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The job of the rockets isn't to gain height (that is a side effect), unless you are working with a sub-orbital design. The job is to accelerate the payload to
by derekp7 14y ago
The job of the rockets isn't to gain height (that is a side effect), unless you are working with a sub-orbital design. The job is to accelerate the payload to around Mach 25 or so. The reason this is challenging is that you not only have to accelerate the payload, you also have to accelerate the fuel, fuel tanks, engines, and other associated hardware too. Which is why the promising techs mostly revolve around using less fuel (such as rail launch for the initial oomph), or something like scram jet for the first stage (they part of the fuel and reaction mass comes from the atmosphere for the first few Mach). Or something that has very small fuel amounts, and only caries reaction mass (nuclear with H2O as the reaction mass).
- learc83 14y agoHowever, air density is lower the higher you go--allowing for easier acceleration.
- danielweber 14y agoThis is the big deal with air-launch or dirigible-launch or whatever you call it. The air resistance of the atmosphere is significant. IIRC, it takes 9 km/s of delta-v to get from ground to LEO. LEO is only orbitting at ~7.7 km/s, though, so you can save at least 1 km/s of fuel. As others have said, schemes like Stratolaunch or SpaceShipOne also get horizontal velocity from the mothership.
- zevets 14y agoIt's not that high. A lot of the losses in take-off aren't the consequence of drag. I think I saw on wikipedia that the aerodynamics losses are only 1-200 m/s of the delta-v budget.
- danielweber 14y agoI surely don't want to use this site as gospel (especially since I can't pull up a better version of that graph) but it claims that the losses are around 25% at sea level. http://www.nasaspaceflight.com/2012/07/commercial-shows-reignited-interest-air-launch-system/ http://www.nasaspaceflight.com/2012/07/commercial-shows-reig... Going by raw math, if the Earth were a perfect magic sphere with no atmosphere, you could attain "orbit" at ground-level by accelerating to 7.90 km/s. Then you could Hohmann to LEO with one 0.06 km/s burn to transfer up and another 0.06 km/s burn to achieve a circular orbit, so that's only 8.02 km/s in an ideal world. [Oh, I forgot that you start with about 0.40 km/s from rotation if you are near the equator.] Oh, finally found a nice reference, on Wiki: http://en.wikipedia.org/wiki/Low_Earth_orbit#Human_use http://en.wikipedia.org/wiki/Low_Earth_orbit#Human_use The delta-v needed to achieve low earth orbit starts around 9.4km/s. Atmospheric and gravity drag associated with launch typically adds 1.5–2.0 km/s to the delta-v launch vehicle required to reach normal LEO orbital velocity of around 7.8 km/s (28,080 km/h). It doesn't break those out, unfortunately.
- lutorm 14y agoThey aren't independent, though. "Gravity drag" comes from the fact that you need to put in 1G just to remain at constant speed. It's thus most efficient to have as high acceleration as possible. Unfortunately the aerodynamics work in the opposite direction: you don't want to go fast because that increases aero losses and there's a limit to how fast you can go in the lower atmosphere without breaking up. So the combined effects probably mean that the net loss is quite a bit larger than you could achieve if you only had to minimize one of them.
- jessriedel 14y agoI did the back-of-the envelope calculation a while ago, and if I recall, 5-10% of the energy used goes toward altitude, with the rest going towards horizontal velocity (modulo a bit of air resistance).
- alpine 14y agoActually, the job of the rocket is to to increase the potential energy of the payload, so using a He balloon to put some distance between the rocket/payload package and the ground is an entirely sensible proposition.
- stcredzero 14y agoMost of it is kinetic energy.
- alpine 14y agoMost of what is kinetic energy? Its job? A rocket that reaches escape velocity from Earth can have zero kinetic energy.
- CamperBob2 14y agoHow so?
- alpine 14y agoA rocket at infinite distance from the Earth will have infinite PE and can have zero KE.
- stcredzero 14y agoMost of what the rocket has to increase to take an object from the earth's surface to LEO. If you calculate the potential energy of a rocket in LEO and compare it to its kinetic energy, you'll find the latter is much larger. (I've already done this and posted it here.)
- lutorm 14y agoSomething with exactly escape velocity from a body always has zero total energy. That's the definition of escape velocity.