
What is the distance of the star (in km) from which a light takes $3\times {{10}^{9}}$ year to reach the earth? (speed of light is $c=3\times {{10}^{8}} m/s$)
Answer
500.1k+ views
Hint:Distance between the star and earth is given in the question and speed of light takes to reach the earth is also given in the question so, we use the simple concept of distance and get the answer.
Formula used:
$\text{Distance}=\text{Speed} \times \text{Time}$
Complete step by step answer:
The distance between the star and the earth is too large, so the time to reach the light in the year. When a light travel form star to earth it takes time $T=3\times {{10}^{9}}\,year$ and we know the speed of light $c=3\times {{10}^{8}}\,m/s$ then we put this information in the formula of distance.
$\text{Distance}=\text{Speed} \times \text{Time}$
We put the time in this formula in second then we have to change time in second. We have to use this information to convert units of time. First of all, we have to change the year into the days then days into the hour then hour into the seconds.
$1\,year=365\,days$, $1\,day=24\,hr$, $1\,hr=3600\,\sec $
Convert into days
$T=3\times {{10}^{9}}\times 365\,days$
Convert into hour
$T=3\times {{10}^{9}}\times 365\times 24\,hr$
Convert into seconds
$T=3\times {{10}^{9}}\times 365\times 24\times 3600\,\sec $
After applying multiplication rule, we get
$T=9.460\times {{10}^{16}}\sec $
We put the value of time and speed in the formula
$\text{Distance}=\text{Speed} \times \text{Time}$
$\Rightarrow \text{Distance}=(3\times {{10}^{8}})\times (9.460\times {{10}^{16}})\,m$
$\Rightarrow \text{Distance}=2.838\times {{10}^{25}}\,m$
According to the question, distance should be in kilometres. So, we have to change the units of distance from meters into kilometers. We use the concept of this formula to change the unit of distance.
$1\,m={{10}^{-3}}\,km$
$\Rightarrow \text{Distance}=2.838\times {{10}^{25}}\times {{10}^{-3}}\,km$
$\therefore \text{Distance}=2.838\times {{10}^{22}}\,km$
So, the distance between the star and earth is $2.838\times {{10}^{22}}\, km$.
Note:In this problem distance is $2.838\times {{10}^{22}}\,km$ which is too large so, the light travel with speed $3\times {{10}^{8}}\,m/s$ and reach on the earth after a long time $T=3\times {{10}^{9}}\,year$. We should know the concept of changing the units like metre to kilometre, days to hour and hour to seconds.
Formula used:
$\text{Distance}=\text{Speed} \times \text{Time}$
Complete step by step answer:
The distance between the star and the earth is too large, so the time to reach the light in the year. When a light travel form star to earth it takes time $T=3\times {{10}^{9}}\,year$ and we know the speed of light $c=3\times {{10}^{8}}\,m/s$ then we put this information in the formula of distance.
$\text{Distance}=\text{Speed} \times \text{Time}$
We put the time in this formula in second then we have to change time in second. We have to use this information to convert units of time. First of all, we have to change the year into the days then days into the hour then hour into the seconds.
$1\,year=365\,days$, $1\,day=24\,hr$, $1\,hr=3600\,\sec $
Convert into days
$T=3\times {{10}^{9}}\times 365\,days$
Convert into hour
$T=3\times {{10}^{9}}\times 365\times 24\,hr$
Convert into seconds
$T=3\times {{10}^{9}}\times 365\times 24\times 3600\,\sec $
After applying multiplication rule, we get
$T=9.460\times {{10}^{16}}\sec $
We put the value of time and speed in the formula
$\text{Distance}=\text{Speed} \times \text{Time}$
$\Rightarrow \text{Distance}=(3\times {{10}^{8}})\times (9.460\times {{10}^{16}})\,m$
$\Rightarrow \text{Distance}=2.838\times {{10}^{25}}\,m$
According to the question, distance should be in kilometres. So, we have to change the units of distance from meters into kilometers. We use the concept of this formula to change the unit of distance.
$1\,m={{10}^{-3}}\,km$
$\Rightarrow \text{Distance}=2.838\times {{10}^{25}}\times {{10}^{-3}}\,km$
$\therefore \text{Distance}=2.838\times {{10}^{22}}\,km$
So, the distance between the star and earth is $2.838\times {{10}^{22}}\, km$.
Note:In this problem distance is $2.838\times {{10}^{22}}\,km$ which is too large so, the light travel with speed $3\times {{10}^{8}}\,m/s$ and reach on the earth after a long time $T=3\times {{10}^{9}}\,year$. We should know the concept of changing the units like metre to kilometre, days to hour and hour to seconds.
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