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Question: Arrange the following lengths in descending order of their magnitudes: Light year, parsec, and Astro...

Arrange the following lengths in descending order of their magnitudes: Light year, parsec, and Astronomical unit.

Explanation

Solution

We know the distance traveled by light in one year is called light-year(ly). It is used to measure large distances. A parsec is a distance at which one astronomical unit subtends an angle of one second of arc, and the Astronomical unit measures the distance between the earth and the sun. A light-year is also related to astronomy and parsec. For the arrangement of these lengths in descending order, we calculate the values of each.

Complete step by step answer:
We know that one light-year is equal to the distance that light travels in a year.
1ly=c×1yr\therefore 1ly = c \times 1yr
Here c is the speed of light in a vacuum, which is equal to 3×108  ms3 \times {10^8}\;\dfrac{m}{s} and 1 ly means one light year. Hence on simplifying the above relation, we get,
1ly=3×108  m/s×365×24×60×60s\Rightarrow 1ly = 3 \times {10^8}\;m/s \times 365 \times 24 \times 60 \times 60s.
1ly=9.46×1015m\Rightarrow 1ly = 9.46 \times {10^{15}}{\rm{m}}
The relation between light-year (ly) and astronomical unit (A.U)
1(ly)=9.46×1015m1({\rm{ly) = 9}}{\rm{.46}} \times {\rm{1}}{{\rm{0}}^{15}}m …… (I) and,
1A.U=1.5×1011m{\rm{1A}}{\rm{.U = 1}}{\rm{.5}} \times {\rm{1}}{{\rm{0}}^{11}}m …… (II)
Here 1 A.U. means 1 astronomical unit.
On dividing equations(I) and (II), we get:
1ly1A.U=9.456×1015m1.5×1011m\dfrac{{1{\rm{ly}}}}{{1{\rm{A}}{\rm{.U}}}} = \dfrac{{9.456 \times {{10}^{15}}m}}{{1.5 \times {{10}^{11}}m}}
1ly1A.U=6.3×104\Rightarrow \dfrac{{1{\rm{ly}}}}{{1{\rm{A}}{\rm{.U}}}} = 6.3 \times {10^4}
1ly=6.3×104A.U\therefore 1{\rm{ly = 6}}{\rm{.3}} \times {\rm{1}}{{\rm{0}}^4}{\rm{A}}{\rm{.U}}
The relation between the light-year (ly) and parsec
Again 1ly=9.46×1015m1{\rm{ly = 9}}{\rm{.46}} \times {\rm{1}}{{\rm{0}}^{15}}m …… (III)
And 1parsec=3.1×1016m{\rm{1parsec = 3}}{\rm{.1}} \times {\rm{1}}{{\rm{0}}^{16}}m…… (IV)
On dividing equations (III) and (IV), we get
1parsec1ly=3.1×1016m9.46×1015m\dfrac{{1{\rm{parsec}}}}{{1{\rm{ly}}}} = \dfrac{{3.1 \times {{10}^{16}}m}}{{9.46 \times {{10}^{15}}m}}
1parsec1ly=3.28\dfrac{{1{\rm{parsec}}}}{{1{\rm{ly}}}} = 3.28
1parsec=3.28ly\therefore 1{\rm{parsec = 3}}{\rm{.28ly}}
Suppose we saw the actual value of light year, parsec, A.U. (Astronomical unit). In that case, we will observe that parsec has greater value than a light-year, and a light year is greater than the Astronomical unit.
Therefore arrange in descending order of their magnitudes parsec>light year>astronomical unit.
  1016>1015>1011\Rightarrow \;{10^{16}} > {10^{15}} > {10^{11}}

Note: The lengths of a light-year, parsec, and astronomical unit are the same and scientifically proven throughout the universe. When we calculate the large distances and very small distances, we use non-S.I units of lengths for such measurements.