B Canceling Orbital Motion

  • Thread starter metastable
  • Start date
24,759
6,190
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.
 
"highest speed relative to earth"
"least fuel expended"
 

russ_watters

Mentor
18,502
4,714
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.
 
Last edited:
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.
 
24,759
6,190
"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

Mentor
18,502
4,714
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.
 
24,759
6,190
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.
 
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?
 
24,759
6,190
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

Mentor
18,502
4,714
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.
 
24,759
6,190
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

Mentor
18,502
4,714
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.
 
24,759
6,190
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

Staff Emeritus
Science Advisor
Insights Author
Gold Member
3,360
1,004
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.
 
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

Mentor
18,502
4,714
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

Science Advisor
7,573
962
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.
 
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."
 
24,759
6,190
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.
 

Janus

Staff Emeritus
Science Advisor
Insights Author
Gold Member
3,360
1,004
^re: C = A * ~2.99... = A * ~3 -- So where did I do my math wrong?
To do the problem correctly, you have to use conservation of energy. The equation for finding the gravitational potential per unit mass anywhere inside a sphere of uniform mass is
$$ E = \frac{2}{3}\pi G p (r^2-3R^2) $$
where p is the density, r is the distance from the center, R is the radius of the mass and G the universal gravitational constant.
Given that for a spherical body,

$$ M = \frac{4 \pi R^3}{3} $$

we can make this

$$ E = \frac{GM}{2} (r^2-3R^2)$$

Given that we are looking for the difference between the surface where r=R and the Center where r=0, we are looking for the difference between

$$E= -\frac{GM}{R}$$

and

$$E= \frac{3}{2}\frac{GM}{R}$$

For the Earth, GM = 2.987e14 m3/ s2

and if we calculate out the difference we get 31255879.6 joules/kg

Since this is the energy that is converted to kinetic energy for the falling body, and KE = mv^2/2,

Then

$$ v = \sqrt { 2 (31255879.6)} $$
( we can ignore the "m" as we are looking for "energy per unit mass".

which equals 7906.4 m/s.

orbital velocity for an object just at the surface of the Earth is

$$ v= \sqrt {\frac{GM}{r}} $$

which equals 7906.4 m/s

This is no accident as

$$ -\frac{GM}{R} - \left (- \frac{3}{2}\frac{GM}{R} \right ) = \frac{GM}{2R} $$

Thus

$$ v = \sqrt{ 2 \frac{GM}{2R}} = \sqrt {\frac{GM}{r}}$$
 
which equals 7906.4 m/s
Isn’t this the average speed from side to side through the earth? If it’s an average there should be a min velocity and a max velocity. The min is starting from 0m/s at the surface so if the avg speed from side to side is 7906.4m/s, then what is the maximum that gives that average?
 
24,759
6,190
Isn’t this the average speed from side to side through the earth?
No, it's the maximum speed--the speed the object has at the instant it is passing the center of the Earth, when all of the potential energy it had at the surface (where it was at rest) has been converted to kinetic energy.
 
No, it's the maximum speed--the speed the object has at the instant it is passing the center of the Earth, when all of the potential energy it had at the surface (where it was at rest) has been converted to kinetic energy.
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
which equals 7906.4 m/s.
space station orbital period: 92.68min

" The traveler would pop up on the opposite side of the Earth after a little more than 42 minutes."

42min = 2520 seconds
earth radius=3,963 miles

earth diameter = 2 * 3,963 miles = 12755660.54 meters

I think I figured out where I went wrong, the roughly correct avg speed is:

12755660.54 meters earth diameter / 2520 seconds across diameter = 5061m/s avg
^the right answer

The mistaken formula I had previously used was:

(2*radius of earth = 6371000 meters) / 1145 seconds across radius = 11128.38m/s avg
^the wrong answer
 
I count at least 3 differences between the "galactic barycenter plunge" vs "fall-through-the-earth" scenarios:
-the galaxy is more of a disc than a sphere
-the matter within the solar systems radius with respect to the galactic barycenter is not a uniform density
-there is a disproportionately large over-density very close to the center of the galaxy in the form of a supermassive black hole
 

Physics Forums Values

We Value Quality
• Topics based on mainstream science
• Proper English grammar and spelling
We Value Civility
• Positive and compassionate attitudes
• Patience while debating
We Value Productivity
• Disciplined to remain on-topic
• Recognition of own weaknesses
• Solo and co-op problem solving

Hot Threads

Top