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Q 2.20: The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsecs? How much parallax would this star (named Alpha Centauri) show when viewed from two locations of the Earth six months apart in its orbit around the Sun?

Answers (1)

Given, Distance of the star from the solar system = 4.29 ly (light years)
1 light year is the distance travelled by light in one year.

(Note: Light year is a measurement of distance and not time!)
(a) 1 ly = (3 \times 10^8)ms^{-1} \times (365 \times 24 \times 60 \times 60)s = 94608 \times 10^{11} m
4.29 ly = 405868.32 \times 10^{11} m
We know, 1 parsec = 3.08 \times 10^{16} m

\therefore 4.29 ly = \frac{405868.32\times10^{11}}{3.08\times10^{16}}= 1.32 parsec

Now,

(b) ? = d/D\Theta = d/D

& d = 3 \times 10^{11} m; D = 405868.32 \times 10^{11} m

\theta = (3\times10^{11})/(405868.32\times10^{11} )= 7.39\times10^{-6} rad

Also, We know 1\ sec = 4.85 \times10^{-6} rad

\therefore 7.39\times10^{-6} rad = \frac{7.39\times10^{-6}}{4.85 \times10^{-6}} = 1.52''

Posted by

Safeer PP

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