Some help with rocket equations

The more appropriate group for this question is probably sci.space.tech, however, moderation seems to take forever over there.
My question is about relative amounts of fuel mass required to double the maximum altitude of a rocket. A friend of mine told me that in order to double SpaceShipOne's X-Prize winning altitude (~360,000 feet) it would take 40x as much fuel. This seems a tad high to me, put I was unable to reduce the algebra to something manageable. If someone with sharper skills could help me out it would be much appreciated.
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snipped-for-privacy@gmail.com wrote:

It's close. Assume 20x as a minimum.
SS1 was low mass fraction.
The better question is "what mass fraction", and the answer would be 85 - 90%.
Jerry
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That's way off.
The potential energy of altitude is simply U=m*g*h, so you need to deliver a minimum of twice the energy per unit mass to get twice as high.
You need more than twice the propellant, however, since you have to get the rocket up to a higher speed, and it also takes more structure to contain the propellant. As a rough guess, 3x the propellant mass should do it.
Since the portential enery equals the kinetic energy: U = K = 1/2*m*v2.
SSO reached Mach 3.5. Orbital velocity is M~24. The energy difference is (24/3.5)^2 ~ 47x the energy per unit mass to get it into orbit. 40x the original propellant would get it close to orbital velocity, but not all the way.
In either event it would be a one way trip. SSO would most likely burn up if it went much higher since the reentry velocity and stagnation temperature would overheat and fail the simple composites used in SSO.
Bob Krech
snipped-for-privacy@gmail.com wrote:

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Bob Krech wrote:

Good explanation. Two additional factors:
1. you need extra fuel to lift the extra fuel that you'll be burning later 2. Even though SS1 starts off at 50kft, a lot of the fuel it expends is lost to drag. Thrust late in the burn is a lot more effective that thrust at the beginning of the burn.
That 3x propellant mass estimate would be pretty close if SS1 was a rocket, but SS1 isn't very aerodynamically efficient compared to most rockets.
If you were to double SS1's fuel capacity and thrust, add 2400kg for the additional fuel, and keep everything else unchanged (including burn time), I'm calculating an apogee around 350km. After that, you'll start seeing limited returns.
Dave
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Dave
Once you're over ~50 km, you're essentially out of the atmosphere and SS1 behaves more like a rocket than a plane on ascent, and aerodynamic efficiency is really out of the picture for the SS1 flight profile.

http://scaled.com/projects/tierone/logs-WK-SS1.htm you will see that at burnout SS1 was traveling at Mach 3.09 @ 213,000kft. This is only an EAS (Equivalent Air Speed) of 24 knots so the aerodynamic drag force at this altitude and velocity can be ignore compared with the motor propulsive force. http://www.aerospaceweb.org/design/scripts/atmosphere/ In fact, the control surfaces are useless here and cold gas thrusters were used for attitude control.
To double the vertical altitude of SS1, you would most likely want to add the additional propellant (and casing weight) and design the burn profile to kept the same thrust to weight ratio as the original flight profile but double the burn time. This would require somewhat more than a factor of 2 increase in propellant mass, that's why I estimated a factor of 3.
For determining vertical altitude, potential energy is approximately the correct measure of energy required for the mission since horizonal velocity increments gain nothing. For orbital insertion, however, it's totally different. It is the horizontal velocity that must be incremented, and the horizonal kinetic energy far overwhelms the vertical potential enegy required to gain altitude.
The design and construction limitations of SS1 will not allow either doubling of the altitude nor achievement of orbit for other than a 1 way mission because the vehicle could not withstand the increase reentry temperatures due to the higher reentry velocities.
Bob Krech
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It does indeed take very little fuel to change orbital dynamics ONCE in orbit or far enough away from the Earth's surface. You see most of the WORK of a rocket engine/motor goes into over coming frictional momentum losses with the atmosphere and over coming the Earth's attraction. The analogy of an atom's structure can be LOSELY applied to this. Electrons closer to the nucleus are more tightly bound in their energy states as opposed to those who are further away from the nucleus. Electrons are actual in distributions around the nucleus and don't actually ORBIT the nucleus. In Quantum Physics, these are referred to as shells and there are equations that try to "best guess" the location of the electron based on probabilities. But alas, we know that Heisenberg's uncertainty principle clearly shows that we can NEVER take an actual measurement of the electron :) And yes I am mixing Gravitational Force (weak force) with Nuclear Force (strong force), I just need the analogy.
So, ~95 percent off the energy/work goes into just getting off the ground and up to an altitude where an orbital path can be reached. Once in orbit, if fuel allows, it takes slight perturbations/impulses (2 actually) to change and then maintain the new orbit. The new orbit has a higher energy of course, but each successive "burn" takes less and less fuel as you are getting further and further away from the Earth's gravitational pull. This is why geosynchronous satellites don't need very much fuel to push them into "on-station" orbits and then once they are no longer needed, nudge them into higher "off-station" orbits.
One way to think about this is through orbital and earth escape velocities: It requires a mass of any size/density to reach a speed of 17,800 mph to stay in orbit at ~90 mi. Now for that very same mass, it requires 25,000 mph to have enough velocity to escape the Earth's influence entirely. One can clearly see that the speed needed in NOT doubled as its only 7,200 mph more. Why, because most of the work has ALREADY been done just getting to orbit.

