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This question already has an answer here:

A term I have heard is the "water was like class". Meaning that the water had 0 waves. Could a rocket be modified to take off a horizontal surface starting slowly increasing throttle on each stage? First using hydrofoil skids which would release when the ground effect wings lift the rocket then dropped when the wings and air breathing stage when it no longer creates lift? Can more thrust come from launching parallel to water?

All the rocket would have to do is fly strait while the curvature of the Earth pulls away as the rocket slowly increases speed.

How much more thrust come from the Rocket exhaust being angled down into the water a little?

enter image description here

Can a flock of birds crash a rocket?

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marked as duplicate by Antzi, Ingolifs, peterh, Muze, Jan Doggen Apr 2 at 7:31

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    $\begingroup$ Can confirm water in the open sea is like glass in that it is dangerous, but certainly not flat. Please do some research before posting. I think your actual question is 'is there any advantage in a horizontal launch' and less about launching from water (other options include land based rails or dry lake beds on wheels), which is very very different en.wikipedia.org/wiki/Sea_Dragon_(rocket) $\endgroup$ – GremlinWranger Mar 31 at 1:24
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    $\begingroup$ At low altitude the main job of a rocket is to get the rocket out of most of the atmosphere so that it can really start accelerating without incredible drag. Only once you are outside of most of the atmosphere can you really start accelerating. Unless you can add some reason why there would be some advantage, this this seems like an absolutely 100% bad idea without any reason to consider it. $\endgroup$ – uhoh Mar 31 at 1:51
  • $\begingroup$ This idea is "for the birds" ;-) which is probably why you are linking to a question about birds? $\endgroup$ – uhoh Mar 31 at 1:52
  • $\begingroup$ You should look into trajectory design. Unfortunately, I can't suggest a good reference about this topic atm. $\endgroup$ – Everyday Astronaut Apr 1 at 19:24
  • $\begingroup$ Uh oh. Somebody doesn't play KSP ;) Go 1000m/s before you exit the lower atmosphere and you'll over heat and die. When they say, "Throttling back for max Q", that's what they're preventing. And spending time fighting gravity instead of accelerating is bad. Thrust isn't what we're after. Thrust-to-weight and total delta-v, that's what we need, because wings that could provide more lift than drag, once you near orbital velocity (at a high enough altitude where you should be doing so), are imaginary. $\endgroup$ – Mazura Apr 24 at 10:41
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The biggest problem with this question is the scale of that diagram, to give you an idea:

  • The earth is 63.7x the height of the atmosphere in radius.
  • This means that leaving the earth horizontally will encounter much more atmosphere than shown.

A plotting of a circular model of the earth with atmosphere/without using equations shows this:

Earth plus atmosphere.

Wolfaram Alpha Scaled Plot Here

But, more importantly, we want the intercept of where you'll be horizontally leaving the atmosphere...

enter image description here

So, basically, you're going to choose to travel through 1133.22km of atmosphere instead of 100km. You're doing 11.3322x the work to do what you've said (simply put, not including massive complications from prolonging a bunch of other stress on the materials). The goal is to minimize stress on the materials, if you've chosen to prolong stress to any substance you're jeopardizing the mission.


Once again, I am an amateur, let me know if any of this is wrong. Note, this information is simplified:

  • Does not include earth rotation.
  • This assumes the earth is a sphere, not true.
  • As JCRM pointed out the atmosphere is a gradient, so the amount of total atmosphere passed through is significantly less than 11.3322x.
    • You are increasing time at max Q though, because the gradient is more spread out.
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    $\begingroup$ your numbers are wrong as it treats the atmosphere as an on/off thing, but the jist is fine $\endgroup$ – JCRM Apr 1 at 21:38

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