Question

The nearest star to our solar system is 4.29 light years away. How much is this distance in terms of parsers? How much parallax would this star (named alpha centauri) show when viewed from two locations of earth six months apart in its orbit around the sun?

Answer 1

Answer:

The distance of the star is,=4.29 ly

We know that one light year is the distance travelled by light in one year.

1 light year = 3×10  

8

×365×24×60×60  

                  =94608×10  

11

m

Therefore, the distance in 4.29 ly can be written as:

4.29 ly=405868.32×10  

11

 m

We know that 1 Parsec is also the unit of distance and its value is:

1 parsec=3.08×10  

16

m

Now, to express the distance in terms of parsec.

4.29ly=  

3.08×10  

16

405868.32×10  

11  

=1.32 parsec

The diameter of the Earth's orbit is,

d=3×10  

11

m

The angle made by the star on the Earth's orbit can be given by:

θ=  

405868.32×10  

11

3×10  

11  

=7.39×10  

−6  rad

But, the angle covered in 1sec=4.85×10  

−6

rad

So, the parallax in viwing the star at two different positions of the Eart's orbit:

for, 7.39×10  

−6

rad=  

4.85×10  

−6

7.39×10  

−6

   

=1.52"

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,

$Θ = \frac{l}{r}$

= 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.