look at the ratios of energies at Low Earth Orbit (LEO) and Escape Velocity Orbit (EVO). I am ignoring the gravitation potential energy by assuming my point mass to be very small. I just want my point mass to orbit and then leave the local earth system.
E sub LEO = .5m(17,800mph)^2 , E sub EVO = .5m(25,000mph)^2
E sub LEO .5m(17,800mph)^2 (17,800mph)^2 ----------------- = ------------------- = ------------------- = 0.507 E sub EVO .5m(25,000mph)^2 (25,000mph)^2
Which => (implies) that the energy in the orbit, needed to escape the Earth's gravity well is about twice needed to orbit it at ~ 90 mi. Actually be neglecting the potential it came out higher. The true value is ~1.41 times more energy. If you set Energy initial to Energy final, you will have a far more accurate result. ( Kinetic + Potential) sub i (Kinetic + Potential) sub f. Just set Potential final = zero. This forces a situation where the object is infinitly far from Earth (the object has escaped the Earth's gravitational influence and is now under the pull of the moon, sun, galaxy,ect.)
The percentage difference in orbital velocity to escape is 28.8%. So you see that most of the HARD work is done just reaching orbit. After that, based on your mass, you will have to carry enough fuel to make the changes. So now you have to consider the mass of object and the mass of the available fuel on board. You have to expel enough fuel to push your mass to the velocity needed to reach the desired orbit.
For further reading/study, I recommend the following book:
(Amazon.com product link shortened)18181464/sr=8-1/ref=sr_8_xs_ap_i1_xgl14/002-1141490-4868863?v=glance&s=books&nP7846
by George P. Sutton. Very good book and well worth the investment.
The equations you need to study are conservation of energy laws, Newton's law of Gravity, and perhaps a bit of Kepler's laws if you chose to see the "big" picture :)
Good luck with your endevors.
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snipped-for-privacy@juno.com wrote:

I'm going to disagree with that statement. The potential energy required to get to an altitude of 100 miles (neglecting the decrease in gravitational acceleration) is:
m*g*h = 1,582,385 J/kg
The kinetic energy of an object in orbit is:
.5*m*v^2 = 32,589,849 J/kg
The kinetic energy of an object in orbit is MUCH greater than the potential energy needed to get there. If you've ever looked at a launch schedule, you'll see that the vast majority of a rocket's burn is excecuted well over the atmosphere. That's because it takes A LOT more energy to get up to orbital velocity than it does to get up past the atmosphere. Secondly, atmospheric drag only accounts for less than about 30% of the total energy of a launch.

Actually, you're almost DOUBLING the kinetic energy an object by accelerating it from 17,800mph to 25,000mph. Secondly, using orbital velocity as a baseline isn't exactly realistic, since orbit is a VERY high energy state.
When you're in orbit, you've overcome very little of the earth's gravitational well, but you have a lot of kinetic energy. Think of it this way: if you were to hold an object 500,000 miles over earth, then drop it, it would hit the ground at over 7 miles per second. And that's just from the conversion of gravitational potential energy to kinetic energy.
Hope that helps, Dave
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dave.harper wrote:

Yes, that is why I neglected the potential energy to make the calculation a bit more straight forward. I also took a mass to be VERY small, so that potential wouldn't be much of a factor.

Ummm the vast majority of a rockets burn is ued up reaching orbital altitude ... I do believe and I will check and post this, but the first stage of a Saturn V contains more fuel than the rest of the vehicle .. let me google the Sat V specs ... yes here it is ... the first stage burns the most propellant ... for the first 37 miles of the flight:
Size: 111 m (363 ft) Payload to orbit: 129,300 kg (285,000 lb) Payload to Moon: 48,500 kg (107,000 lb) Manufacturer: Boeing Co. (prime) 1st stage: five F-1 engines Propellants: RP-1 (kerosene) and liquid oxygen Total thrust: 33,360,000 newtons (7,500,000 lb) Manufacturer: Rocketdyne 2nd stage: five J-2 engines Propellants: liquid hydrogen and liquid oxygen Total thrust: 5,560,000 newtons (1,250,000 lb) Manufacturer: Rocketdyne 3rd stage: one J-2 engine Thrust: 1,112,000 newtons (250,000 lb)
So I am a bit confused by your statement above where you point out that the majority of the rocket's burn is above the atmosphere ... please elaborate.

