Can Canceling Orbital Motion Improve Rocket Efficiency?

[DaveC426913]

If what you are really asking is what is the most efficient way to thrust to get from orbit around a the center of a gravity well, such as the galaxy center, to falling straight into it, then your answer is correct: thrust against your orbital velocity, not toward the object/gravity well.

Ah. Like the shuttle, deorbiting, does a burn directly opposed to orbital motion.

 

[PeterDonis]

Any rest frame >30,000km/s with respect to trapped electron on Earth with the least possible propellant

In other words, you don't actually care about your speed relative to the galactic barycenter, or relative to anything else except Earth?

In that case, pretty much everything said so far in this thread has been a waste of time.

 

[metastable]

I would say yes, in only that one direction (toward the galactic center). Now what? What can you do with that other than getting pulled apart by a black hole?

To potentially reach other destinations along that vector.

 

[russ_watters]

To potentially reach other destinations along that vector.

No, it doesn't help you do that except for an extremely short flyby or spectacular crash.

 

[metastable]

No, it doesn't help you do that except for an extremely short flyby or spectacular crash.

I envisioned a very long duration mission with a series of flybys, with potential for gravity assist maneuvers along the way.

 

[PeterDonis]

I envisioned a very long duration mission

A mission where?

 

[russ_watters]

I envisioned a very long duration mission with a series of flybys, with potential for gravity assist maneuvers along the way.

Like the Voyager probes - sure.

 

[metastable]

A mission where?

As far from Earth as possible in the least time with the least energy.

 

[russ_watters]

As far from Earth as possible in the least time with the least energy.

Please note that those are competing parameters that need to be specified in order for the answer to be meaningful. Most distance and least time are literally the inverse of each other.

I feel like you are being purposely vague because you think it's helpful. It's not.

 

[metastable]

Please note that those are competing parameters that need to be specified in order for the answer to be meaningful. Most distance and least time are literally the inverse of each other.

I feel like you are being purposely vague because you think it's helpful. It's not.

Sorry for the sloppy language. I'm not aware of any other methods that could in theory get a craft to the same arbitrary 30,000km/s velocity relative to the Earth's surface using less fuel, so I wondered if anyone here knew of such a method?

 

[DaveC426913]

If you plan to coast to the galactic centre, you're still going to have a heckuva time dodging all the stellar gravity wells you pass through.

Perhaps a more oblique approach would achieve the desired effect. Blast north, out of the galactic plane. Then all the mass of the galaxy will be pulling you in the same direction.

EDIT: Ah. If the goal is to facilitate fly-bys of other destinations, then leaving the galactic plane will be ... counter-productive.

 

[russ_watters]

Sorry for the sloppy language. I'm not aware of any other methods that could in theory get a craft to the same arbitrary 30,000km/s velocity relative to the Earth's surface using less fuel, so I wondered if anyone here knew of such a method?

I suppose not, but please note that the acceleration will be really slow after the rocket stops firing. If I did the calc right, it's 1/1000th of a g, so it would take about a thousand years to reach that speed.

 

[PeterDonis]

As far from Earth as possible in the least time with the least energy.

As @russ_watters pointed out, you can't have all three of these at once. You need to pick two. I assume "least energy" is one, so that leaves either "most distance" or "least time", but you have to pick one.

 

[PeterDonis]

I'm not aware of any other methods that could in theory get a craft to the same arbitrary 30,000km/s velocity relative to the Earth's surface

And this is yet a fourth criterion "highest speed", in addition to "most distance", "least time", and "least energy". And you can still only have two. Which two?

 

[metastable]

And this is yet a fourth criterion "highest speed", in addition to "most distance", "least time", and "least energy". And you can still only have two. Which two?

I'm not aware of any other methods (besides a ~215km/s engine boost from Earth surface to cancel the vehicle's galactic orbital motion) that can get a 10kg vehicle mass (containing a trapped electron) launched to >30,000km/s with respect to a trapped electron orbiting earth, using less propellant energy.

 

[PeterDonis]

I'm not aware of any other methods (besides a ~215km/s engine boost from Earth surface to cancel the vehicle's galactic orbital motion) that can get a 10kg vehicle mass (containing a trapped electron) launched to >30,000km/s with respect to a trapped electron orbiting earth, using less propellant energy

This doesn't answer the question I and others are asking. You have given four criteria: "most distance", "least time", "least fuel expended", and "highest speed relative to Earth". You can only have two of them. Which two?

Either answer that question or this thread will be closed for being too vague.

 

[metastable]

"highest speed relative to earth"
"least fuel expended"

 

[russ_watters]

I'm not aware of any other methods (besides a ~215km/s engine boost from Earth surface to cancel the vehicle's galactic orbital motion) that can get a 10kg vehicle mass (containing a trapped electron) launched to >30,000km/s with respect to a trapped electron orbiting earth, using less propellant energy.

