Thought this might tickle a few brains here. 1950 movie.
http://www.youtube.com/watch?v=DsisGSBlQqo
Gone to the moon, BRB(Destination Moon)

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Gone to the moon, BRB(Destination Moon)
Evil is evil, no matter how small
I knew it sounded familiar, but it wasn't until I started looking at it and reading the intro credits that it came to me ... Destination Moon, by R A Heinlein! I wasn't aware it had been made into a movie, but it was nice to see he was one of the three screen writers that did the conversion from the novel.
I'll have to watch this when I'm at home (at work right now).
I'll have to watch this when I'm at home (at work right now).
Re: Gone to the moon, BRB(Destination Moon)
Try this. Four trailers inside.kunkmiester wrote:Thought this might tickle a few brains here. 1950 movie.
http://www.youtube.com/watch?v=DsisGSBlQqo
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I don't know about ISP (I don't grok it yet), but the other day I worked out that you need at least about 12km/s delta V to get to lunar orbit (ie, to get into a similar orbit to the moon, not orbit around the moon (eg, leading/trailing trojan): that will be a little extra).
Orbital velocity in LEO (150km) is about 7821m/s.
"Orbit" at 10km is about 7907m/s (yes, I know very well you can't orbit at 10km).
Transfer orbit from 10km is 7949m/s@10km and 7779m/s@150km. Since orbit is impossible at 10km, just use its velocity as delta v to get to LEO (plus some, thus the "at least"), and the circularization delta v is about 42m/s, so about 8000m/s (rounding) + atmospheric drag
Lunar orbit velocity is about 1019m/s.
Transfer orbit from LEO to Lunar is 10967m/s@150km and 186m/s@Lunar
Delta V is about 3979m/s (ie, about 4000m/s).
So total delta V to get to Lunar orbit is 8000 (+ drag losses) + 4000 = 12000m/s + drag losses.
This ignores any effects of the moon on the orbit and EarthLEO transfer is very sloppy. Transfer orbits are Holman (sp?)
BTW: trip time is a shade over 5 days.
Equations used:
vcirc = sqrt (GM/r)
vtrans = sqrt ([2GM/(r1+r2)]*[r1r2/r]) where r1 and r2 are the two circular orbits and r is either r1 or r2 depending on which velocity you want.
G=6.67e11
M=5.98e24kg
Re=6.37e6m
Rl=3.84e8m
Orbital velocity in LEO (150km) is about 7821m/s.
"Orbit" at 10km is about 7907m/s (yes, I know very well you can't orbit at 10km).
Transfer orbit from 10km is 7949m/s@10km and 7779m/s@150km. Since orbit is impossible at 10km, just use its velocity as delta v to get to LEO (plus some, thus the "at least"), and the circularization delta v is about 42m/s, so about 8000m/s (rounding) + atmospheric drag
Lunar orbit velocity is about 1019m/s.
Transfer orbit from LEO to Lunar is 10967m/s@150km and 186m/s@Lunar
Delta V is about 3979m/s (ie, about 4000m/s).
So total delta V to get to Lunar orbit is 8000 (+ drag losses) + 4000 = 12000m/s + drag losses.
This ignores any effects of the moon on the orbit and EarthLEO transfer is very sloppy. Transfer orbits are Holman (sp?)
BTW: trip time is a shade over 5 days.
Equations used:
vcirc = sqrt (GM/r)
vtrans = sqrt ([2GM/(r1+r2)]*[r1r2/r]) where r1 and r2 are the two circular orbits and r is either r1 or r2 depending on which velocity you want.
G=6.67e11
M=5.98e24kg
Re=6.37e6m
Rl=3.84e8m
Isp is best thought of as the measure of rocket efficiency. Technically it is the velocity of the rocket exhaust products divided by gravitational acceleration. For example, H2/LOX has an exhaust velocity of ~4500m/s, and a theoretical max Isp of just over 450 seconds ((4500m/s)/(9.80665m/s^2)).taniwha wrote:I don't know about ISP (I don't grok it yet),
The Rocket Equation:
dv = Isp*(G)*ln(Mo/Mf)
dv = deltav, the change in velocity you want the rocket to deliver
Isp = Specific Impulse, units denoted in seconds
G = the Earth's gravitational acceleration, 9.80665m/s^2
ln = natural log
Mo = Mass original, gross mass prior to rocket ignition
Mf = Mass final, rocket mass after the target dv is achieved
Closer to 16,000m/s. Both the Moon and Mars oddly require almost the same dv to land on, Earth surface to alien surface.taniwha wrote:but the other day I worked out that you need at least about 12km/s delta V to get to lunar orbit (ie, to get into a similar orbit to the moon, not orbit around the moon (eg, leading/trailing trojan): that will be a little extra).
You also need to add in gravitational drag, the Earth's gravitational field trying to drag you back down. Both atmo and grav drag together add a penalty of about 2000m/s to the baseline LEO orbital velocity. The average design dv to reach LEO is between 9500 and 10,000m/s, depending on where you launch from and what altitude of "LEO" you want to achieve.taniwha wrote:Orbital velocity in LEO (150km) is about 7821m/s.
"Orbit" at 10km is about 7907m/s (yes, I know very well you can't orbit at 10km).
Transfer orbit from 10km is 7949m/s@10km and 7779m/s@150km. Since orbit is impossible at 10km, just use its velocity as delta v to get to LEO (plus some, thus the "at least"), and the circularization delta v is about 42m/s, so about 8000m/s (rounding) + atmospheric drag
http://en.wikipedia.org/wiki/File:Deltavs.svgtaniwha wrote:So total delta V to get to Lunar orbit is 8000 (+ drag losses) + 4000 = 12000m/s + drag losses.
Hohmann.taniwha wrote:This ignores any effects of the moon on the orbit and EarthLEO transfer is very sloppy. Transfer orbits are Holman (sp?)
Vae Victis
djolds1 wrote:Isp is best thought of as the measure of rocket efficiency. Technically it is the velocity of the rocket exhaust products divided by gravitational acceleration. For example, H2/LOX has an exhaust velocity of ~4500m/s, and a theoretical max Isp of just over 450 seconds ((4500m/s)/(9.80665m/s^2)).taniwha wrote:I don't know about ISP (I don't grok it yet),
The Rocket Equation:
dv = Isp*(G)*ln(Mo/Mf)
As I thought, Isp alone isn't enough to answer kunkmiester's question. I just didn't have the confidence.
djolds1 wrote:Closer to 16,000m/s. Both the Moon and Mars oddly require almost the same dv to land on, Earth surface to alien surface.
Very interesting, and a little surprising, though it agrees with the figures quoted in Apollo 13 (just watched it for the 3rd time, Japanese subs mentioned 17000+m/s on reentry). Note, however, my figures aren't for landing on the moon, or even going into orbit around it.
djolds1 wrote:You also need to add in gravitational drag, the Earth's gravitational field trying to drag you back down. Both atmo and grav drag together add a penalty of about 2000m/s to the baseline LEO orbital velocity. The average design dv to reach LEO is between 9500 and 10,000m/s, depending on where you launch from and what altitude of "LEO" you want to achieve.
Does using the parameters of an elliptical orbit from ground not take gravitational drag into account? Indeed, it doesn't take the atmospheric drag into account.
BTW, interesting chart. I got a LEO to lunar distance orbit deltav of 3979m/s, so an extra 100200m/s for lunar landing (orbit?) seems reasonable considering the gravitational field enroute to the moon gets quite interesting.
djolds1 wrote:Hohmann.
Thanks for that.

