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Patrick

Sun Nov 26 2017, 03:26AM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

Im looking at the math and physics of orbiting objects and im trying to figure out mass fractions of the rockets components and the end result.

I drew this based on what i think i know on gravity, magnetism and electrostatic fields.

If i understand the "Gravity well" idea, the energy lifting one Kg near earth surface a constant velocity / height "unit" is more energy expensive than the same unit change higher up for the same Kg.

If so, i conclude that 1Kg leaving the ground at 0 speed, 0 altitude costs alot, a whole lot more than lifting a Kg from a higher elevation even with 0 horizontal component, such as a rockoon or X-15.

Carbon_Rod

Sun Nov 26 2017, 07:33AM

Registered Member #65 Joined: Thu Feb 09 2006, 06:43AM
Location:
Posts: 1155

If I recall, there is a mathematical proof someplace stating a 3 to 4 stage rocket had the most efficient payload capability.

The part I could never figure out was how people compensated for the change in the center-of-pressure relative to the vertical change in the center-of-gravity during the engine burn. In this regard, each stage's mass distribution must be carefully designed to keep the remaining body above stable, and thus the theoretical ideal model looks very different from what will actually work. Additionally, the models for solid and liquid fuel are usually inversely related, so one is left to wonder if both the Shuttles engine systems running for the initial part of the missions improved stability.

This is why I am impressed every time SpaceX lands a reusable rocket, as the physics become less stable as the fuel reaches empty in these vehicles. I wouldn't personally risk my dog on those flights, but it is interesting to think about when bored.

klugesmith

Sun Nov 26 2017, 08:53PM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1714

>>If i understand the "Gravity well" idea, the energy lifting one Kg near earth surface a constant velocity / height "unit" is more energy expensive than the same unit change higher up for the same Kg.

It's not clear what change you are talking about. Delta v or delta h?

Delta v view: A rocket stage directly delivers Î”v to the rest of the vehicle. Same value, no matter how high you are or how fast you start. The engine's total impulse (delta momentum) divided by the accelerated mass (not-yet-burned fuel + dry weight of the stage + the rest of the vehicle). Concurrently with the rocket burn, the vehicle is accumulating some Î”v downward from gravity and backward from drag.

Delta h view. Lifting a mass against gravity, from h1 to h2 with no net velocity change, takes energy proportional to h2-h1. To account for the small change in "g" between sea level and LEO altitudes, it can be convenient to express the heights as "geopotential heights" (a term common in atmospheric science).
Since your sketch uses miles:

h_geometric h_geopotential
  0  0
 20 19.9
 40 39.6
 60 59.1
 80 78.4
100 97.5

In this context, the term "gravitational well" makes me think of solar-system orbital mechanics. When closely passing a planet (or entering / changing / escaping from orbit), the Î”v from rocket engine is most efficient when added at or around periapsis. Deep in the well, when the ballistic velocity is maximum.

I think that model isn't particularly useful for ground-to-orbit problems and ballistic missile problems. The height scale is small compared to planet radius, so gravity field is approximately uniform. Along the trajectory, the change in g's direction could be more important than changes in its magnitude.

The primitive 1-dimensional finite-element model above is based on a spherical, nonrotating Earth. In practice, geopotential height calculations include the latitude-dependent variation in g. The direction of g is always vertical (exactly normal to a level surface), by definition, but that's generally different from a line to the Earth's center.

Patrick

Sun Nov 26 2017, 10:47PM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

Ok let be try to clear up my end goal of understanding.

How much more energy is needed to push 1kg to orbit from the ground, than if that 1kg started at 120,000 ft and went to the same orbit.

it seems like theres an enormous exponential pain / cost to start at the earths surface.

klugesmith

Mon Nov 27 2017, 04:50AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1714

Well let's take a quantitative look, but start with a guess. Next year's launch of the Parker Solar Probe, which is not exceptionally massive, will use a Delta IV Heavy rocket. Here's one of the three booster cores:

Then after about seven gravity assists from close passes over Venus, the probe can get deep into the Sun's gravity well -- about 9 solar radii away from the surface. Now suppose there were a 120,000-foot-high mountain in Florida, and the probe could be launched from its summit. I bet it would still need a Delta IV Heavy, or similar class rocket.

