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(A) Hubble's Law is given by $v = H 	imes r$,where $v$ is the velocity,$H$ is the Hubble constant,and $r$ is the distance.$H = \frac{v}{r} = \frac{86

Vedclass · Accepted Answer

$2 	imes 10^{-5} \ km \ s^{-1} \ ly^{-1}, \ 1.49 	imes 10^{10} \ 	ext{years}$

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(A) Hubble's Law is given by $v = H 	imes r$,where $v$ is the velocity,$H$ is the Hubble constant,and $r$ is the distance.$H = \frac{v}{r} = \frac{8600 \ km \ s^{-1}}{430 	imes 10^6 \ ly} = 2 	imes 10^{-5} \ km \ s^{-1} \ ly^{-1}$.The age of the universe $t_0$ is given by $t_0 = \frac{1}{H}$.To calculate the age in years,we convert the distance $r$ from light-years to kilometers: $1 \ ly \approx 9.46 	imes 10^{12} \ km$.$r = 430 	imes 10^6 	imes 9.46 	imes 10^{12} \ km = 4.0678 	imes 10^{21} \ km$.$t_0 = \frac{r}{v} = \frac{4.0678 	imes 10^{21} \ km}{8600 \ km \ s^{-1}} \approx 4.73 	imes 10^{17} \ s$.Converting seconds to years: $t_0 = \frac{4.73 	imes 10^{17}}{3600 	imes 24 	imes 365.25} \approx 1.49 	imes 10^{10} \ 	ext{years}$.

$A$ galaxy is observed to be moving with a velocity of $8600 \ km \ s^{-1}$. If it is at a distance of $430$ million light-years from us,the Hubble constant and the corresponding age of the universe are respectively:

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