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

Explanation:

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Distance of the star from the solar system = 4.29 ly

1 light year is the distance travelled by light in one year.

1 light year = Speed of light × 1 year

= 3 × 108 × 365 × 24 × 60 × 60 = 94608 × 1011 m

∴ 4.29 ly = 405868.32 × 1011 m

∵ 1 parsec = 3.08 × 1016 m

∴ 4.29 ly = 405868.32 × 1011 / 3.08 × 1016 = 1.32 parsec

Using the relation,

θ = d / D

where,

Diameter of Earth’s orbit, d = 3 × 1011 m

Distance of the star from the earth, D = 405868.32 × 1011 m

∴ θ = 3 × 1011 / 405868.32 × 1011 = 7.39 × 10-6 rad

But, 1 sec = 4.85 × 10–6 rad

∴ 7.39 × 10-6 rad = 7.39 × 10-6 / 4.85 × 10-6 = 1.52“

Answer 2

Answer:1.32 parsec; 1.58''

Explanation:1 Ly= 3×10⁸ m/s×31.5×10⁶ m (1 year=31.5×10⁶ sec.)

therefore, 4.29 Ly= 4.29×3×10⁸×31.5×10⁶ m

= 405.8×10¹⁴ m

1 parsec= 3×10¹⁶ m

therefore, total distance in parsec,

=  405.8×10¹⁴ m/3×10¹⁶ m= 1.32 parsec

Now, parallax angle of Alpha Centauri,θ,

= basis, b/distance between Alpha Centauri and Earth

basis= position of Earth six month's apart= 3×10¹¹

therefore, θ=  3×10¹¹/405.8×10¹⁴ m

=7.4⁻⁶

1''= 4.85×10⁻⁶ rad

∴ parallax angle, θ,

= 7.4⁻⁶/4.85×10⁻⁶ rad

= 1.58'' . Answer.