Back to your question.

First, suppose the lift to orbital altitude (say, 160 km) and the horizontal acceleration are done separately.
Lifting energy is m * g * h. In round numbers, taking g to be 10 m/s^2, the 1-kg payload needs 1.6 MJ from sea level or 1.2 MJ from h = 40 km.
Then it needs almost 8 km/s of horizontal velocity for a circular orbit. Kinetic energy is m * v^2 / 2; for 1 kg that's about 32 MJ.

We'll use a bit more precision in a model to illustrate some economies in rocketry, contrasting velocity addition with height addition.

The geopotential heights of our orbit and a balloon at 120,000 feet are about 156.1 and 36.4 km. Taking g to be 9.8, the lift energies for 1 kg are 1.53 and 1.17 MJ. If we ignore atmospheric drag, the imparted velocity requirements are 1.75 and 1.53 km/s. (178 vs 156 seconds to coast up to peak altitude. I think at least one of Gerry Bull's HARP gun projectiles went higher than that in the 1960's.)

The real circular-orbit velocity at 160 km is only about 7.81 km/s (KE = 30.5 MJ/kg).
Earth rotation speed at the equator is about 0.46 km/s (0.11 MJ/kg for payload in storage at launch site). We can boost that to orbital velocity with a rocket delta-v of only 7.35 km/s (27.0 MJ/kg, a savings of almost 10%).

Actual optimum launch trajectories are a curve that manages the lifting, horizontal acceleration, and getting out of dense air ASAP within limits on aerodynamic loading. It's the last part where balloons, or even fixed-wing aircraft, can help substantially -- unless you still need a million-pound rocket.

Another benefit: first-stage engine nozzles could be optimized for lower ambient pressure, instead of having to balance sea-level thrust with near-vacuum thrust.

This non-rocket-scientist tried converting the total orbit energy (kinetic + potential) to equivalent delta-v, then subtracting the earth-rotation v. Resulting rocket delta-v values are 7.54 and 7.49 km/s, for launches from sea level and from 120,000 feet.

Plasma

Tue Nov 28 2017, 09:25AM

Registered Member #61406 Joined: Thu Jan 05 2017, 11:31PM
Location:
Posts: 268

I can't explain these formula at this point, you'll probably work them out.
energy = 0.5*R0*g0*((R0+2*h)/(R0+h))

R0 = 6374km,earth radius
G0 =9.806m/sec
h = height above sea level m

Subtiting has of 1meter and 100,000meters,to get energy difference.

Velocity = R0*sqrt((2*g0)/(R0+h)

Fg0= G*m1*m2/R^2
m1 = 5.974*10^24kg
m2 rocket mass
Green = 6.670*10-^11 m/kgsec
Ryan = Ryan difference between centre of earth and rocket centre of mass.

Copressure is composated by thrust vector ing
Hope it helps

Edit some tables, don't know the accuracy
50%fuel 2.7km/sec delta vee
75% 5.1
80% 6.3
90% 9
95% 11.6

For a rocket of 400Isp and 3.9km/sec exhaust velocity.
you need 7.6km/sec delta vee to get into earth orbit.

Patrick

Wed Nov 29 2017, 02:32AM

Registered Member #2431 Joined: Tue Oct 13 2009, 09:47PM
Location: Chico, CA. USA
Posts: 5639

ok let me ask differently.

lets say it takes a Delta 4 of 500,000 kg to lift 20,000 kg to orbit from ocean level.

now i want to lift the same 20,000 kg to orbit from 120,000 feet. How much less mass could the Delta 4 weigh now?

alot of that 500,000 kg is mass burned just to get the rest of the mass to 120,000 feet. Right ?

