Proof that our solar system will leave the milky way.

In summary, the conversation discusses the escape velocity needed for a solar system to stay in orbit around the Milky Way, Kepler's third law, and the unphysical problem of trying to calculate the Hubble rate in "per second" units.
  • #1
zeromodz
246
0
Let us assume Ho = 71 1 / s


In order for our solar system to stay in orbit around the milky way, the escape velocity must be greater than the recessional.

Resc > Rrec
(2GM / R)^1/2 > HoR

Now let's refer to Kepler's third law to substitute in for the mass.

T^2 = 4π^2R^3 / GM
M = 4π^2R^3 / GT^2((2G / R) * (4π^2R^3 / GT^2))^1/2 > HoR
(8π^2R^2 / T^2)^1/2 > HoR
πR√(8) / T > HoR
π√(8) / T > Ho <------------------------- FINAL EQUATION!

What do you think? the units are dimensionally correct. If the inequality is satisfied, the object will stay in orbit forever by the means of the expanding universe. We can actually rearrange this even more to give..

π√(8) / Ho > T
0.125 s > T <-------------- This must be satisfied! Given Ho = 71 1 / s

Obviously, our solar system doesn't take 0.125 s to orbit the galactic center. Therefore, we should be moving away from it.
 
Last edited:
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  • #2
You lost a factor of root(R) in there
 
  • #3
zeromodz said:
Let us assume Ho = 71 1 / s

That's not correct. It's many orders of magnitude (Mpc/km) off.
 
  • #4
There is just something wrong with posing the problem already. We do not move away from the galactic center because pf the Hubble flow (expansion of the universe). That is only correct on large scales (to be more precise: on scales over which the mean density is constant). So even if the numbers were correct, it is an unphysical problem.
 
  • #5
zeromodz,

Vanadium and Harcel are right.

However in part you did something clever, namely you tried to write the Hubble rate in "per second" units.

You wanted to write it down in terms of "1/s" or s-1

But you got the value wrong. It is not 71 (seconds)-1.

71 is much too big, like Vanadium says "by many orders of mag."

I think the Hubble parameter, as an inverse time, is about

1/(14 billion years)

That would be one over a huge number of seconds.

We could calculate what it is using the google calculator, just to see.
I have to go out. back later.

Back now. I just typed this into google
"71 km/s per megaparsec" (without the quotes)
and it told me the "per second" value of the Hubble rate. It said

2.3 x 10-18 per second

or what is the same, 2.3 x 10-18 Herz

Google has this built-in calculator which is good at converting unfamiliar units, so you don't have to look up the conversion factor. You don't have to do anything but type in the expression you want to have it evaluate.
If you need any help typing stuff in the right format so it will be recognized. Ask.
 
Last edited:
  • #6
I think its safe to say we don't stay in one place in space AT ALL! (We are constantly drifting)
 
  • #7
Dude111 said:
I think its safe to say we don't stay in one place in space AT ALL! (We are constantly drifting)

Yes, but what does this have to do with us leaving the Milky Way?
 
  • #8
marcus said:
zeromodz,

Vanadium and Harcel are right.

However in part you did something clever, namely you tried to write the Hubble rate in "per second" units.

You wanted to write it down in terms of "1/s" or s-1

But you got the value wrong. It is not 71 (seconds)-1.

71 is much too big, like Vanadium says "by many orders of mag."

I think the Hubble parameter, as an inverse time, is about

1/(14 billion years)

That would be one over a huge number of seconds.

We could calculate what it is using the google calculator, just to see.
I have to go out. back later.

Back now. I just typed this into google
"71 km/s per megaparsec" (without the quotes)
and it told me the "per second" value of the Hubble rate. It said

2.3 x 10-18 per second

or what is the same, 2.3 x 10-18 Herz

Google has this built-in calculator which is good at converting unfamiliar units, so you don't have to look up the conversion factor. You don't have to do anything but type in the expression you want to have it evaluate.
If you need any help typing stuff in the right format so it will be recognized. Ask.

71, 2.3 x 10-18 ... very close! ;)

Thanks for the info about the google calculator, I had no idea it was that impressive.
 

What evidence suggests that our solar system will leave the Milky Way?

Scientists have observed that the Milky Way is on a collision course with the Andromeda galaxy, and as a result, our solar system will eventually be ejected from the Milky Way.

When will our solar system leave the Milky Way?

According to current estimates, it will take approximately 4 billion years for the Milky Way and Andromeda to collide, but it could take up to 6 billion years for our solar system to fully leave the Milky Way's gravitational pull.

Will our solar system be destroyed when it leaves the Milky Way?

No, our solar system will not be destroyed when it leaves the Milky Way. It will continue to exist and orbit around the center of the galaxy it is ejected into.

What will happen to the planets in our solar system when we leave the Milky Way?

The planets in our solar system will continue to orbit around the sun, but they will also be pulled towards the center of the new galaxy we are ejected into. This could potentially change the orbits and climates of the planets.

How will the ejection from the Milky Way affect life on Earth?

It is difficult to predict exactly how the ejection from the Milky Way will affect life on Earth, but it is likely that it will cause significant changes in the environment and potentially lead to the extinction of many species. However, it is also possible that life will adapt and continue to thrive in the new galaxy.

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