B Canceling Orbital Motion

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Would accelerating a rocket in a vector that "cancels" earth's orbital motion relative to the center of the milky way in some cases improve the energy efficiency of a rocket for a given amount of change in velocity relative to the earth?
 

jim mcnamara

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I think it is called the local standard of rest. See this article. Note that you are talking about high velocities, if I correctly understood.

 

Ibix

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Would accelerating a rocket in a vector that "cancels" earth's orbital motion relative to the center of the milky way in some cases improve the energy efficiency of a rocket for a given amount of change in velocity relative to the earth?
No, in short. Earthbound observers are free to regard themselves as at rest, so from their perspective there's nothing to choose between the rockets.

Note that the gravitational effects of the Sun and the galaxy will modify that slightly. A rocket moving towards galactic centre will be slightly accelerated while one moving away will be slightly decelerated. I don't think this is what you are talking about, however.
 
Note that the gravitational effects of the Sun and the galaxy will modify that slightly. A rocket moving towards galactic centre will be slightly accelerated while one moving away will be slightly decelerated. I don't think this is what you are talking about, however.
I am talking about using enough rocket propellant to cancel the rocket's orbital motion around the galaxy after lifting off from the earth. Next I want to look at the maximum possible change in velocity with respect to earth this rocket can reach, while plummeting towards the center of the galaxy.
 

Ibix

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Next I want to look at the maximum possible change in velocity with respect to earth this rocket can reach, while plummeting towards the center of the galaxy.
You mean you intend to burn until your orbital velocity is zero, then pick a direction and burn again until you are out of propellant? Then you want to know of your final velocity with tespect to the Earth depends on the direction you picked?

If so, take the simple example of carrying enough propellant to accelerate back to the Sun's orbital velocity from your zero orbital speed state. Trivially your speed compared to the Sun may be anywhere between zero (if you accelerate in the direction of galactic rotation) and twice the orbital speed (if you accelerate in the opposite direction).
 
You mean you intend to burn until your orbital velocity is zero, then pick a direction and burn again until you are out of propellant?
No I want to burn until my orbital velocity is zero, and then not burn anymore. Will my speed with respect to Earth continue to increase as I accelerate towards the barycenter of the galaxy? I want to know what speed I could reach with respect to the surface of the earth before passing the galactic barycenter.
 

DaveC426913

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I want to know what speed I could reach with respect to the surface of the earth before passing the galactic barycenter.
That's going to be much trickier than you expect.

The galaxy is an extended gravitational object; you can't treat it as a point.
  1. At any given time, almost as many stars are behind you as in front of you. At that number will increase the closer you get to the core.
  2. The effect of local stars within a few dozen light years of your path will likely have a greater impact on your trajectory than the pull of the core, which is tens of thousands of light years away.
 
That's going to be much trickier than you expect.

The galaxy is an extended gravitational object; you can't treat it as a point.
1] Almost as many stars are behind you as in front of you.
2] The effect of local stars with a few dozen light years of your path will have a greater impact on your trajectory than the pull of the core, which is tens of thousands of light years away.
To calculate the "anomalous" acceleration with respect to the earth by an object on a trajectory with no galactic orbital velocity in earths approximate galactic vicinity, could I look at the acceleration of an object at earth's orbital radius from the sun with no orbital motion, and then compare it to the solar systems orbital velocity around the galaxy?

Assuming I don't get "too close" to any of the nearer stars in my path, isn't the initial ~215km/s more than likely sufficient to ensure escape velocity with these?
 
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I think it is called the local standard of rest.
The "local standard of rest" as defined in the article you linked to is not "at rest relative to the galactic barycenter". It is "at rest relative to the average motion of stars in the vicinity of the solar system", which, since all of those stars are orbiting the galactic barycenter, will be something like "at rest relative to the average orbital velocity around the galactic barycenter in the vicinity of the solar system".
 
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If I understand the linked article correctly the necessary initial velocity from Earth is ~215,919.24m/s...
It's about 230 km/s according to Wikipedia:


The error in this calculation is probably fairly significant, since I don't think we have a very precise position for the galactic barycenter. So 215 km/s is probably within the current error bars.
 
