Question

A star is 1.45 parsec away. How much parallax will this star show when viewed from two locations of the earth six months apart in its orbit around the sun?

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#Answer with Explanation​

Answer 1

Parallax Angle

Parallax Angle refers to the angle formed between the two apparent positions of an object when observing from two different positions.

The Earth revolves around the Sun with an orbital period of 12 months (1 year). The mean distance between the Earth and the Sun is 1 AU(Astronomical Unit).

At six months apart, the Earth will be at diametrically opposite positions in its orbit. So, the distance between two positions of the Earth becomes 2 AU.

Now, consider a distant star. When we observe the Star from Earth six months apart, we find a difference in the direction we see the star in. The two positions form a specific angle. This is the Parallax Angle, which we want to find out in this question.

See the diagram attached.

We will consider the standard formula, as seen in the second image attached.

$\huge\boxed{\theta = \dfrac{l}{r}}$

The $l$ in the image, and in reality, is a curve. However, objects like Stars are really far away and the angle $\theta$ is very very small. So, the curve represented by length $l$ can be safely approximated by a straight line.

Here, it is represented as the distance between positions of Earth 6 months apart.

$\rule{300}{1}$

We have a couple of standard astronomy constants: Parsec (pc) and Astronomical Unit (AU).

$\sf 1 \: \textsf{pc} = 3.086 \times 10^{16} \: m\\ \\ 1 \: \textsf{AU} = 1.496 \times 10^{11} \: m$

Let's now look at our data:

$l = 2 \: AU = 2\times 1.496\times 10^{11} \: m = 2.992 \times 10^{11} \: m \\ \\ r = 1.45 \: pc = 1.45 \times 3.086 \times 10^{16} \: m = 4.4747\times 10^{16} \: m$

We need the Parallax Angle. Let us call it $\theta$.

We can obtain it by our simple formula:

$\displaystyle \theta = \frac{l}{r} \\\\\\ \implies \theta = \frac{2.992 \times 10^{11}}{4.4747\times 10^{16}} \approx 6.686 \times 10^{-6} \: radians \\\\\\ \implies \huge \boxed{\sf\theta \approx 6.686 \times 10^{-6} \: \textsf{rad}}$

So, we obtained the answer in radians. However, since the notion of degrees, minutes and seconds is so much more comprehensible than a radian, let's change the units.

$\displaystyle\pi \: rad = 180^{\circ} \\\\\\ \left[1^{\circ} =60\,' \text{ and } 1\,' = 60\,'' \right] \\ \\ \\ \implies \pi \: rad = (180 \times 60 \times 60)\, '' \\\\\\ \implies 1\: rad = \left(\dfrac{180\times 60\times 60}{\pi}\right)'' \\\\\\ \implies \theta = 6.686\times 10^{-6} \: rad = \left(\frac{6.686\times 10^{-6}\times 180\times 3600}{\pi}\right)'' \\\\\\ \implies \huge \boxed{\theta\approx1.379\,''}$

As we see, the Parallax Angle is indeed extremely small!! It is hardly just 1 second!

This just shows us the scale of the Universe! Stars are so far apart. The angles subtended get so small.

Finally, we have our answer:

$\Large \boxed{\boxed{\sf \theta \approx 6.686\times 10^{-6} \textsf{ rad} \approx 1.379\,''}}$

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EXTRA INFO

The answer has finished. This is the Extra Info Section. Read on if you wish to know more!

1) Astronomical Unit (AU)

The Astronomical Unit is a unit of distance.

The mean distance between the Sun and Earth is defined as 1 AU.

The Earth's orbit is not perfectly circular. It is slightly elliptical. So, at some points in its orbit, the Earth is closer to the Sun than other points.

The point in the Earth's orbit when it is closest to the Sun is called the Perihelion. At this point, the Earth is about 147 million kilometres from the Sun.

The point in the Earth's orbit when it is farthest from the Sun is called the Aphelion. At this point, the Earth is about 153 million kilometres from the Sun.

The mean distance is around 150 million kilometres, and this is defined as 1 AU.

[See the third image attached]

Precisely, it is 149.6 million kilometres.

$\sf 1 \textsf{ AU} = 1.496 \times 10^{11} \textsf{ m}$

2) Parsec (pc)

Parsec is another unit of length.

A parsec is defined as the distance at which an arc length of 1 AU subtends an angle of 1 second.

[See the fourth image attached]

Here, as seen, we have an arc length of 1 AU, and the angle subtended as 1''. On calculating from:

$\sf\tan 1'' = \dfrac{\textsf{1 AU}}{\textsf{1 pc}}$

Since 1'' is an extremely small angle, we can directly approximate it as:

$\sf\displaystyle 1'' = \frac{\textsf{1 AU}}{\textsf{1 pc}} \\\\\\ \implies \frac{1}{3600} \times \frac{\pi}{180} \: rad =\frac{\textsf{1 AU}}{\textsf{1 pc}} \\\\\\ \implies 1\ pc = \frac{1.496\times 10^{11}\times 3600\times 180}{\pi} \: m \\\\\\ \implies 1\ pc \approx 3.086 \times 10^{16}\ m$

1 parsec is huge! The order of $10^{16}$ shows just this. It is around $10^5$ times the Astronomical Unit.

For comparison, the diameter of Solar System is estimated to be 122 AU! Much less than even a parsec!

The Universe is huge! And this is where we live :)

Answer 2

Explanation:

Parallax Angle refers to the angle formed between the two apparent positions of an object when observing from two different positions.

The Earth revolves around the Sun with an orbital period of 12 months (1 year). The mean distance between the Earth and the Sun is 1 AU(Astronomical Unit).

At six months apart, the Earth will be at diametrically opposite positions in its orbit. So, the distance between two positions of the Earth becomes 2 AU.

Now, consider a distant star. When we observe the Star from Earth six months apart, we find a difference in the direction we see the star in. The two positions form a specific angle. This is the Parallax Angle, which we want to find out in this question

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