Question

The nearest star 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?

Answer 1

As we all know that to measure the distance between two stellar objects, we use the trigonometric method if one object revolves around another object in the space. For example, a star’s apparent movement can be calculated with respect to a more distant and bigger star if the smaller star is revolving round the bigger star.

Complete step by step solution:
Since one light-year is the distance travelled by light in a year. So we will calculate the distance travelled by light in the year in metres.
Since we know that the speed of light is $v = 3 \times {10^8}\;{\text{m/s}}$.
Hence, the distance travelled by light in one year is,
$D = v \times t$
Here $D$ is the distance travelled by light and $t$ is the time.
Substitute $v = 3 \times {10^8}\;{\text{m/s}}$ and $t = 1y$ (year), and it comes out to be,
$D = 3 \times {10^8}\;{\text{m/s}} \times {\text{1year}} \\\ D = 3 \times {10^8}\;{\text{m/s}} \times {\text{365}} \times {\text{24}} \times {\text{60}} \times {\text{60}} \\\ D = 94608 \times {10^{11}}\;{\text{m}} \\\$
Therefore, 4.29 ly can be written as,
${D_1} = 4.29 \times 94608 \times {10^{11}}\;{\text{m}} \\\ {D_1} = 405868.32 \times {10^{11}}\;{\text{m}} \\\$
The angle on earth’s orbit made by the star is given by,
$\theta = \dfrac{d}{{{D_1}}}$…… (i)
Here, d is the diameter of the earth’s orbit and D1 is the distance travelled by the light.
Substitute $d = 3 \times {10^{11}}\;{\text{m}}$ and ${D_1} = 405868.32 \times {10^{11}}\;{\text{m}}$ in equation (i) to find the value of $\theta$.
$\theta = \dfrac{{3 \times {{10}^{11}}}}{{405868.32 \times {{10}^{11}}\;{\text{m}}}} \\\ \theta = 7.39 \times {10^{ - 6}}\;{\text{rad}} \\\$

Since the angle captured in $1\;{\text{s = 4}}{\text{.85}} \times {\text{1}}{{\text{0}}^{ - 6}}\;{\text{rad}}$ therefore, the parallax P in observing the stars in two distinct location is,
$P =\dfrac{7.39 \times 10^{ - 6}}{4.85 \times 10^{-6}}$
$P=1.52’’$

$\therefore$ The parallax in viewing the star at two different positions of the Earth's orbit is 1.52’’

Note:
The distance between two positions of the earth in the span of six months is equal to two astronomical units. We can also notice that over a course of the year some stars move a very small amount relative to other stars. The stars that are closer seem to move but the stars that are far away don’t seem to move. Actually all stars are moving through space but much more slowly than parallax so we don’t come to notice it.