Well the Spaceship One had an orbit ... it just failed to put enough energy into its orbit to miss the Earth ...

Newton's law for Gravity - F = GMm/r*r as you leave the Earth the force of Gravity falls off VERY quickly. Remember, gravity is a WEAK force. If you don't believe this ... just stand up! Notice that the nuclear forces of your floor easily over come the force of gravity (your weight) pressing down upon it). Gravity is a very weak force in the universe.

I hope that helps too :)
Lunarlos
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snipped-for-privacy@juno.com wrote:

Well, you quoted that about 95% of the energy/work goes into getting it off the ground. But in actuality it's a small fraction of the total lauch weight. I'm confused why you now say you "neglected" it?
Also, increasing mass has the same effect on both potential and kinetic energy. If you increase the mass, the total kinetic energy needed and total potential energy needed increase by the same factor.

<specs cut>

Staging changes things around a little, since you drop off mass a lot quicker as you loose stages... (this means you require less fuel, and thus less energy, than you would a comparable single-stage rocket).
However, the first stage operates for only 150 seconds. And it gets it to an ALTITUDE of 37 miles (and a speed of 6,000 mph), which is well over any significant drag-inducing atmosphere. The second stage then operates for another 360 seconds, getting it to about 15,000mph. The third stage burns for a short while to get it to 18,000mph. So the vast majority of the burn occurs well over the atmosphere (i.e. where drag is very low).

Failed by A LOT. It would have needed ATLEAST 25 times as much energy to get enough velocity to get into orbit.

...yes, but it's also the furthest acting when compared to the rest of the fundamental forces. Although it drops off relatively quickly, it doesn't drop off quite as fast as you might think. Remember, 'r' for us is already 6000+ km. Here's a table:
Alt(mi) g(m/s) 0 9.8 100 9.3 200 8.9 300 8.5 400 8.1 500 7.7 600 7.4 700 7.1 800 6.8 900 6.5 1000 6.2 2000 4.3 3000 3.2 4000 2.4 5000 1.9 6000 1.5 7000 1.3 8000 1.1 9000 0.9 10000 0.8
The gravity operating on most LEO spacecraft is still about 80% to 90% as much as it is here on the ground. Even at an altitude of 1000 miles, it's still over 60% of what it is on the ground.
Dave
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I see this is getting into a pissing match ... and you win. I just don't have the time for it ... sorry. Nothing I have said is false or has been proven to be in error. You wish to split hairs, then you can have at it. The underlying physics I have described landed man on the moon. I am not sure what point you are trying to make, but frankly ... I really don't care :)
Maybe you should stop talking about it, and build your own space vehicle? I have, and actually placed equipment into space. You are very very wrong on alot of your assumptions. Gravity is NOT the furthest acting force ... this is why we have sent objects out to the Sun's Heliosphere ( Voyager 1 passed into it a few weeks ago, and will stay in it another 10 years or so). Gravity doesn't act at a distance ... you are VERY wrong on that statement. Please study Einstein's Special Theory of Relativity and if you comprehend it, you will see that Gravity doesn't act over great distances like Newton thought. If the Sun should disapear, it will take 8.5 minutes for us to feel the effects and be thrown out of orbit. It doesn't take place instantaneously. Also, the universe is expanding, and unless there is some great amount of "dark matter" found it will continue to expand until enthalpy does its job and everything grows cold ...
Oh well, good luck ... I just don't want to pull teeth anymore ...
Good Luck
Lunarlos
P.s. your table proves my point. Thank you. Over very short distances, 10k mi is VERY short when speaking in terms so Lunar, Solar, and Stellar distances, the force of Gravity falls off VERY quickly. At least on this we agree! :)
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snipped-for-privacy@juno.com wrote:

Voyager is still within the suns gravitational influence. It is at escape velocity, so it will never come back, but that doesnt' mean gravity has dissapeared.

Gravity propogates at the speed of light, yes. What does that have to do with the distance gravity has an effect at? The planets are held in the solar system by the suns gravity acting at a distance. The sun is held in the galaxy by gravity acting at even greater distances. The galaxy is being pulled toward its neighbor, Andoromeda, and a supercluster of galaxies by gravity acting at even greater distances.

