# Homework Help: Hubbkles Constant and the age of the universe Its not very long apparentely

1. Nov 18, 2008

### TFM

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

Neglecting other elements of the cosmological model, the inverse of the Hubble constant, $$1/H_0$$, tells us the time since the Big Bang - the age of the universe. If $$H_0 = 70$$km/sec/Mpc, then how old is the universe?

2. Relevant equations

Given above

3. The attempt at a solution

Apparently the answer is shorter then the time you probably took reading the title...

I think the problem lies in a conversion error. I have converted the constant into km/sec/km, which gives the large answer of $$2.16*10^{21}$$ then into m/sec/m, which gives $$2.16*10^{24}$$. this gives an inverse, and thus the age of the universe, to be about $$*10^{-24}$$. Where have I gione wrong? as stated above, I am sure its a conversion error, bu can't see where?

Any ideas?

TFM

2. Nov 18, 2008

### Jack_O

You need to divide H(0) by Mpc, not multiply (check the units). You should get a correct value for the age of the universe with this correction.

3. Nov 18, 2008

### D H

Staff Emeritus
Part of the problem here is the mixed units. Obviously a more rational system of units is needed. :tongue2:

H0 = 21.7637665 (mil/fortnight)/megafurlong

Using the handy conversion 1 furlong=7920000 mil,

\aligned H_0 &= 21.8\,\text{(mil/fortnight)/megafurlong} \\ &= 21.8\,\frac{\text{mil}}{\text{fortnight}\cdot \text{megafurlong}} \cdot\frac{1\,\text{megafurlong}}{10^6\,\text{furlong}} \cdot\frac{1\,\text{furlong}}{7920000\,\text{mil}} \\ &= \frac{21.8}{7.92\cdot10^{12}}\,\text{fortnight}^{-1} = 2.75\cdot10^{-12}\,\text{fortnight}^{-1} \endaligned

You should have a bunch of units canceling one another (i.e., appearing in both the numerator and denominator) whenever you are doing unit conversions. In this case, the furlongs, megafurlongs, and mils all cancel.

You can do the same using the slightly less insane (km/sec)/megaparsec units typically used in the Hubble constant.

4. Nov 19, 2008

### TFM

So I have 70, and should divide it by:

$$1 Mpc = 3.06 x 10^{19} km$$

this gives me:

$$2.29*10^{-18}$$

Giving the age to be:$$4.37*10^{17}$$

but this was done through using the Hubble Constant in km/s/km, do I need to convert to m, or will this do?

TFM

5. Nov 19, 2008

### D H

Staff Emeritus
Where did you get that value? 1 megaparsec is 30.8568×1018 km.

[soapbox]Try to get in the habit of always carrying the units with you.[/soapbox]

Years. You differ slightly from the standard value of Hubble time because you used a slightly off conversion factor for km/Mpc.

The kilometers cancel: 2.29×10-18 km/sec/km = 2.29×10-18 sec-1.

6. Nov 19, 2008

### TFM

That value was given on the website which the work was from (SDSS)

Wouldn't the value I calculated be in seconds, though, since I worked it out into km/s/km?

TFM

7. Nov 19, 2008

### D H

Staff Emeritus
Yes it would, but you didn't say 2.29×10-18 sec-1, you just said 2.29×10-18. You similarly omitted the units on the Hubble time: You said the time is 4.37×1017 rather than 4.37×1017 seconds. The reason I am being so pedantic here about units is that failing to pay careful attention to units is one of the key reasons students get wrong answers. Failing to pay careful attention to units is exactly what led you to the wrong answer in the first post and made you ask us for help.

8. Nov 19, 2008

### TFM

Re: Hubbkles Constant and the age of the universe...Its not very long apparentley...

Okay, I definitely agree with you there...

I assume it will be best to convert the answer into years, for a more manageable answer?

TFM