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## Main Question or Discussion Point

Hubble's law states that the speed of a distant object is proportional to it's distance (this implies an expanding universe). By finding the distance and speed (via redshift) of various distant galaxies you can produce a graph displaying the linear relationship. The gradient of the line is Hubble's constant.

In school I've been told that Hubble's constant is the inverse of the age of the universe, this doesn't make sense to me, because in order to arrive at that solution you must first assume that the velocity of a moving body is constant, and that defies the entire point of Hubble's law. What sounds better to me is something along these lines:

If Velocity = Hubble's constant × Displacement, and a galaxy 470 Mpc (1 Mpc = 3 × 10

Now this doesn't really seem that complicated at first glance, just find the time it takes for a body who's velocity is proportional to its displacement to reach its current position from originally being at the same spot as us (for that means that everything was at the same spot, ie. The Big Bang)

So, using calculus which is at the greatest extent of my yr 12 knowledge I came up with this little differential equation:

[tex]V = Hs \\ \frac{ds}{dt} = Hs \\ \int \frac{1}{s} \ ds = \int H \ dt \\ \ln|s| + c = Ht \\ t = \frac{\ln|s| + c}{H}[/tex]

The problem is: 1. this doesn't actually answer the question, 2. I have no idea what I've actually

In school I've been told that Hubble's constant is the inverse of the age of the universe, this doesn't make sense to me, because in order to arrive at that solution you must first assume that the velocity of a moving body is constant, and that defies the entire point of Hubble's law. What sounds better to me is something along these lines:

If Velocity = Hubble's constant × Displacement, and a galaxy 470 Mpc (1 Mpc = 3 × 10

^{19}km) away is travelling at 30000 km/s, what's the age of the universe.Now this doesn't really seem that complicated at first glance, just find the time it takes for a body who's velocity is proportional to its displacement to reach its current position from originally being at the same spot as us (for that means that everything was at the same spot, ie. The Big Bang)

So, using calculus which is at the greatest extent of my yr 12 knowledge I came up with this little differential equation:

[tex]V = Hs \\ \frac{ds}{dt} = Hs \\ \int \frac{1}{s} \ ds = \int H \ dt \\ \ln|s| + c = Ht \\ t = \frac{\ln|s| + c}{H}[/tex]

The problem is: 1. this doesn't actually answer the question, 2. I have no idea what I've actually

*done*; integrating has something to do with area under the curve, but which curve? I started with a straight line graph and ended up with a differential equation. By substituting in H and s you cannot get a value because of that constant sitting in there. So I was wondering if anyone here knows how to get an answer to this problem. I dunno, can you integrate definitely? Can you use common sense or context to eliminate the constant? Is there a better way of going about it? Am I completely and utterly wrong?