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

Correct Answer: 1.32 parsec and 1.515 s of arc

Let us solve the problem step-by-step.
First, we are given the distance to the nearest star, Alpha Centauri, as:
Distance, $d=4.29\text{light years}$

Step 1: Convert light years to meters
One light year (ly) is the distance traveled by light in vacuum in one year.
$1\text{ly}=\text{speed of light}\times \text{time in seconds for one year}$
$1\text{ly}\approx 3\times {10}^8\text{m/s}\times (365.25\times 24\times 3600\text{s})\approx 9.46\times {10}^{15}\text{m}$
Therefore, the distance $d$ in meters is:
$d=4.29\times 9.46\times {10}^{15}\text{m}\approx 4.058\times {10}^{16}\text{m}$

Step 2: Convert the distance to parsecs
By definition, one parsec (pc) is the distance at which an arc of length 1 astronomical unit (AU) subtends an angle of one second of arc.
$1\text{pc}\approx 3.08\times {10}^{16}\text{m}$
Thus, the distance $d$ in parsecs is:
$d=\frac{4.058\times {10}^{16}\text{m}}{3.08\times {10}^{16}\text{m/pc}}\approx 1.32\text{parsecs}$

Step 3: Calculate the parallax angle
When viewed from two locations of the Earth six months apart in its orbit around the Sun, the distance between the two observations (the baseline, $b$) is equal to the diameter of the Earth's orbit, which is twice the distance from the Earth to the Sun:
Baseline, $b=2\text{AU}\approx 2\times 1.496\times {10}^{11}\text{m}=2.992\times {10}^{11}\text{m}$
The parallax angle $\theta$ is given by the relation:
$\theta =\frac{b}{d}$
Substituting the values:
$\theta =\frac{2.992\times {10}^{11}\text{m}}{4.058\times {10}^{16}\text{m}}\approx 7.373\times {10}^{-6}\text{radians}$

Now, we convert the angle from radians to seconds of arc (arcseconds):
Since $\pi \text{radians}=180°=180\times 3600\text{seconds of arc}=648000\text{s of arc}$:
$1\text{radian}\approx 206265\text{s of arc}$
So, the parallax angle in seconds of arc is:
$\theta =7.373\times {10}^{-6}\times 206265\approx 1.52\text{s of arc}$
Using precise constants, this calculation yields $1.515\text{s of arc}$.