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Sagot :
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 = [tex]H_{0}[/tex] x d
[tex]H_{0}[/tex] = 73 km/sec/Mpc
where, Mpc = Mega Parsec = 1 Mpc = 3.086 x [tex]10^{19}[/tex] km
We can use Hubble constant to tell the age of universe.
t = d/v
t = d/( [tex]H_{0}[/tex] xd)
t = 1/[tex]H_{0}[/tex]
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 [tex]H_{0}[/tex] = 73 km/sec/Mpc to [tex]H_{0}[/tex] = 700 km/sec/Mpc
then the age of universe will be:
t = 1/[tex]H_{0}[/tex]
first convert the units of new [tex]H_{0}[/tex] into 1/s
[tex]H_{0}[/tex] = (700) x (/3.08 x [tex]10^{19}[/tex] )
[tex]H_{0}[/tex] = 227.27 x[tex]10^{-19}[/tex] = 2.27 x [tex]10^{-21}[/tex] 1/s
So,
Age of universe will be:
t = 1/[tex]H_{0}[/tex] = 1/2.27x[tex]10^{-21}[/tex] 1/s
t = 2.27 x [tex]10^{21}[/tex] 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
The universe is "4.4×10²⁰ seconds" old.
Given value:
Hubble Constant,
- [tex]H_o = \frac{700 \ km/s}{Mpc}[/tex]
We know,
- [tex]Mpc = 3.086\times 10^{19} \ km[/tex]
By substituting the value of "Mpc" in Hubble constant, we get
→ [tex]H_o = \frac{700}{3.086\times 10^{19}}[/tex]
[tex]= 227\times 10^{-19} \ 1/s[/tex]
[tex]= 2.27\times 10^{-21} \ 1/s[/tex]
The Hubble's time will be:
→ [tex]H_o = \frac{1}{t}[/tex]
or,
→ [tex]t = \frac{1}{H_o}[/tex]
[tex]= \frac{1}{2.27\times 10^{-21}}[/tex]
[tex]= 4.4\times 10^{20} \ seconds[/tex]
Thus the above approach is right.
Learn more about Hubble Constant here:
https://brainly.com/question/13691927?
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