# Proof that our solar system will leave the milky way.

## Main Question or Discussion Point

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 lets 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.

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## Answers and Replies

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You lost a factor of root(R) in there

Staff Emeritus
2019 Award
Let us assume Ho = 71 1 / s
That's not correct. It's many orders of magnitude (Mpc/km) off.

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.

marcus
Gold Member
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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.

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I think its safe to say we dont stay in one place in space AT ALL!! (We are constantly drifting)

I think its safe to say we dont 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?

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.