Plasma

Wed Nov 29 2017, 06:22AM

Registered Member #61406 Joined: Thu Jan 05 2017, 11:31PM
Location:
Posts: 268

It's all about velocity, at sea level and low orbit the force of gravity that the rocket is fitting against is less, more power goes to acceraltion. Compare 3627km radius add 100 even squared at a high value number doesn't make much difference, say exhaust velocity is related to temperature above 20000K the increase in speed is minimal, but the energy is still there.
Will lookup a table that shows at 1MKelvin small trips around the solar system is possible.

I will let you know what I'm working on, is water and hydroxide get hydro pumped into Al, it makes H2, that gets heated buy electricity in a format, and levels as Hydrogen.
at 1200K and k-1 the ISP is Max 1600,plug that into delta vee

Uspring

Thu Nov 30 2017, 11:02AM

Registered Member #3988 Joined: Thu Jul 07 2011, 03:25PM
Location:
Posts: 711

Patrick wrote:

alot of that 500,000 kg is mass burned just to get the rest of the mass to 120,000 feet. Right ?

Yes, but be aware, that at this altitude the rocket will have considerable speed. The energy due to this speed will be much larger than the energy needed to lift the rocket up to this altitude.

Some notes:
1) The reduction of gravitational pull on the rocket due to its altitude is almost negligible. The force obeys the inverse square law, where the distance in the law is measured from the center of the earth. For the earth surface, this is 6370km, for 40km altitude it is 6410km. Not much difference there.

2) As klugesmith calculated, the energy to propel the rocket to its orbital velocity is very much larger, than the energy needed to lift the rocket to its orbital height. In principle it would be an advantage to launch a rocket almost horizontally in order to gain orbital speed as fast as possible. Air drag prohibits this, though. At 40km height, that becomes feasible.

3) During the usual vertical takeoff a considerable amount of thrust is wasted just to keep the rocket aloft. In the extreme case, where the thrust is just barely above the weight of the rocket, the engines would burn away without the rocket going anywhere. Rockets aren't designed this way, but some of the effect is left.

An optimal launch trajectory is almost vertical initially and going horizontal as soon as the air drag allows. Due to the complexity the calculation can be done only numerically and I can't give a quantitative answer to the how much less fuel is required for a high altitude launch. AFAIK, note 3) accounts for the major advantage of a high altitude launch. I.e. high altitude -> almost horizontal launch -> less waste for staying aloft.

klugesmith

Fri Dec 01 2017, 06:09AM

Registered Member #2099 Joined: Wed Apr 29 2009, 12:22AM
Location: Los Altos, California
Posts: 1714

Uspring wrote ...
... calculation can be done only numerically and I can't give a quantitative answer to the how much less fuel is required for a high altitude launch.

I agree, and think it's not really that hard an exercise. Here's a report by someone who did a good job of homebrew simulation and writing about it.

I think there's even a link to his Matlab code. Being rusty with Matlab, and so far unable to find a ready-to-run simulation, I'd probably do it as an Excel spreadsheet. Can't spare the time to learn Python.

Maybe a dozen or 20 formula columns, and up to 1000 timesteps of 1 second each. Set the fuel mass-flow rate (which you can throttle) and figure the thrust using Isp as a coefficient. Simple exponential model of air density vs height; density and speed give you the dynamic pressure; that and frontal area and drag coefficient give you the drag force. Remaining mass times g gives you the gravity force. Raul figures motion in two axes (upward and downrange), and lets natural "gravity turning" define the arc. A term I'd never heard of before this week.

The simulation can be checked by using real numbers from well-documented rockets, and (we hope) matching the well-reported milestones of actual ascents to orbit. Better yet, matching second-by-second tabular data, like this example from STS-121:

At the top is a link to an even richer data set. Do we have a reader good enough with OCR tools, or Web searching, to turn those images into machine-readable numbers?

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