# Homework Help: Recessional velocity! Help

1. Dec 20, 2009

### jackyjack

It has been found that, on average; galaxies are moving away from Earth at a speed that is propotional to their distance from Earth and that speed is called: Recessional Speed. Recessional speed v of a galaxy a distancce r from Earth is given by v = H.r;
where H = 1.58e-18 (1/s) is called Hubble constant. There are 2 galaxies that have distances from Earth are: 5.00e22 (m) nd 2.00e25 (m). If the galaxies at each of these distances had traveled at their recessional speeds, how long ago would they have been at our location?

2. Dec 20, 2009

### Stonebridge

You need to show your attempt at this, and explain your difficulty.

3. Dec 20, 2009

### jackyjack

oh sorry; I'm a newbie. Firstly, i think that if the velocity is proportional with distance so there's must be a acceleration around here. Then, I calculate that acceleration by the equation: vt^2 - v0^2 = 2as. Next, by using the equation:
delta(x) = v0x.t + (a.t^2)/2 with v0x=0 and the acceleration above to derive the time t. But I'm not sure this is the exact outcome of the problem because it's kind of irrelevant.

4. Dec 20, 2009

### Stonebridge

The question is saying that if these galaxies have travelled at this constant speed, how long would they have taken to get to where they are.
You know how far away they are - given in the question, and you can work out their speeds from the Hubble constant. It gives speed in terms of this distance from Earth.

By the way: welcome to Physics Forums.

5. Dec 20, 2009

### ideasrule

Once you get an answer, make sure it's roughly equal to the age of the universe (14 billion years). I'll leave it to you to figure out why.

6. Dec 21, 2009

### jackyjack

Thank you for the greeting.

So your idea is that those galaxies will make it to our location without acceleration? But the problem states that their speed is proportional to the distance; therefore, the speed will change according to time t and that is where acceleration come from.

7. Dec 21, 2009

### jackyjack

I have also had your idea in my mind when this problem came up, but my answer is extremely large (it's about 25 billion years!).

8. Dec 21, 2009

### Stonebridge

You have to assume that any individual galaxy has been travelling at this constant speed in order to work out the answer to the problem.
This Hubble calculation is not about acceleration, strange as it may seem.
It may also be worth looking up the latest value for the Hubble constant to compare it to the one given here. It is usually given in a different unit (not 1/s).

9. Sep 17, 2011

### ohms law

Hello folks,
I have almost exactly this problem for homework.

What I did was figure out that:

a) r=5.00x10^22m gives: 7.9x10^4 m/s
b) r=2.00X10^25m gives: 3.16x10^7 m/s

so, since d=rt (d: distance, r: rate, t: time) d/r = t
5.00x10^22m / 7.9x10^4 m/s = 2.00x10^10 s (20 billion years)
2.00X10^25m / 3.16x10^7 m/s = 2.00x10^10 s (20 billion years)

Since they both come out the same, and it's near enough to 14 billion years to seem reasonable, I think this is the correct answer. If I missed anything I'd appreciate it if someone could let me know. Thanks!