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the "anomalous" acceleration with respect to the earth by an object on a trajectory with no galactic orbital velocity in earths approximate galactic vicinity
I don't know what you mean by "anomalous acceleration".
 
I don't know what you mean by "anomalous acceleration".
I thought the solar system doesn't fall towards the galactic barycenter thanks to its orbital motion. If I eliminate the orbital motion at the same radius, I expect to plummet toward the center of the galaxy. Isn't the effect of the initial acceleration from earth, to cancel the galactic orbital motion with engines (215km/s), plus the additional acceleration towards the center of the galaxy from gravity, more total velocity change with respect to the Earth's surface than the initial change in velocity provided by the engines?
 
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the initial acceleration from earth, to cancel the galactic orbital motion with engines (215km/s)
This isn't an acceleration as that term is usually used; it's a total change in relative velocity over some finite time. "Acceleration" means the instantaneous rate of change of velocity.

the additional acceleration towards the center of the galaxy from gravity
Have you tried to estimate the magnitude of this? And how much it would affect the trajectory over the time it would take for a total velocity change of the magnitude you're postulating?
 
Have you tried to estimate the magnitude of this? And how much it would affect the trajectory over the time it would take for a total velocity change of the magnitude you're postulating?
Based on the approximate location of Sagittarius A* I thought the craft might be able to reach close to the speed of light, but I had my doubts so I wasn't sure.

 
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Based on the approximate location of Sagittarius A* I thought the craft might be able to reach close to the speed of light
If the craft actually falls into the black hole, its speed relative to stationary "hovering" observers in its vicinity will indeed approach the speed of light as it approaches the hole's horizon.

In a curved spacetime, there is no well-defined concept of "relative velocity" for objects that are far distant from each other. So there isn't really a well-defined speed of an object approaching Sagittarius A* "relative to Earth".

If we were in a hypothetical galaxy of the size and overall mass of the Milky Way that did not have a black hole at its center, then the velocity of an object that fell freely from a distance similar to that of the solar system, to the barycenter, relative to a stationary observer at the barycenter, would be much smaller than the speed of light. The simplest way to get a rough order of magnitude estimate of that speed is that it will be the same as the orbital speed around the barycenter of an object in a free-fall orbit at the same distance from the barycenter--in other words, around 215 km/s. As above, there isn't really a well-defined "speed relative to Earth" for an object at the barycenter, but again, a rough estimate would just be the vector sum of 215 km/s in two perpendicular directions (radial along the Earth-barycenter line, and tangential to that line), which is 215 times ##\sqrt{2}##.
 
To calculate the initial acceleration of an object in free fall (no orbital motion) from the height of the solar system towards the milky way's barycenter, can I use a formula based on:

Circular Orbital Speed at Earth Distance From Solar System Barycenter = A
Instantaneous Acceleration of Object in Free Fall at Earth Distance From Solar System Barycenter = B
Circular Orbital Speed at Solar System Distance From Galactic Barycenter = C
Instantaneous Acceleration of Object in Free Fall at Solar System Distance From Galactic Barycenter = D

^Trying to solve for D
 
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To calculate the initial acceleration of an object in free fall (no orbital motion) from the height of the solar system towards the milky way's barycenter
You would just use the standard Newtonian gravity formula, plugging in the mass of the Milky Way and the distance from its barycenter. (Strictly speaking, it would just be the mass of the Milky Way that is closer to the center than the solar system, and there might be some correction because the Milky Way is not spherically symmetric, but these corrections are small and hard to estimate anyway. It depends on how accurate you want your answer to be.)
 

DaveC426913

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Assuming I don't get "too close" to any of the nearer stars in my path
Avoid them as you will, the nearer stars are going to be mere dozens of light years away. That's about 3 orders of magnitude closer than the core.

Since gravitational pull falls off as the square of the distance, the pull of one star at the core will be 6 orders of magnitude (1,000,000 times) less than a near star. In orders words, a million stars at the core will have only as much effect on your craft is a single star a dozen light years away.

