#### Question

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 × 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 meters?

#### Solution 1

Diameter of Earth’s orbit = 3 × 10^{11} m

Radius of Earth’s orbit, *r* = 1.5 × 10^{11} m

Let the distance parallax angle be 1”= 4.847 × 10^{–6} rad.

Let the distance of the star be *D.*

Parsec is defined as the distance at which the average radius of the Earth’s orbit subtends an angle of 1.

:. We have `theta = r/D`

`D=r/theta = (1.5xx10^(11))/(4.847xx10^(-6))`

`=0.309 xx 10^(-6) ~~ 3.09xx10^(16)m`

Hence, 1 parsec ≈ 3.09 × 10^{16} m.

#### Solution 2

From parallax method we can say

θ=b/D,where b=baseline ,D = distance of distant object or star

Since, θ=1″ (s) and b=3 x 10^{11 }m

D=b/20=3 x 10^{11}/2 x 4.85 x 10^{-6} m or D=3 x 10^{11}/9.7 x 10^{-6} m =30 x 10^{16}/9.7 m

= 3.09 x 10^{16} m = 3 x 10^{16} m.