What does that have to do with anything? Now I feel your just trying to sound smart, which the 'built your own space vehicle' reinforces.

Gravity follows the inverse square law. This means it falls of very quickly, but never completely dissapears. Gravity from stars in other galaxies is acting on you right now, the force is just too small to notice, or even measure.
The fact that the earths gravitational pull is negligible at interplanetary distances (definatly not lunar, or the moon wouldn't be in orbit!) has little to do with getting Space Ship One into orbit in the first place, of course. To reach those distances one must first reach escape velocity, which takes even more DeltaV than reaching orbit.
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snipped-for-privacy@juno.com wrote:

I didn't think this was getting into a pissing match. There's been no name-calling or insults. Just because we disagree on some points doesn't make it a pissing contest.

The main reason I responded to your first post was because you stated the following:

Do you still believe your statement to be true? (See our previous math)

You made your own space vehicle? Or made equipment that went up on one? There's hundreds of thousands of people in the U.S. that have been a part of making equipment that has gone into space (me included). Making equipment that goes into space and designing the rocket to get there are not the same thing.

I'm open to that possibility. Please let me know what incorrect assumptions you think I've made?

Since most objects in the universe are electrically neutral, it effectively is. The strong, weak, and electromagnetic forces do not influence distant objects as far as gravity.

First, gravity does act at a distance. If it didn't, the please explain our spiral galaxy. If gravity didn't act at a distance, it wouldn't spin, and would fly apart.
Secondly, the SPECIAL theory of relativty deals with reference frames and relative velocity. I think you mean the GENERAL theory of relativity. Regardless, GToR doesn't state that gravity doesn't act at great distances.

Correct, this statement deals with the speed of gravity. How does this support your statements?

Again, correct. It's expanding because it's reached escape velocity (from a simple viewpoint). How does this support your statements?
Dave
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snipped-for-privacy@juno.com wrote:

Its effects are measureable (via galactic cluster mechanics and gravitational lensing) over millions of light years. What force is further acting?

Ummmm, gravity does act at a distance, as you pointed out previously, Gm1m2/r^2.
The heliosphere boundaries (heliopause) are not gravity related. Somewhat ironically, the heliosphere is created by magnetic and electrostatic (ionized solar wind) force interactions.

Special Relativity concerns velocity issues; time dilation, Lorentz contractions, relativistic mass, etc, within inertial, or non-accelerating, reference frames. Gravitational fields induce accelerations (forces) and are not treated in Special Relativity.
Perhaps you mean General Relativity, which treats accelerating reference frames and gravity. It does not limit the "range" of gravitational (space-time distortion) effects.

Enthalpy is a thermodynamic system state variable; a measure of something, not a process. You are most probably referring to the 2nd Law of Thermodynamics.
Indeed, missing "dark matter" would tend to counteract the expansion, because of its GRAVITATIONAL effect upon the entire universe (large distances). Unfortunately, there is some reason to believe the expansion rate is increasing, hence there is not enough dark matter to be found, if true. Try searching the web for "Dark Energy", somewhat of a misnomer.

Compared to what? Most forces and field strengths, from spherical or point sources, fall off in an inverse square fashion. What force falls off less rapidly over comparable distances? Saying gravity falls off "very quickly" begs for a comparison. I say it falls off at the universally (sic) average rate; k/r^2.
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Gary Bolles

