the width.
Symmetry 10y ago
With a wider, shorter rocket and lower pressure you could also use wider engine bells, for more efficient thrust.

TallGuyShort 10y ago

When you're dealing with LEO velocities, the difference between the speed of the aircraft they launched with (significantly <M1) and a drifting balloon is relatively negligible considering the benefit of start at the edge of the atmosphere inexpensively, I would think. Definitely a new technique that would have some hard problems, but not fundamentally infeasible, I would think.