Question

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?

Answer 1

Hii dear,

# Answer-

a) Distance = 1.32 parsec

b) parallax = 7.39×10^-6 rad

# Explaination-

a) Given is distance of star from solar system D = 4.29 ly

But 1 ly = 0.307 parsec

Hence,

D = 4.29×0.307 = 1.32 parsec

(Also D = 4.06×10^16 m)

b) The two locations of observation are 6 months apart.

Hence, distance between 2 locations = diameter of earth's orbit = 3×10^11 m

Now,

parallax = d/D

parallax = (3×10^11)/(4.06×10^16)

parallax = 7.39×10^-6 rad

Hope you got your answer...

Answer 2

We know that:-

$1 \: light \: year \: = 9.46 \times 10 ^{15} m$

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

According to the given problem:-

$\: distance \: = 4.29 \: light \: years \\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = 4.29 \times 9.46 \times 10 ^{15} m\\ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: = \frac{4.29 \times 9.4 \times 10 ^{15} }{3.08 \times 10 ^{16} } parsec \\ \: \: \: \: \: \: \: \: \: \: = 1.32 \: parsecs$

l = 3*10^11m , r = 4.29*9.46*10^5m

Now, Θ = l / r

= 3*10^11 / 4.29*9.46*10^15

= 0.0739*10 ^ (-4)

= 7.39*10^(-6)radian

Therefore , 7.39*10^(-6)radian is the Answer.