Cosmology Question: Expansion of the Universe

1. Feb 25, 2008

Moneer81

1. The problem statement, all variables and given/known data

So the other day in my cosmology class we talked about whether smaller objects like a solar system, a star or a galaxy expand with the rest of the universe. While the answer is obviously no, since local gravitational forces can overcome that expansion, our professor thought it would be kinda amusing the calculate the effect of that expansion on an object like a human being if we were to expand along with the universe.

So we're asked to find out how long it would it take us to expand by 1mm, given the current value of the Hubble parameter of Ho = 72km/s/Mpc, and assuming for the sake of this problem that the Hubble parameter is constant even if the time we find is very large.

We're also have to find how tall we have to be for that expansion to take one year.

I know it looks like a silly problem but I am just having a hard time finding a place to start thinking about it.

Thanks for taking the time.

2. Feb 25, 2008

Moneer81

Not takers?

3. Feb 25, 2008

CaptainQuasar

I'm not a cosmologist but I'll take a complete shot in the dark at it.

It seems like the Hubble parameter value means that something which is one megaparsec in size will grow to become one megaparsec and 72km in size after one second. Is that right?

If that's true it seems like you'd have to rig up an algebraic equality with ratios, the

$$\frac{a}{b} = \frac{c}{d}$$

kinda thing, or an expression with calculus to account for compound growth

Hope I'm not misunderstanding and ridiculously oversimplifying the problem.

4. Feb 25, 2008

kdv

Hubble's law is $$v= \frac{\Delta x}{\Delta t} = H_0 d$$

So you simply need to isolate delta x. This gives by how much the distance between two objects which were initially a distance "d" apart will have increased after a time delta t. (This assumes of course that delta t is not so large that H_0 will have changed value appreciably during that time interval)

5. Feb 25, 2008

kdv

The logic is correct. And it gives the same equation that I provided in my post.