The nearest star to our solar system is 4.29 light years away. How mcuh is this distance in terms of par sec ? How mcuh parallax would this star show when viewed from two locations of the earth six months apart in its orbit around the sun?

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Solution :
As we know, 1 light year `=9.46xx10^(15) m` <br> `:. 4.29` light years `=4.29xx9.46xx10^(15)=4.058xx10^(16) m` <br> Also, 1 parsec `=3.08xx10^(16) m` <br> `:. 4.29` light years `=4.508xx10^(16)//3.80xx10^(16)=1.318` parsec `=1.32` parsec. <br> As a parsec distance subtends a parallax angle of 1" for a basis of radius of Earth's orbit around the sun (r). In present problem base is the distance between two location of the Earth six months apart in its orbit around the Sun = dimeter of Earth's orbit `(b = 2r)`. <br> `:.` Parallax angle subtended by 1 parsec distance at this basis =2 second (by definition of parsec). <br> `:.` Parallax angle subtended by the star Alpha Centauri at the given basis `theta =1.32 xx 2=2.64`.

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