Can Canceling Orbital Motion Improve Rocket Efficiency?

In summary: There's no guarantee that the entire journey will be without encounters with stars, and even if it is, the journey will still take a long time.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...Assuming you can avoid any nearby stars, the initial speed is more than likely sufficient.
  • #1
metastable
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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?
 
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  • #3
metastable said:
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.
 
  • #4
Ibix said:
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.
 
  • #5
metastable said:
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).
 
  • #6
Ibix said:
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.
 
  • #7
jim mcnamara said:
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.

If I understand the linked article correctly the necessary initial velocity from Earth is ~215,919.24m/s...

244566

https://astrosociety.org/edu/publications/tnl/71/howfast.html
 
  • #8
metastable said:
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.
 
  • #9
DaveC426913 said:
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 Earth's 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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  • #10
jim mcnamara said:
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".
 
  • #11
metastable said:
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:

https://en.wikipedia.org/wiki/Galactic_year
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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  • #12
metastable said:
the "anomalous" acceleration with respect to the Earth by an object on a trajectory with no galactic orbital velocity in Earth's approximate galactic vicinity

I don't know what you mean by "anomalous acceleration".
 
  • #13
PeterDonis said:
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?
 
  • #14
metastable said:
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.

metastable said:
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?
 
  • #15
PeterDonis said:
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.

 
  • #16
metastable said:
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}##.
 
  • #17
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
 
  • #18
metastable said:
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.)
 
  • #19
metastable said:
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.

PeterDonis said:
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.
 
  • #20
DaveC426913 said:
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.
 
  • #21
Ibix said:
Whether or not a factor of four matters depends why @metastable wants to know.

metastable said:
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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  • #22
PeterDonis said:
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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  • #23
Ibix said:
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.
 
  • #24
Ibix said:
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...
 
  • #25
russ_watters said:
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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  • #26
Ibix said:
I think metastable has asked several different questions.
My original question is about efficiency... does such a trajectory improve the energy efficiency of the rocket defined as change in velocity from a fixed point (trapped electron?) on earth?
 
  • #27
metastable said:
does such a trajectory improve the energy efficiency of the rocket defined as change in velocity from a fixed point (trapped electron?) on earth?

"Improve the energy efficiency of the rocket" is a strange way to look at it when you're just using the rocket for a single burn (to negate the 215 km/s velocity of the solar system relative to the galactic barycenter) and then the rocket is done and everything else is just free fall.
 
  • #28
PeterDonis said:
"Improve the energy efficiency of the rocket" is a strange way to look at it when you're just using the rocket for a single burn (to negate the 215 km/s velocity of the solar system relative to the galactic barycenter) and then the rocket is done and everything else is just free fall.

Perhaps, but my meaning is efficiency compared to other potential trajectories. For example, suppose I have a 30,000km/s target relative to trapped electron on earth, the implications of the original question are: will any other trajectories get me there faster with less impulse energy?
 
  • #29
metastable said:
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?
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.
 
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  • #30
metastable said:
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.
This is the part that's confusing: it doesn't just matter where you are going from, it also matters where you are going to. Unless you are actually talking about travel to the galaxy center, this technique doesn't do anything for you. It doesn't help you get to Mars, for example. Your description of what you want is unclear enough that could be hidden in it...
 
  • #31
russ_watters said:
This is the part that's confusing: it doesn't just matter where you are going from, it also matters where you are going to.
The destination I had in mind: anywhere that eventually gets me to (for example) 30,000km/s relative to a trapped electron on earth, with the least amount of propellant.
 
  • #32
metastable said:
Perhaps, but my meaning is efficiency compared to other potential trajectories. For example, suppose I have a 30,000km/s target relative to trapped electron on earth, the implications of the original question are: will any other trajectories get me there faster with less impulse energy?
Get you where? Versus what alternative trajectory?
 
  • #33
metastable said:
The destination I had in mind: anywhere that eventually gets me to (for example) 30,000km/s relative to a trapped electron on earth, with the least amount of propellant.
That's a speed, not a destination.
 
  • #34
russ_watters said:
Get you where? Versus what alternative trajectory?
Any rest frame >30,000km/s with respect to trapped electron on Earth with the least possible propellant.
 
  • #35
metastable said:
Any rest frame >30,000km/s with respect to trapped electron on Earth with the least possible propellant.
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?
 
<h2>1. How does canceling orbital motion improve rocket efficiency?</h2><p>By canceling orbital motion, the rocket is able to achieve a more direct and efficient trajectory towards its destination. This reduces the amount of fuel needed for course corrections and minimizes the effects of atmospheric drag, resulting in improved efficiency.</p><h2>2. Can canceling orbital motion save fuel?</h2><p>Yes, canceling orbital motion can save fuel as it allows the rocket to follow a more direct path towards its destination, reducing the need for frequent course corrections and minimizing the effects of atmospheric drag.</p><h2>3. What are the potential drawbacks of canceling orbital motion?</h2><p>Canceling orbital motion requires precise calculations and adjustments to the rocket's trajectory, which can be challenging and may increase the risk of errors. Additionally, canceling orbital motion may not always be possible or practical depending on the mission objectives.</p><h2>4. How does canceling orbital motion affect the speed of the rocket?</h2><p>Canceling orbital motion can increase the speed of the rocket as it allows for a more direct and efficient trajectory towards its destination. However, the overall speed of the rocket will also depend on other factors such as the amount of thrust and the weight of the payload.</p><h2>5. Are there any real-life examples of canceling orbital motion to improve rocket efficiency?</h2><p>Yes, there are several real-life examples of canceling orbital motion to improve rocket efficiency. One notable example is NASA's Apollo 13 mission, where the astronauts had to cancel their initial orbital trajectory and make a direct return to Earth due to an oxygen tank explosion. This required precise calculations and adjustments to the trajectory to ensure a safe and efficient return.</p>

1. How does canceling orbital motion improve rocket efficiency?

By canceling orbital motion, the rocket is able to achieve a more direct and efficient trajectory towards its destination. This reduces the amount of fuel needed for course corrections and minimizes the effects of atmospheric drag, resulting in improved efficiency.

2. Can canceling orbital motion save fuel?

Yes, canceling orbital motion can save fuel as it allows the rocket to follow a more direct path towards its destination, reducing the need for frequent course corrections and minimizing the effects of atmospheric drag.

3. What are the potential drawbacks of canceling orbital motion?

Canceling orbital motion requires precise calculations and adjustments to the rocket's trajectory, which can be challenging and may increase the risk of errors. Additionally, canceling orbital motion may not always be possible or practical depending on the mission objectives.

4. How does canceling orbital motion affect the speed of the rocket?

Canceling orbital motion can increase the speed of the rocket as it allows for a more direct and efficient trajectory towards its destination. However, the overall speed of the rocket will also depend on other factors such as the amount of thrust and the weight of the payload.

5. Are there any real-life examples of canceling orbital motion to improve rocket efficiency?

Yes, there are several real-life examples of canceling orbital motion to improve rocket efficiency. One notable example is NASA's Apollo 13 mission, where the astronauts had to cancel their initial orbital trajectory and make a direct return to Earth due to an oxygen tank explosion. This required precise calculations and adjustments to the trajectory to ensure a safe and efficient return.

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