the universe is expanding

[logistic_guy]

here is the question

Estimate the rate of which hydrogen atoms would have to be created, according to the steady-state model, to maintain the present density of the universe of about \(\displaystyle 10^{-27}\) kg/m\(\displaystyle ^3\), assuming the universe is expanding with Hubble constant \(\displaystyle H_0 = 70\) km/s/Mpc.


my attemb
the steady-state model is another way to think of the expansion of the universe
in contrast to the big bang theory, in the steady-state model we don't have beggning and end
this question is related to cosmological theory and can be solved by Hubble's law
\(\displaystyle v = H_0D\)
\(\displaystyle v: \) recessional velocity
\(\displaystyle H_0: \) Hubble's constant
\(\displaystyle D: \) the proper distance from the galaxy to the observer

if i treat \(\displaystyle D\) as the radius of the universe
then its volume is \(\displaystyle V = \frac{4}{3}\pi D^3\)

i can't think of a way to get the hydrogen atoms involved in this two formulas

 

[Agent Smith]

Gravity?

 

[logistic_guy]

Gravity?

thank

i think \(\displaystyle 9.8\) m/s\(\displaystyle ^2\)

 

[Agent Smith]

I meant H (through gravity, because it has mass) will/may(?) affect the Hubble constant, by generating an opposing attractive force to the expansion.

 

[logistic_guy]

I meant H (through gravity, because it has mass) will/may(?) affect the Hubble constant, by generating an opposing attractive force to the expansion.

i don't understand what you're saying

the only mass in the question is the mass of the hydrogen atoms
\(\displaystyle 1\) hydrogen atom have mass \(\displaystyle m_H = 1.67 \times 10^{-27}\) kg

 

[Agent Smith]

i don't understand what you're saying

the only mass in the question is the mass of the hydrogen atoms
\(\displaystyle 1\) hydrogen atom have mass \(\displaystyle m_H = 1.67 \times 10^{-27}\) kg

What is there to not understand? It's possible that I misunderstood your question (highly likely). If I'm correct you want to know how to work Hydrogen into your equation about the expansion of the universe. You mention the Hubble "constant" which looks like the rate of expansion of the universe. Gravity will surely(?) affect the Hubble constant by producing an opposing attractive force. Hydrogen will exert gravity (it has mass).

Hubble constant [imath]H = f(\text{mass of Hydrogen in the universe})[/imath]. That's how the situation appears to me.

 

[logistic_guy]

What is there to not understand? It's possible that I misunderstood your question (highly likely). If I'm correct you want to know how to work Hydrogen into your equation about the expansion of the universe. You mention the Hubble "constant" which looks like the rate of expansion of the universe. Gravity will surely(?) affect the Hubble constant by producing an opposing attractive force. Hydrogen will exert gravity (it has mass).

Hubble constant [imath]H = f(\text{mass of Hydrogen in the universe})[/imath]. That's how the situation appears to me.

Hubble constant give the speed of a galaxy \(\displaystyle 3.09 \times 10^{19}\) km away

\(\displaystyle 3.09 \times 10^{19}\) km \(\displaystyle =\) Mpc \(\displaystyle = 1\) megaparsec \(\displaystyle = 3.26\) million light years

one new idea is to use the density given in the question

\(\displaystyle \rho = \frac{m}{V} = \frac{m_H N}{V}\), where \(\displaystyle N\) is the number of hydrogen atoms

i can't see how this will help. i'm slowly but surely building some useful formulas

 

[Agent Smith]

Hubble constant give the speed of a galaxy \(\displaystyle 3.09 \times 10^{19}\) km away

\(\displaystyle 3.09 \times 10^{19}\) km \(\displaystyle =\) Mpc \(\displaystyle = 1\) megaparsec \(\displaystyle = 3.26\) million light years

one new idea is to use the density given in the question

\(\displaystyle \rho = \frac{m}{V} = \frac{m_H N}{V}\), where \(\displaystyle N\) is the number of hydrogen atoms

i can't see how this will help. i'm slowly but surely building some useful formulas

That's a good line of thought. I hope you can complete the quest and show us your formula (if you want to). At astronomical scales ... stuff gets weird.

Density of H = [imath]\rho[/imath].