alpha2cen Posted August 6, 2012 Share Posted August 6, 2012 (edited) If the air density of the Earth surface were 1/2 of the current air density, how much energy would be required to escape to the space by rocket? Gravity is the same as the current gravity. A) same energy B) more energy C) less energy D) others why? Edited August 6, 2012 by alpha2cen Link to comment Share on other sites More sharing options...
CaptainPanic Posted August 6, 2012 Share Posted August 6, 2012 A. Same energy. Well, if we are really strict, then the reduced friction would make it a little bit easier, so you could argue that it's C (less energy), but it's a negligible bit less. 1 Link to comment Share on other sites More sharing options...
imatfaal Posted August 6, 2012 Share Posted August 6, 2012 (edited) Less energy. The main calculation is to do with gravity and will not be affected - but making it easier for the rocket to slip through the atmosphere will make a difference. as the rocket functions on equal and opposite forces and doesn't rely on pushing on anything, and unlike a jet it doesnt rely on an air flow into the engine - the difference in drive would be zero/minimal as far as I could see. Whether the actual mechanics of a currently used rocket would work in half atmospheric pressure I have no clue about - my guess would be no. Edited August 6, 2012 by imatfaal xposted with the CaptP Link to comment Share on other sites More sharing options...
John Cuthber Posted August 6, 2012 Share Posted August 6, 2012 "Whether the actual mechanics of a currently used rocket would work in half atmospheric pressure I have no clue about - my guess would be no. " My guess is yes- because they do. Shortly after take-off the rocket does travel through air of half the normal density, and it works just fine. It's also interesting to consider why it would take slightly less energy. To a good approximation the viscosity of air is independent of the pressure. The reduced drag is due to the reduced "thickness" of the atmosphere. (the depth of the "ocean of air" so to speak.) But quite a bit of the acceleration to escape velocity takes place when the rocket has already left (most of) the atmosphere so the effect is much smaller than you might expect and depends on the velocity vs height profile. Of course there's the other way to halve the atmosphere's density: you can double it's temperature. That would raise the viscosity of the air and increase the thickness/ depth of the atmosphere so the rocket would travel further through more viscous air and both factors would increase the energy required. However the change in energy required would be small compared to the total energy. 2 Link to comment Share on other sites More sharing options...
CaptainPanic Posted August 6, 2012 Share Posted August 6, 2012 Although this is not my field of expertise, I always understood that Drag is dependent on whether it is turbulent or not. In turbulent (high velocity) motion, like a rocket through air, the drag is a function of the air density (and that is a function of pressure). In laminar (low velocity) motion, the drag is a function of the viscosity, as described in Stokes' Law. But Stokes' Law is not valid for rocketry, because in rocketry you will have a high Reynold's number (Reynold's number is a measure of how laminar/turbulent something is). Still, the point stands that the drag of the atmopsphere will account for only a small fraction of the total energy required for getting into orbit. Link to comment Share on other sites More sharing options...
Enthalpy Posted August 7, 2012 Share Posted August 7, 2012 At a small rocket like Vega, the atmospheric drag consumes 100m/s or less. At a bigger rocket, it's under 50m/s. Compare with some 8200m/s minimum to reach orbit in a theoretical transfer, or rather 9500m/s because you want to accelerate moderately so the payload survives. That's why rockets are not given nice smooth shapes. Any launcher has a fairing with short conical shapes to save height and mass, not streamlined shapes like an airglider. Thinner atmosphere, or none at all, would bring additional advantages, though. First, it would allow a better trajectory that gets earlier inclined. Second, it would allow bigger nozzles (...when feasible!) that expand to less than 1 bar and make the rocket engine more efficient - and this brings a bigger advantage than the reduced atmospheric drag. But this can be obtained through other means. A few launchers ignite at altitude, being dropped from an aeroplane. The main advantage is to save a spaceport and launch from any place, especially outside national territory and its regulations. But it does not justify the development of an aircraft, so current designs take an existing B-52, White Knight or F-4; as well, a big launcher is just too heavy for an aircraft (Ariane V: 800t, Airbus 380: 650t max take-off. 1 Link to comment Share on other sites More sharing options...
John Cuthber Posted August 8, 2012 Share Posted August 8, 2012 Another advantage to starting with just half as much air in the way is that you wouldn't need to make the pointy end so strong in order to push the air out of the way. You could make the rocket lighter and that would save significant fuel. http://en.wikipedia.org/wiki/Max_Q (You may be amused to hear that I recently learned about that on a ride at the Kennedy Space Centre.) Link to comment Share on other sites More sharing options...
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