Question

Approximately, What is the value of the Hubble Constant, as measured by scientists? Hypothetically, if the value of the Hubble Constant were 700 km/s/Mpc, what would this imply about the age of our universe?

Answer 1

Answer:

The current value of the Hubble's constant = 73 km/sec/Mpc.

t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc

Explanation:

The current value of the Hubble's constant = 73 km/sec/Mpc. However, recent discoveries in the cosmology contradicts the idea of Hubble constant as being fixed. Some scientists are not agreeing on this value and the debate is going on.

Hubble law states that how fast universe is expanding or in other words, galaxies are expanding separating with a speed directly proportional to the distance of galaxies to the earth.

Hence,

v is directly proportional to d

where, v = apparent velocity

d = distance

if we equate velocity and distance then there comes Hubble constant.

v = $H_{0}$ x d

 $H_{0}$ = 73 km/sec/Mpc

where, Mpc = Mega Parsec = 1 Mpc = 3.086 x $10^{19}$ km      

We can use Hubble constant to tell the age of universe.

t = d/v

t = d/( $H_{0}$ xd)

t = 1/$H_{0}$

Scientist calculated the age of universe by using Hubble constant, which is 13.4 billion years.

Now, if we hypothetically change the value of Hubble constant,

from $H_{0}$ = 73 km/sec/Mpc to $H_{0}$ = 700 km/sec/Mpc

then the age of universe will be:

t = 1/$H_{0}$

first convert the units of new $H_{0}$ into 1/s

$H_{0}$ = (700) x (/3.08 x $10^{19}$ )

$H_{0}$ = 227.27 x$10^{-19}$  = 2.27 x $10^{-21}$ 1/s

So,

Age of universe will be:

t = 1/$H_{0}$ = 1/2.27x$10^{-21}$ 1/s

t = 2.27 x $10^{21}$ s

t = 71.9 trillion years

t = 71.9 trillion years will be the new age of universe if the Hubble constant = 700 km/s/Mpc