4 ms·

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

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.
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.

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.

Someone 15y ago