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Question : 19 of 33
Marks: +1, -0
The principle of ‘parallax’ is used in the determination of distances of very distant stars. The baseline AB is the line joining the Earth’s two locations six months apart in its orbit around the Sun. That is, the baseline is about the diameter of the Earth’s orbit ≃ 3 × 1011^{11} m. However, even the nearest stars are so distant that with such a long baseline, they show parallax only of the order of 1” (second) of arc or so. A parsec is a convenient unit of length on the astronomical scale. It is the distance of an object that will show a parallax of 1” (second) of arc from opposite ends of a baseline equal to the distance from the Earth to the Sun. How much is a parsec in terms of metres?
Solution:  
Here, the diameter of the Earth’s orbit =3×1011 m= 3 \times 10^{11} \text{ m},
Therefore, distance of the sun from the earth =3×10112=1.5×1011 m= \frac{3 \times 10^{11}}{2} = 1.5 \times 10^{11} \text{ m}
Thus, length of the baseline,b=1.5×1011 mb = 1.5 \times 10^{11} \text{ m}
Also, the parallax,
θ=1′′=1∘60×60\theta = 1'' = \frac{1^{\circ}}{60 \times 60}
=160×60×π180= \frac{1}{60 \times 60} \times \frac{\pi}{180}
=4.85×10−6 rad= 4.85 \times 10^{-6} \text{ rad}
Now, S=bθS = \frac{b}{\theta}
By definition,
When b=1.5×1011 mb = 1.5 \times 10^{11} \text{ m} and θ=1′′,S=1\theta = 1'', S = 1 parsec
∴1\therefore 1 parsec =bθ= \frac{b}{\theta}
=1.5×10114.85×10−6= \frac{1.5 \times 10^{11}}{4.85 \times 10^{-6}}
=3.09×1016 m≈3×1016 m= 3.09 \times 10^{16} \text{ m} \approx 3 \times 10^{16} \text{ m}
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