Question

A star is $1.45parsec$ light years away. How much parallax would this star show when viewed from two locations of the earth six months apart in its orbit around the sun?

• A. $2Parsec$
• B. $0.725Parsec$
• C. $1.45Parsec$
• D. $2.9Parsec$

Answer 1

The correct option is A $2Parsec$

One light year $=$ speed of light $\times$ one year

or $1ly=3\times {10}^8\times (24\times 3600)=94608\times {10}^{11}m$

So, $4.29ly=4.29\times (94608\times {10}^{11})=4.058\times {10}^{16}m$

As $1$ parsec $=3.08\times {10}^{16}m$

(Parsec is a unit of length used to measure large distances to objects outside our Solar System)

Thus, $4.29ly=\frac{4.058\times {10}^{16}}{3.08\times {10}^{16}}=1.32$ parsec

Now angular displacement , $\theta =\frac{d}{D}$

where $d=$ diameter of earth's orbit $=3\times {10}^{11}m$ and

$D=$ distance of star from the earth $=4.058\times {10}^{16}m$

So, $\theta =\frac{3\times {10}^{11}}{4.058\times {10}^{16}}=7.39\times {10}^{-6}$ rad

As $1sec=4.85\times {10}^{-6}rad$ so, $7.39\times {10}^{-6}rad=\frac{7.39\times {10}^{-6}}{4.85\times {10}^{-6}}=1.52sec$