# How fast is the universe expanding at the farthest edge we can see?

Aug 3, 2016

I surmise that the rate is uniform now, from there to here..

#### Explanation:

Not-easily-understandable-to-the-commonalty:,

The average expansion rate of the universe is the Hubble

constant

${H}_{0}$ = 71 km / sec / megaparsec

and the age of our universe is

13.77 billion years, nearly..

For making it relatively, easier to understand, this can be

interpreted as

(UNIT DISTANCE)/(UNIT DISTANCE/second

= 1 km /1 km / 1 second

= (1 AU / 1 AU / `second

= 1 parsec / 1 parsec /second

= Hubble constant ${H}_{9}$

= 71 km / s / megaparsec ....

Choosing befitting units,

Hubble constant X Age = 1.

Dimension-wise,

[H_0]=T^(-1)] and [Age] $= T$.

The farthest globular cluster is surmised to be at a distance of

about 13.77 billion light years .

time of 13.77 billion years, taken by light from the source, to reach

us.

Perhaps, the rate of expansion was very much faster when our

universe was ( in the mathematical sense of epsilon) incredibly

small, at its center, about 13.77 billion years ago..

I surmise that the rate is uniform now.

We can derive, the Hubble constant ${H}_{0} = 71$ km//s/megaparsec

from the the age 13.77 billion years and vice versa. .

See how this is done.

The reciprocal of age

= 1/(13..77 billion years)

1/((13.77 X ${10}^{9}$)(365.25)(24)(3600) seconds

=.1/(4.3455 X ${10}^{16}$ seconds)

Now,

1 mega parsec/ 1 mega parsec

${10}^{6}$ X 206365 AU/1 mega parsec

= (10^6 X 206365)(149597871) km/1 mega parsec

= 3.0857 X ${10}^{19}$ km/ 1 mega parsec.

So, ${H}_{0}$

= 1/age , in km/mega parsec / sec is

= (3.0857 X ${10}^{19}$)/(4.3455 X $10$16)

= 71..009