Parallaxes and Distances

\[d_\text{pc} = \frac{1}{p_\text{arcsec}}\text{ pc}\quad\text{[Eq. (15.3)]}\]

and

\[1\text{ pc} = 3.26\text{ ly} = \text{206,265 au}\quad\text{[Eq. (15.4)]}\]

Note: If you feel uncomfortable with working with fractions, variables, or unit conversion, you should review the following tutorials before working through this tutorial:

Sections

Introduction

Distances to astronomical objects can be one of the most difficult and important quantities to measure. One cannot simply stretch a tape measure out to the star Vega for example (or even to the Moon!). The first and most fundamental way to measure distances to the nearby stars is by making use of the well-known phenomenon of parallax. Parallax occurs when one observes an object from two different locations. The line-of-sight from one location to the object differs from the line-of-sight from the other location. How much the line-of-sight changes depends on how far apart the two locations are. The fact that our eyes are separated is one of the ways in which we perceive distance (depth perception).

Imagine looking at a nearby tree from two spots located 30 meters apart. In one of the locations the tree may appear to be aligned with a distant building, but from the other location the same tree might be aligned with a distant hillside. If one of the observing points remained the same, but the other was now separated by 50 m, a different object would align with the tree. On the other hand, suppose that the two original observing positions are used, but a much more distant tree is the target of the observations. In such a case it might be the same building is aligned with the tree from both viewing locations, but the exact spots on the building would be in different locations on the building’s wall. The greater the separation of observing locations, the greater the shift in the line-of-sight. On the other hand, the more distant the object being observed, the smaller the shift.

Figure 15.19

Figure 15.19 illustrates stellar parallax by observing a star from two different locations in Earth’s orbit, six months apart. The line-of-sight to the star on January 1 differs from the line-of-sight to the star on July 1. In this ideal case, a right triangle can be constructed by drawing lines from Earth to the Sun, from the Sun to the star, and from the star back down to Earth. Because the distance from the Sun to the star is much, much greater than the distance from Earth to the Sun, the triangle is very long and skinny (in other words, the distance \(d\) is much greater than the distance from Earth to the Sun (\(1\text{ au})\), and the angle \(p\) is tiny. Using the mathematics of trigonometry and making a good approximation because the parallax angle \(p\) is so small gives Equation (15.3) at the top of the page, assuming that \(p\) is expressed in units of arcseconds (there are \(3600\) arcseconds in one degree and \(\text{1,296,000}\) arcseconds in a complete circle). Since the distance from Earth to the Sun is always about \(1\text{ au}\), that length of the triangle is essentially constant, as a result, measuring \(p\) in arcseconds immediately gives \(d\) in parsecs (pc).

The star with the largest parallax angle is the closest star to the Sun – Proxima Centauri; that parallax angle is only \(0.77\text{ arcsec}\).

Examples

  1. How far is Proxima Centauri from our Sun, expressed in parsecs, light-years, and astronomical units?
    Since the parallax angle is \(p_\text{arcsec} = 0.77\text{ arcsec}\), Equation (15.3) tells us that \[d_\text{pc} = \frac{1}{0.77\text{ arcsec}} = 1.3\text{ pc}.\]
    Since \(1\text{ pc} = 3.26\text{ ly}\), the distance to Proxima Centauri in light-years is \(d = 1.3\text{ pc}\times3.26\text{ ly/pc} = 4.2\text{ ly}\), implying that light takes 4.2 years to reach us from Proxima Centauri (we see that star 4.2 years in the past).
    It is also the case that \(1\text{ pc} = \text{206,265 au}\). Therefore the distance to Proxima Centauri can also be expressed as \(d = 1.3\text{ pc} \times \text{206,265 au}/\text{pc} = \text{268,000 au}\). Proxima Centauri is \(268,000\) times farther from the Sun than Earth is.
  2. If Martian astronomers measured the parallax angle to Proxima Centauri from their planet, would their parallax angle be larger or smaller than the parallax angle measured from Earth?
    The distance of Mars from the Sun is \(1.52\text{ au}\) compared to Earth’s \(1\text{ au}\). The larger separation of Mars from the Sun means that the parallax angle would be larger as well. In fact, the parallax angle for any star as seen from Mars would \(1.52\) times larger than the parallax angle from Earth. In the case of Proxima Centauri, the parallax angle would be \(0.77\text{ arcsec}\times 1.52 = 1.17\text{ arcsec}\).
  3. Would it be easier to measure parallax angles from Mercury or from the orbit of Saturn? Why?
    Neglecting any environmental challenges, the baseline from Saturn is much greater than it is for Mercury (\(9.58\text{ au}\) compared to \(0.39\text{ au}\), or about \(17\) times larger). Consequently the parallax angle would by \(17\) times larger from Saturn compared to Mercury. The larger the baseline, the larger the parallax angle, and the easier it is to measure the angle.
  4. Astronomers on Earth measure the parallax angle of a newly-discovered star to be \(0.0037\text{ arcsec}\). How far is the star from our Sun in units of parsecs and light-years?
    \[d_\text{pc} = \frac{1}{p_\text{arcsec}} = \frac{1}{0.0037\text{ arcsec}} = 270\text{ pc}\].
    Since \(1\text{ pc} = 3.26\text{ ly}\), the distance is \(270\text{ pc}\times3.26\text{ ly}/\text{pc} = 880\text{ ly}\).

Quiz Yourself

Try the following exercises to make sure that you really do understand the material just presented. By the way, computers and many calculators write scientific notation in “computer-speak,” where \(10\) raised to an integer (either positive or negative) replaces \(\times 10^\text{exponent}\) with \({\rm E(exponent)}\) or \({\rm e(exponent)}\); in other words, \(3.14159 \times 10^9\) is represented by \(3.14159\text{E}9\). If your calculator has the \(\Large{\hat{ }}\) symbol, then you would enter the number as \(3.14159 \times 10{\Large{\hat{ }}}9\). If your calculator has a \(10^x\) button then you would enter the number as \(3.14159\times9\,\,10^x\). If you have having problems with entering numbers in scientific notation, read the manual or help screens, or ask your instructor.

(Answers to the problems are given below)

  1. The star Kruger 60 has a parallax angle of \(0.25\text{ arcsec}\). What is the distance to Kruger 60, as measured in parsecs and light-years?
  2. Ross 128 has a parallax angle of \(0.3\text{ arcsec}\). How far is Ross 128 from the Sun, in units of parsecs and light-years?
  3. The parallax angle of Antares is \(0.0054\text{ arcsec}\). Calculate its distance in parsecs and light-years.
  4. Sirius A, the brightest star in the night sky, has a parallax angle of \(0.38\text{ arcsec}\). What is one major reason that Sirius A is so bright?
Answers
  1. \(4\text{ pc} = 13\text{ ly}\)
  2. \(3.3\text{ pc} = 10.8\text{ ly}\)
  3. \(185\text{ pc} = 604\text{ly}\)
  4. The distance to Sirius A is only \(2.6\text{ pc}\) or \(8.6\text{ ly}\). As a result, Sirius A is one of the closest stars in the sky to Earth (there are only five other stars that are closer).
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