So, we've fixed one constraint (target speed) and optimized for another (energy). Just be aware that this approach takes more time and achieves less distance/time (average speed for the same distance or time) than just using the engines the whole way.

[edit]
Please note: this scenario is just a limiting case of the Hohmann transfer orbit. It uses the least energy, but is not the fastest way - in speed or time - to get somewhere.

 

[metastable]

I'd plan to burn all remaining fuel on board (if any) right before arrival near the barycenter to take advantage of the oberth effect.

 

[PeterDonis]

"highest speed relative to earth"
"least fuel expended"

Ok, then, as I said before, pretty much everything that's been said in this thread has been a waste of time. You could have just asked: "How can I give a rocket the highest speed relative to Earth for the least fuel expended?" Or, equivalently, "Given a fixed allowance of rocket fuel, how would I maximize the rocket's speed relative to Earth?" Or, exchanging which one is the constraint, "Given a fixed desired speed relative to Earth, how can I achieve that speed with the least rocket fuel expended?", which seems to be how you are viewing it.

 

[russ_watters]

I'd plan to burn all remaining fuel on board (if any) right before arrival near the barycenter to take advantage of the oberth effect.

Ehhhhh k.

 

[PeterDonis]

I'd plan to burn all remaining fuel on board (if any) right before arrival near the barycenter to take advantage of the oberth effect.

If there is a black hole at the center of the galaxy and you want to achieve 30,000 km/s relative to the galactic barycenter (note that "relative to Earth" has no meaning for this case for reasons I gave earlier), you will need to aim very, very carefully to achieve that desired target speed as you pass the hole without falling into it. (And that is leaving out the other issue that @DaveC426913 brought up earlier.)

If there is no black hole at the center of the galaxy, I don't think it's possible to achieve 30,000 km/s at all. The galaxy's gravity well without a black hole is not deep enough by a couple of orders of magnitude.

 

[metastable]

Sorry for the sloppy language. I'm not aware of any other methods that could in theory get a craft to the same arbitrary 30,000km/s velocity relative to the Earth's surface using less fuel, so I wondered if anyone here knew of such a method?

I suppose not, but please note that the acceleration will be really slow after the rocket stops firing. If I did the calc right, it's 1/1000th of a g, so it would take about a thousand years to reach that speed.

If there is no black hole at the center of the galaxy, I don't think it's possible to achieve 30,000 km/s at all. The galaxy's gravity well without a black hole is not deep enough by a couple of orders of magnitude.

I'm confused because If we look at all three of these statements I see a conflict... if the acceleration is 1/1000g and it takes 1000 years to reach 30,000km/s, would I pass the barycenter after less than 1000 years?

 

[PeterDonis]

if the acceleration is 1/1000g and it takes 1000 years to reach 30,000km/s

You're confusing two different kinds of acceleration.

If you have a rocket that is providing sufficient thrust to accelerate at 1/1000 g, it can keep providing that thrust as long as you have fuel. So if you have enough fuel, it will eventually accelerate you to 30,000 km/s relative to your starting point. (At that speed relativistic effects are still pretty small so a Newtonian calculation is fine; at higher speeds you would need to use the relativistic rocket equations.)

But you're talking about a case where the "acceleration" is really free fall along a geodesic in the gravitational field of the galaxy. This "acceleration" is not constant (it gets smaller as you get closer to the galactic center, is zero at the galactic center, and will start to decelerate you once you fly past the galactic center and are climbing out the other side), and the speed it can get you to (if there isn't a black hole at the center) is limited by the depth of the galaxy's gravity well.

 

[russ_watters]

You're confusing two different kinds of acceleration.

If you have a rocket that is providing sufficient thrust to accelerate at 1/1000 g, it can keep providing that thrust as long as you have fuel. So if you have enough fuel, it will eventually accelerate you to 30,000 km/s relative to your starting point. (At that speed relativistic effects are still pretty small so a Newtonian calculation is fine; at higher speeds you would need to use the relativistic rocket equations.)

But you're talking about a case where the "acceleration" is really free fall along a geodesic in the gravitational field of the galaxy. This "acceleration" is not constant (it gets smaller as you get closer to the galactic center, is zero at the galactic center, and will start to decelerate you once you fly past the galactic center and are climbing out the other side), and the speed it can get you to (if there isn't a black hole at the center) is limited by the depth of the galaxy's gravity well.

So...could you check my math on that then please. I calculated an acceleration of 0.011 m/s^2 based on our centripetal acceleration around the galaxy center. At that acceleration we'd achieve the target speed in 1000 years and barely move on a galactic scale.