As you approach the core, more and more of the galaxy's mass will be behind you. Your acceleration will slow.

You would just use the standard Newtonian gravity formula, plugging in the mass of the Milky Way and the distance from its barycenter. (Strictly speaking, it would just be the mass of the Milky Way that is closer to the center than the solar system,
I assert that it is a significant amount - enough to result in a very wrong answer.
 

Ibix

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I assert that it is a significant amount - enough to result in a very wrong answer.
If I'm interpreting tables 2 and 3 here correctly, mass inside our current radius is 7.3×1010 solar masses and the total mass is 30×1010 solar masses. That would make @PeterDonis' approximate number a factor of four high. Whether or not a factor of four matters depends why @metastable wants to know.
 
Whether or not a factor of four matters depends why @metastable wants to know.
Would accelerating a rocket in a vector that "cancels" earth's orbital motion relative to the center of the milky way in some cases improve the energy efficiency of a rocket for a given amount of change in velocity relative to the earth?
I wondered if it would be the most efficient method for a craft to reach arbitrarily high velocities in terms of propellant energy expended to reach a given change in velocity relative to a fixed point on the earth's surface.

milkyway.jpg
 
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If we were in a hypothetical galaxy of the size and overall mass of the Milky Way that did not have a black hole at its center, then the velocity of an object that fell freely from a distance similar to that of the solar system, to the barycenter, relative to a stationary observer at the barycenter, would be much smaller than the speed of light. The simplest way to get a rough order of magnitude estimate of that speed is that it will be the same as the orbital speed around the barycenter of an object in a free-fall orbit at the same distance from the barycenter--in other words, around 215 km/s. As above, there isn't really a well-defined "speed relative to Earth" for an object at the barycenter, but again, a rough estimate would just be the vector sum of 215 km/s in two perpendicular directions (radial along the Earth-barycenter line, and tangential to that line), which is 215 times √22\sqrt{2}.
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
 
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If I'm interpreting tables 2 and 3 here correctly, mass inside our current radius is 7.3×1010 solar masses and the total mass is 30×1010 solar masses. That would make @PeterDonis' approximate number a factor of four high.
If one used the "whole galaxy" number, yes. I had not realized there was that much mass outside the solar system's radius. But since the mass inside the solar system's radius is known, one could just use that in the Newtonian formula instead to get a more accurate value for initial acceleration from rest towards the galactic center.
 

russ_watters

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If I'm interpreting tables 2 and 3 here correctly, mass inside our current radius is 7.3×1010 solar masses and the total mass is 30×1010 solar masses. That would make @PeterDonis' approximate number a factor of four high. Whether or not a factor of four matters depends why @metastable wants to know.
No, I agree with @DaveC426913 here, I think it's worse than that. What matters is the fraction of the mass that's in the central black hole. Everything else is "above" you after you've past.

For example:
This has been modeled for the inside of the Earth...googling tells me g remains roughly constant until about 0.5r (the outer core), then drops linearly to 0.

For SGT-A, I'm seeing a mass of 4x10^6, or a difference of 5 orders of magnitude. I'd wager that for practical purposes that makes the galaxy a uniform density disk, with a g that drops linearly until you get very close to the black hole.

...but this will primarily depend on the distribution of dark matter, since the solar system is almost entirely (95%) dark matter.

...edit: apparently the dark matter is not evenly distributed and this is an active area of study...
 

Ibix

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No, I agree with @DaveC426913 here, I think it's worse than that. What matters is the fraction of the mass that's in the central black hole. Everything else is "above" you after you've past.
I think metastable has asked several different questions. He asked about the initial acceleration due to gravity in #17, and I think that's what Peter was answering. In which case his approach, modified by the number for the mass inside the Sun's orbit, is correct.

If you want to know the velocity, as he asks elsewhere, then sure I agree with you that it isn't so straightforward. But if we're getting near black holes, Peter's previous comments about "velocity relative to the Earth" not being a well-defined quantity apply.
 

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