4 ms·

Considering that 30 kilometers is almost 100,000 feet you have a lot air you do NOT have to cross. This takes less (solid) fuel and is thus cheaper.

darklajid 15y ago
I already stated my ignorance, so - I'll believe whatever claims are made here. But - my mind desperately tries to fight that and tells me about a _huuuge_ machine that needs to be maintained, and consumes fuel as well. And getting to 30km straight up in a rocket shape instead of going up in a loooong slope powered by someone else seems expensive, in terms of energy used.
- marvin 15y ago
  It's not about energy, it's about momentum and energy/weight ratio. Due to the energy density of chemical fuels, the vast majority of a space rocket's fuel is actually used to carry the _rest_ of the rocket's fuel to the altitude at which it will be burned. I'm not good with rocket physics, but I'm pretty sure that gaining a few kilometers of altitude and 1000 km/h of kinetic energy will have a large impact on the total amount of fuel/propellant required to reach orbit. Is someone in here good enough with rocket physics to comment on this? Of course, it would be even better if they could go to business-jet altitudes or above (15000 meters instead of the 9000 quoted in the article), and I am guessing from the number and size of the engines that this is what they are actually aiming at. But this is just idle speculation on my part.
- bulte-rs 15y ago
  I clicked reply during the time you wrote your own reply I guess. The fact that going up a slope with a jet-engine is preferred is all due to costs. Jet-engines are just cheaper to operate (and easier to supply with fuel). Perhaps more energy is required in total; but this energy is released in a much cheaper way.
  - dalke 15y ago
    Bear in mind that jets carry only fuel and use air for the oxidizer while rockets have to carry fuel and oxidizer with them.
- sp332 15y ago
  If you have to travel through air, you might as well use a wing to get some lift. That way the air is useful, instead of just slowing down your rocket.
- mikeash 15y ago
  747s are cheap to operate compared to rockets. You can buy a ticket on one for less than a thousand bucks. Commercial airlines collectively operate them by the thousand. Airplanes are really cheap in both money and fuel compared to rockets. The technology is more mature, the energies involved are much less, and involves much more finesse, as you gain lift from manipulating the air rather than simply smashing your way through it. A system which can substitute an airplane for a rocket for a decent chunk of the ascent can pull off a big win. Rockets also suffer from painful exponential effects. If you add weight, you need to add fuel. But adding fuel adds weight, so you need to add more fuel. That adds more weight, etc. Removing a small portion of the trip can have big consequences for weight and fuel. Fun fact: the Saturn V first stage "only" got the rocket to about 1/3rd of its final altitude and speed despite making up 3/4ths of the rocket's total mass. Airplanes actually experience the same exponential effect, but because the fraction of the total mass spent on fuel is so small, it's much less significant.
Someone 15y ago
Most of the air, in fact. If you look at http://en.wikipedia.org/wiki/File:Comparison_US_standard_atmosphere_1962.svg http://en.wikipedia.org/wiki/File:Comparison_US_standard_atm..., you will see that air density is way lower at 30km. That is important because a lot of the energy expenditure for rockets is spent overcoming air resistance, instead of for lifting the rocket against gravity (that also is what makes the idea of a space elevator so enticing). Once you are at 30km, getting anywhere is relatively cheap.
- alphaBetaGamma 15y ago
  I'm not sure this is correct. Low earth orbit [1] is ~200km in altitude, and ~7km/s. So to get in this orbit you must provide an energy of 2 MJ/Kg to get 200km high, and 25 MJ/Kg to get to orbital velocity. So the vast majority of the energy expenditure is to get speed, not altitude. I would be suprized if air drag in the first 30Km is the bulk of the energy expenditure. [1] http://en.wikipedia.org/wiki/Orbital_speed#Earth_orbits http://en.wikipedia.org/wiki/Orbital_speed#Earth_orbits
  - Mvandenbergh 15y ago
    That's absolutely correct, but launching above most of the atmosphere means that you don't have to design the rocket to take atmospheric loads. Light-weighting a rocket has a pretty big effect on the payload / rocket ratio which is good. Taking X kg off the weight of a rocket typically means that you can take X/k (where k < 1) off as well because that was additional structure, fuel, or engine capacity that is no-longer needed to support what you've already removed. Of course, now that the rocket is even lighter, there's another round of mass reduction. Don't worry kids, the series converges in a few iterations, but the point is that removing any weight from a rocket is a huge deal.
  - Someone 15y ago
    The best I could find is <http://www.rocketmime.com/rockets/rckt_eqn.html> http://www.rocketmime.com/rockets/rckt_eqn.html>. If, in that example, I adjust air density, I get: - 1.2 = normal density: - top speed 118 m/s - powered height 100 m - total height 335 m - 0.05 = a guess at the density at height: - top speed 150 m/s - powered height 113 m - total height 972 m Back to the problem at hand: if you were to fire this rocket from 30 km up instead of from sea level, it would get about the same potential energy, but about double the kinetic energy. Kinetic energy, as you indicate, being the more important factor, I think it helps if you manage to bring a rocket up high using some other method than a rocket (which, I think, can only be efficient if it accelerates rapidly) However, I am not sure this is correct. Feel free to correct me.

