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  2. Use the principle of ‘parallax’. 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 × 10^11m

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Use the principle of ‘parallax’. 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 × 10^11m

2.19 The principle of ‘parallax’ in section 2.3.1 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 \times 10^{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?

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The diameter of Earth’s orbit = 3 \times 10^{11} m
\therefore The radius of Earth’s orbit, r = 1.5 \times 10^{11} m
Let the distance parallax angle be = 4.847 \times 10^{-6} rad.
Let the distance between earth and star be R.
(Parsec is the distance at which average radius of earth’s orbit
subtends an angle of 1{}''.)
We have     \Theta = r/R    (Analogous to a circle, R here is the radius, r is the arc length and \Theta is the angle covered ! )

R = \frac{r}{\Theta } = \frac{1.5 \times 10^{11}}{4.847 \times 10^{-6}} = 0.309 \times 10^{17}

= 3.09 \times 10^{16}\ m

Hence, 1 parsec = 3.09 \times 10^{16}\ m.

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