Question

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

Answer 1

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 .

So, the age of our universe is surmised to be about this

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