darklajid 15y ago

I already stated my ignorance, so - I'll believe whatever claims are made here. But - my mind desperately tries to fight that and tells me about a _huuuge_ machine that needs to be maintained, and consumes fuel as well. And getting to 30km straight up in a rocket shape instead of going up in a loooong slope powered by someone else seems expensive, in terms of energy used.

marvin 15y ago
It's not about energy, it's about momentum and energy/weight ratio. Due to the energy density of chemical fuels, the vast majority of a space rocket's fuel is actually used to carry the _rest_ of the rocket's fuel to the altitude at which it will be burned. I'm not good with rocket physics, but I'm pretty sure that gaining a few kilometers of altitude and 1000 km/h of kinetic energy will have a large impact on the total amount of fuel/propellant required to reach orbit. Is someone in here good enough with rocket physics to comment on this? Of course, it would be even better if they could go to business-jet altitudes or above (15000 meters instead of the 9000 quoted in the article), and I am guessing from the number and size of the engines that this is what they are actually aiming at. But this is just idle speculation on my part.
bulte-rs 15y ago
I clicked reply during the time you wrote your own reply I guess. The fact that going up a slope with a jet-engine is preferred is all due to costs. Jet-engines are just cheaper to operate (and easier to supply with fuel). Perhaps more energy is required in total; but this energy is released in a much cheaper way.
- dalke 15y ago
  Bear in mind that jets carry only fuel and use air for the oxidizer while rockets have to carry fuel and oxidizer with them.
sp332 15y ago
If you have to travel through air, you might as well use a wing to get some lift. That way the air is useful, instead of just slowing down your rocket.
mikeash 15y ago
747s are cheap to operate compared to rockets. You can buy a ticket on one for less than a thousand bucks. Commercial airlines collectively operate them by the thousand. Airplanes are really cheap in both money and fuel compared to rockets. The technology is more mature, the energies involved are much less, and involves much more finesse, as you gain lift from manipulating the air rather than simply smashing your way through it. A system which can substitute an airplane for a rocket for a decent chunk of the ascent can pull off a big win. Rockets also suffer from painful exponential effects. If you add weight, you need to add fuel. But adding fuel adds weight, so you need to add more fuel. That adds more weight, etc. Removing a small portion of the trip can have big consequences for weight and fuel. Fun fact: the Saturn V first stage "only" got the rocket to about 1/3rd of its final altitude and speed despite making up 3/4ths of the rocket's total mass. Airplanes actually experience the same exponential effect, but because the fraction of the total mass spent on fuel is so small, it's much less significant.

Someone 15y ago

Most of the air, in fact. If you look at http://en.wikipedia.org/wiki/File:Comparison_US_standard_atmosphere_1962.svg http://en.wikipedia.org/wiki/File:Comparison_US_standard_atm..., you will see that air density is way lower at 30km. That is important because a lot of the energy expenditure for rockets is spent overcoming air resistance, instead of for lifting the rocket against gravity (that also is what makes the idea of a space elevator so enticing). Once you are at 30km, getting anywhere is relatively cheap.

alphaBetaGamma 15y ago
I'm not sure this is correct. Low earth orbit [1] is ~200km in altitude, and ~7km/s. So to get in this orbit you must provide an energy of 2 MJ/Kg to get 200km high, and 25 MJ/Kg to get to orbital velocity. So the vast majority of the energy expenditure is to get speed, not altitude. I would be suprized if air drag in the first 30Km is the bulk of the energy expenditure. [1] http://en.wikipedia.org/wiki/Orbital_speed#Earth_orbits http://en.wikipedia.org/wiki/Orbital_speed#Earth_orbits
- Mvandenbergh 15y ago
  That's absolutely correct, but launching above most of the atmosphere means that you don't have to design the rocket to take atmospheric loads. Light-weighting a rocket has a pretty big effect on the payload / rocket ratio which is good. Taking X kg off the weight of a rocket typically means that you can take X/k (where k < 1) off as well because that was additional structure, fuel, or engine capacity that is no-longer needed to support what you've already removed. Of course, now that the rocket is even lighter, there's another round of mass reduction. Don't worry kids, the series converges in a few iterations, but the point is that removing any weight from a rocket is a huge deal.
- Someone 15y ago
  The best I could find is <http://www.rocketmime.com/rockets/rckt_eqn.html> http://www.rocketmime.com/rockets/rckt_eqn.html>. If, in that example, I adjust air density, I get: - 1.2 = normal density: - top speed 118 m/s - powered height 100 m - total height 335 m - 0.05 = a guess at the density at height: - top speed 150 m/s - powered height 113 m - total height 972 m Back to the problem at hand: if you were to fire this rocket from 30 km up instead of from sea level, it would get about the same potential energy, but about double the kinetic energy. Kinetic energy, as you indicate, being the more important factor, I think it helps if you manage to bring a rocket up high using some other method than a rocket (which, I think, can only be efficient if it accelerates rapidly) However, I am not sure this is correct. Feel free to correct me.