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Oh, indeed, but Isp isn't even half the equation. For any specific Isp, you can get from LEO to the Moon so long as you have enough reaction mass.
(reaction_mass+dry_mass)/dry_mass = exp (delta_V/(g*Isp))
Of course, it gets ugly fast as your Isp drops.
And then the power requirements are proportional to Isp squared. More specifically, for F=Me*Ve, ΔV=Ve*log((Mr+Md)/Md), P=Me*Ve*Ve/2 (F=thrust, Me=mass rate of propellant, Ve=velocity of propelant (g*Isp), Mr=reaction mass, Md=dry mass).
Rocket science is easy. Rocket engineering is... not.
(reaction_mass+dry_mass)/dry_mass = exp (delta_V/(g*Isp))
Of course, it gets ugly fast as your Isp drops.
And then the power requirements are proportional to Isp squared. More specifically, for F=Me*Ve, ΔV=Ve*log((Mr+Md)/Md), P=Me*Ve*Ve/2 (F=thrust, Me=mass rate of propellant, Ve=velocity of propelant (g*Isp), Mr=reaction mass, Md=dry mass).
Rocket science is easy. Rocket engineering is... not.

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Heh
Anyway, I figured out why djolds1 was saying my math wasn't enough to account for gravity drag: it's because it takes time to build up that deltav.
I'm running a simulation of a ship going from LEO (150km) to "lunar" orbit (ie, same orbit as the moon, not in orbit around the moon). I set the ship's acceleration to 0.1m/s/s (1/100th of a G). The ship looped around the Earth ~3.75 times before the apogee extended past the moon, and the orbits were still highly circular up to about 3.5. I haven't yet worked out the deltav used.
When I'm done, I'll post the simulation itself (a Blender file).
Anyway, I figured out why djolds1 was saying my math wasn't enough to account for gravity drag: it's because it takes time to build up that deltav.
I'm running a simulation of a ship going from LEO (150km) to "lunar" orbit (ie, same orbit as the moon, not in orbit around the moon). I set the ship's acceleration to 0.1m/s/s (1/100th of a G). The ship looped around the Earth ~3.75 times before the apogee extended past the moon, and the orbits were still highly circular up to about 3.5. I haven't yet worked out the deltav used.
When I'm done, I'll post the simulation itself (a Blender file).
Read up on Bob Truax's '60s concept called "Sea Dragon."taniwha wrote:Rocket science is easy. Rocket engineering is... not.
Rocket engineering is relatively easy. Goldplated elegant and overpriced rocket engineering is expensive.
In part. In rocket engineering, mass expended on the way from "Mo" to "Mf" includes not only fuel burned, but also the mass of rocket stages that have been discarded during flight. Also, the usual rule is lower Isp = higher thrust engines; usual but not always. Orions and Nuclear Salt Water Rockets are dirty mothers that deliver the best of both worlds.kunkmiester wrote:ISP tells you how much fuel you need though, for a given time/thrust, right? And that's what you need to know to get all that delta V. Right?
Vae Victis
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