We're 26,000ly from the galactic center.

 

[PeterDonis]

I calculated an acceleration of 0.011 m/s^2 based on our centripetal acceleration around the galaxy center. At that acceleration we'd achieve the target speed in 1000 years

No, you wouldn't, because the acceleration decreases as you get closer to the galactic center. We're not talking about a rocket providing 0.011 m/s^2 of thrust. We're talking about free fall in a gravity well where the amount of mass beneath you decreases as you fall. Not the same thing.

 

[russ_watters]

No, you wouldn't, because the acceleration decreases as you get closer to the galactic center. We're not talking about a rocket providing 0.011 m/s^2 of thrust. We're talking about free fall in a gravity well where the amount of mass beneath you decreases as you fall. Not the same thing.

It can be assumed constant if we aren't moving much relative to the radius of the galaxy -- like we assume a constant 9.81 m/s² in the vicinity of Earth's surface.

...but let's set that aside for a minute though because I do think I made an error. Here's the calc:

a=V²/r = 230,000m/s / 2469x10¹⁷km = 2.1x10^-10 m/s²

Yeah, I hit the ^ instead of the EE button on my calculator...

Anyway, you're right that that acceleration is so small and takes so long that the constant acceleration assumption fails. I agree this wouldn't get one to the target speed until they got to the vicinity of the central black hole.

 

[PeterDonis]

It can be assumed constant if we aren't moving much relative to the radius of the galaxy

But we are; we're going all the way to the galactic center.

It looks like there was indeed a mistake in his calculation of the acceleration, though.

 

[Janus]

I'm confused because If we look at all three of these statements I see a conflict... if the acceleration is 1/1000g and it takes 1000 years to reach 30,000km/s, would I pass the barycenter after less than 1000 years?

As already pointed out, the acceleration will not be constant. I'll give you an hypothetical example using the Earth's distance from the Center of the galaxy and your value of 215 km/sec for its orbital speed.
For this example, we will assume that the Earth is orbiting a spherical mass of stars with the mass of these stars uniformly distributed throughout, and the radius of said sphere is equal to the radius of the Sun's galactic orbit. The total mass is just enough to produce the required orbital speed
If you killed the Earth's galactic orbital velocity, how fast would it be moving when it reaches the center? . Oddly enough, the answer works out to be 215 km/sec, or its original orbital speed. How long will it take to reach the center? Roughly 57 million years ( or 1/4 of the time it would have taken to complete a full orbit.)

The fact that the stars interior to the Sun's galactic orbit are not distributed in a sphere or uniformly will alter this result, but not by any great magnitude.

 

[metastable]

I'm not sure how to do the actual problem, but I am building up to it by attempting to solve a similar problem:

I take:

1/2 space station orbital period = 2700 seconds
1/2 circumference of Earth meters = 20037500 meters
A = ~7421.29 meters per second = 20037500 meters / 2700 seconds = rough circular orbit velocity
radius of Earth = 6371000 meters
duration of fall from across Earth diameter = 2291 seconds
duration of fall to center of Earth = 1145 seconds
avg velocity from non-rotating surface crossing center = 11127.56 meters/s = B
C = max velocity from non-rotating surface crossing center = assumption 2*B = 22255.12 meters/s

C = A * ~2.99... = A * ~3
A/C= 0.333... = ~1/3

If you killed the Earth's galactic orbital velocity, how fast would it be moving when it reaches the center? . Oddly enough, the answer works out to be 215 km/sec, or its original orbital speed.

^re: C = A * ~2.99... = A * ~3 -- So where did I do my math wrong?

 

[russ_watters]

But we are; we're going all the way to the galactic center.

It looks like there was indeed a mistake in his calculation of the acceleration, though.

It was my calculation...

 

[PAllen]

Haven't checked various approximate calculations here, but if anyone is approximating on the basis of 'stars further away not counting' because of approximate even distribution, that won't work. You need at least an approximate spherical shell for that. An approximate ring or disc beyond some radius are a completely different, more complex case. They cannot be approximately ignored.

 

[PeterDonis]

where did I do my math wrong?

I'm not sure because you just quoted a lot of numbers without saying where they came from or how they were calculated.

 

[metastable]

I'm not sure because you just quoted a lot of numbers without saying where they came from or how they were calculated.

I attempted a comparison of the time for the space station to complete a half orbit with the time it supposedly takes an object to fall through "a hole through the center of the Earth to the other side."

 

[PeterDonis]

I attempted a comparison of the time for the space station to complete a half orbit with the time it supposedly takes an object to fall through "a hole through the center of the Earth to the other side."

Yes, and to do that you just quoted a lot of numbers without saying where they came from or how they were calculated. So I can't tell what, if anything, you did wrong.