In typical rockoon a balloon is the lifting body to the upper atmosphere where a rocket is released and ignited. But I would like to ask about a variation to launch small payloads to space.

Instead of attaching the rocket to some kind of platform pointing upwards, to attach rocket to seaplane pointing downwards. Release the rocket attached to seaplane and let it fall till it reaches speed close to 500m/s. Let's assume that rocket is released at 40km and reaches 500m/s at approximately 10km height. Before it hits the denser layer of atmosphere change the direction and steer the rocket upwards by seaplane. Detach rocket from seaplane and ignite rocket. Use atmospheric drag to slow down seaplane and safely land on water. For safety reason let's assume that balloon is filed with helium and lifted from ship offshore with enough clearance to avoid crash issues.

Is this ever tried in practice, or I am missing something which would make this impossible to realize or there are no benefits? If it was tried what were the disadvantages which cause it impractical?

From my point of view this variation provides several advantages to traditional rockoon:

  • Starting rocket engine at 500m/s is much efficient than at 0 m/s. Eventually rocket will go much higher, maybe even SSTO is possible with hydlolox. At least you can downgrade tree stage rocket to two stage rocket for the same height.
  • You can steer rocket wherever desired with seaplane. Traditional rockoon suffers from drift because ballon cannot be steered, leading to launching rocket in undesired direction.
  • Maneuvering is done before hitting denser layers of atmosphere (90% of atmosphere is till 10km) therefore atmospheric drag for rocket should not be a significant problem. Seaplane with supersonic design should not have an issue moving through denser layers of atmosphere.
  • Most expensive part - seaplane is 100% reusable and can be made to land by autopilot on water.
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    $\begingroup$ There's an outside chance that the plummeting phase might be useful for something like starting a ramjet, but it seems a convoluted way to go about doing things. $\endgroup$ Commented Feb 10, 2020 at 7:55
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    $\begingroup$ I think this is an interesting idea, but I would pitch it a bit differently. The key point here seems to be supersonic air-launch. You'll want to use air-breathing engines here, for better Isp. Supersonic is necessary (because “space is fast away, not far away”) to make the air-launch really worthwhile, but the problem with that is that a supersonic plane would need an insanely long runway to take off with that kind of load, at dangerously high take-off speed. And where to even put the wheerls? –That's where the balloon comes in handy. $\endgroup$ Commented Feb 10, 2020 at 11:59
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    $\begingroup$ Stage 1: release rocket from plane and have it drop under gravity. Stage 2: steer rocket back up using plane. Question: How? If you've dropped the rocket from the plane in stage 1, the two are no longer connected. What additional mechanism are you thinking about which you haven't mentioned in your question, so that the plane can pull the rocket back up? Some kind of long cable connecting them? If so, this is a crucial missing detail in the question. Either way, Antzi's answer is correct; but we can still improve the question. $\endgroup$
    – Graham
    Commented Feb 10, 2020 at 14:43
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    $\begingroup$ I happen to be a commercial seaplane pilot. Max speed of the aircraft I fly is about 110 knots. I'd sure like the opportunity to go supersonic, so where do I apply? Whee! $\endgroup$
    – Stu Smith
    Commented Feb 10, 2020 at 19:02
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    $\begingroup$ Lifting the rocket to 40 km and then let it fall down to 10 km is a waste of energy anyway due to atmospheric drag. It is not possible to convert all downward speed to upward speed by the use of wings. $\endgroup$
    – Uwe
    Commented Feb 12, 2020 at 17:14

1 Answer 1


Before I start my answer:

This is not a rockoon anymore, since there is no balloon part. It's just a quirky air launched rocket.

My answer:


What you are doing is:

  • Phase 1: Climb with a plane
  • Phase 2: Descend with a plane
  • Phase 3: Climb without a plane

So yes, Phase 3 climb will begin with higher speed. However, since it's all energy that was gained from descending, you don't gain anything, because you will have to expend the speed to climb again. Actually you will lose more because of drag when climbing and descending.

You would have been better off by starting your rocket before phase 2.

Edit: ok so p1 is done with a balloon. So it reduces a little complexity on the seaplane and adds a lot of complexity on the ballon. I’m pretty sure it’s going to be even worse...

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    $\begingroup$ As I understood the question, Phase 1 would be to raise the plane/rocket assembly with a balloon to the starting altitude. This makes the method just as convoluted (if not more), but there is the -oon part. $\endgroup$
    – zovits
    Commented Feb 10, 2020 at 9:33
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    $\begingroup$ That's, right ascent would be done by balloon. Maybe that was not obvious from the question but I did mention that for sefty reason helium would be used for filling a balloon. Also I considered that is common knowledge that no jet airplane can reach heights of 40km with cargo attached. Current altitude record is 37650m for MIG-25 without any cargo. Rocket assisted airplanes can go hire but then what is the point. $\endgroup$
    – WOW 6EQUJ5
    Commented Feb 10, 2020 at 9:52
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    $\begingroup$ @zovits rockaeroploon? $\endgroup$ Commented Feb 10, 2020 at 13:08
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    $\begingroup$ @StarfishPrime I'm more inclined to name this BallPlanEt (though BallAnCket would be fine as well) to better reflect the ordering of the components, but I guess OP would be more interested in learning why this is an impractical idea (what size of balloon would be required to lift anything practical), but at the moment I lack the time to do this satisfactorily. $\endgroup$
    – zovits
    Commented Feb 10, 2020 at 13:13
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    $\begingroup$ Wouldn't the increased efficiency of rocket engines at higher velocity make this beneficial, at least if the loss to air resistance isn't too much during the descent? $\endgroup$
    – jpa
    Commented Feb 10, 2020 at 17:53

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