Question 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?
Class XI Physics Units And Measurements Page 37
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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
●therefore 4.29 ly = 405868.32 × 1011 m
● since 1 parsec = 3.08 × 1016 m
● therefore 4.29 ly = 405868.32 × 1011 / 3.08 × 1016 = 1.32 parsec
NowUsing 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
therefore θ = 3 × 1011 / 405868.32 × 1011 = 7.39 × 10-6 rad
But, 1 sec = 4.85 × 10–6 rad
Hence7.39 × 10-6 rad = 7.39 × 10-6 / 4.85 × 10-6 = 1.52"
●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
●therefore 4.29 ly = 405868.32 × 1011 m
● since 1 parsec = 3.08 × 1016 m
● therefore 4.29 ly = 405868.32 × 1011 / 3.08 × 1016 = 1.32 parsec
NowUsing 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
therefore θ = 3 × 1011 / 405868.32 × 1011 = 7.39 × 10-6 rad
But, 1 sec = 4.85 × 10–6 rad
Hence7.39 × 10-6 rad = 7.39 × 10-6 / 4.85 × 10-6 = 1.52"
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