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snipped-for-privacy@juno.com wrote: > > dave.harper wrote: > >> snipped-for-privacy@juno.com wrote: >> >> >>>So, ~95 percent off the energy/work goes into just getting off the >>>ground and up to an altitude where an orbital path can be reached. >> >>I'm going to disagree with that statement. The potential energy >>required to get to an altitude of 100 miles (neglecting the decrease in >>gravitational acceleration) is: >> >>m*g*h = 1,582,385 J/kg >> >>The kinetic energy of an object in orbit is: >> >>.5*m*v^2 = 32,589,849 J/kg > > > Yes, that is why I neglected the potential energy to make the > calculation > a bit more straight forward. I also took a mass to be VERY small, so > that potential wouldn't be much of a factor. > > >>The kinetic energy of an object in orbit is MUCH greater than the >>potential energy needed to get there. If you've ever looked at a >>launch schedule, you'll see that the vast majority of a rocket's burn >>is excecuted well over the atmosphere. That's because it takes A LOT >>more energy to get up to orbital velocity than it does to get up past >>the atmosphere. Secondly, atmospheric drag only accounts for less than >>about 30% of the total energy of a launch. > > > Ummm the vast majority of a rockets burn is ued up reaching orbital > altitude ... I do believe and I will check and post this, but the first > stage of a Saturn V contains more fuel than the rest of the vehicle .. > let me google the Sat V specs ... yes here it is ... the first stage > burns the most propellant ... for the first 37 miles of the flight: > > Size: 111 m (363 ft) > Payload to orbit: 129,300 kg (285,000 lb) > Payload to Moon: 48,500 kg (107,000 lb) > Manufacturer: Boeing Co. (prime) > 1st stage: five F-1 engines > Propellants: RP-1 (kerosene) and liquid oxygen > Total thrust: 33,360,000 newtons (7,500,000 lb) > Manufacturer: Rocketdyne > 2nd stage: five J-2 engines > Propellants: liquid hydrogen and liquid oxygen > Total thrust: 5,560,000 newtons (1,250,000 lb) > Manufacturer: Rocketdyne > 3rd stage: one J-2 engine > Thrust: 1,112,000 newtons (250,000 lb) >
Nothing here backs up your claim that most of the fuel is in the first stage. I am not saying it isn't true, but you make a claim then provide information which you say backs it up. Except that it doesn't. A very poor way to win a debate. Then there is the significant detail that the first stage does more than just lift the rocket up. It also adds a significant fraction of the required orbital velocity.
Not to mention that you started with an energy argument and have now switched to mass. Yet the first stage of the Saturn V uses a low ISP fuel which of course increases the mass in that stage.
Getting back to the claim that 95% of the energy expended is used _just_ to reach orbital altitude rather than orbital velocity, perhaps an example will illustrate the problems with that arguement.
Assume a SSTO ship to keep things simple. The Delta-V budget for LEO typically given is 7.8km/s orbital velocity plus 1.5 to 2 km/s for drag and gravity losses. Also assume that the rocket first expends fuel on the gravity/drag losses and then does the orbital burn. This isn't true but it keeps things simple and actually favors your arguement. Assume SSME exhaust velocity of 4500 m/s.
Mass left after 2 km/s burn:
m1 = m0 e^(-2,000/4500) = 0.64 m0
and after the 7.8km/s burn:
m2 = m1 e^ (-7,800/4,500) = 0.18 m1
36% of the initial launch mass is burned overcoming gravitational/drag losses. 82% of what is left or 52% of the initial launch mass is required for the orbital velocity.
52% vs. 36%
I fail to see anything even vaguely resembling a 20 to 1 ratio.
If you used Saturn V first stage engines that have an ISP of 304s for the 2km/s second burn, the mass fraction would be:
m1 = m0 e^( -2,000/ 304s * 9.8m/s/s) = 0.51
> > So I am a bit confused by your statement above where you point out that > the > majority of the rocket's burn is above the atmosphere ... please > elaborate. > > >>>One way to think about this is through orbital and earth escape >>>velocities: It requires a mass of any size/density to reach a speed of >>>17,800 mph to stay in orbit at ~90 mi. Now for that very same mass, it >>>requires 25,000 mph to have enough velocity to escape the Earth's >>>influence entirely. One can clearly see that the speed needed in NOT >>>doubled as its only 7,200 mph more. Why, because most of the work >>>has ALREADY been done just getting to orbit. >> >>Actually, you're almost DOUBLING the kinetic energy an object by >>accelerating it from 17,800mph to 25,000mph. Secondly, using orbital >>velocity as a baseline isn't exactly realistic, since orbit is a VERY >>high energy state. > > > Well the Spaceship One had an orbit ... it just failed to put enough > energy into its orbit to miss the Earth ... > > >>When you're in orbit, you've overcome very little of the earth's >>gravitational well, but you have a lot of kinetic energy. Think of it >>this way: if you were to hold an object 500,000 miles over earth, then >>drop it, it would hit the ground at over 7 miles per second. And >>that's just from the conversion of gravitational potential energy to >>kinetic energy. >> > > > > Newton's law for Gravity - F = GMm/r*r as you leave the Earth the > force of Gravity falls off VERY quickly. Remember, gravity is a WEAK > force. If you don't believe this ... just stand up! Notice that the > nuclear forces of your floor easily over come the force of gravity > (your weight) pressing down upon it). Gravity is a very weak force in > the universe. >
The difference in the gravitational force from sea level to LEO is small. Mean radius of the Earth is 3959 miles, assume an LEO altitude of 100 miles. So the difference (assuming point sources) is 3959^2/ (3959+100)^2 or 95% of the sea level value.
If gravity falls off so rapidly, why does the ESA claim that the force of gravity at the ISS is 88% of that at sea level?
http://www.spaceflight.esa.int/users/file.cfm?filename c-iss-pe-g
> >>Hope that helps, >>Dave > > > I hope that helps too :) > > Lunarlos >
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