Astronomical Scales – General Structure of the Solar System (Senior 4 Physics) | Ganzaa

General Structure of the Solar System

Lessons

Senior 4 · Physics

General Structure of the Solar System

4 lessons

Lesson 1 of 4

Astronomical Scales

25 min

A football in Amahoro Stadium

Imagine shrinking the Sun to the size of a football, 22 cm across, and placing it on the centre spot of a stadium. On the same scale the Earth is a grain of sand about 2 mm wide, 24 m away, about halfway from the centre spot to the goal; the Moon is a speck 6 cm from the Earth. Neptune, the farthest planet, is a small marble more than 700 m away, well outside the stadium. And the nearest star? Another football more than 6000 km away. Space is mostly empty, and its distances are so large that kilometres become clumsy. In this lesson you meet the units astronomers use, find how far light travels in them, and see how the distance to a star can be measured with a triangle.

Units for astronomical distances

Three units cover the scales of astronomy (Figure 1.1):

the astronomical unit (AU), the average distance from the Earth to the Sun, defined since 2012 as exactly $149\,597\,870\,700\,\mathrm{m} \approx 1.496 \times 10^{11}\,\mathrm{m}$ ; it is used inside the solar system;
the light-year (ly), the distance light travels in a vacuum in one year of 365.25 days: $1\,\mathrm{ly} = c \times t = 3.00 \times 10^{8} \times 3.156 \times 10^{7} = 9.46 \times 10^{15}\,\mathrm{m}$ ;
the parsec (pc), the distance at which 1 AU subtends an angle of one arcsecond (Figure 1.2): $1\,\mathrm{pc} = 3.09 \times 10^{16}\,\mathrm{m} = 3.26\,\mathrm{ly} = 206\,265\,\mathrm{AU}$ .

Because light has a finite speed, $c = 3.00 \times 10^{8}\,\mathrm{m\,s^{-1}}$ , looking at distant objects means looking into the past: we see the Sun as it was about 8 minutes ago, and Proxima Centauri as it was 4.25 years ago.

Figure 1.1: Distances in the solar system and beyond, on a logarithmic scale. Data: NASA.

Definition

The astronomical unit (AU) is the mean Earth–Sun distance, $1.496 \times 10^{11}\,\mathrm{m}$ . A light-year (ly) is the distance light travels in a vacuum in one year, $9.46 \times 10^{15}\,\mathrm{m}$ : a distance, not a time. A parsec (pc) is the distance at which the radius of the Earth's orbit (1 AU) subtends an angle of 1 arcsecond ( $1'' = \frac{1}{3600}$ of a degree); $1\,\mathrm{pc} = 3.09 \times 10^{16}\,\mathrm{m} = 3.26\,\mathrm{ly}$ . Example: Jupiter's mean distance from the Sun, $7.785 \times 10^{11}\,\mathrm{m}$ , is $\frac{7.785 \times 10^{11}}{1.496 \times 10^{11}} = 5.20\,\mathrm{AU}$ .

Worked example 1: how old is the light?

How long does light take to reach us from (a) the Moon, 384 400 km away, (b) the Sun, 1 AU away, (c) Neptune, $4.515 \times 10^{9}\,\mathrm{km}$ from the Sun?

t_{\text{Moon}} = \frac{3.844 \times 10^{8}}{3.00 \times 10^{8}} = 1.28\,\mathrm{s}, \qquad t_{\text{Sun}} = \frac{1.496 \times 10^{11}}{3.00 \times 10^{8}} = 499\,\mathrm{s} = 8.3\,\mathrm{min}

t_{\text{Neptune}} = \frac{4.515 \times 10^{12}}{3.00 \times 10^{8}} = 1.5 \times 10^{4}\,\mathrm{s} = 4.2\,\mathrm{h}

A radio command to a spacecraft at Neptune takes over four hours to arrive, so such spacecraft must be able to act on their own.

Angles in the sky: angular size

We cannot lay a ruler across the Moon, but we can measure the angle it covers in the sky, its angular size. For a body of diameter $D$ at a distance $d$ much larger than $D$ , the angle in radians is

\theta \approx \frac{D}{d} \qquad (\theta \text{ in radians; } 1^\circ = 60' = 3600'')

This "small-angle formula" is the arc-length rule $s = r\theta$ from Unit 7. Astronomers often work in arcminutes ( $^{'}$ ) and arcseconds ( $^{''}$ ), because most angles in the sky are tiny.

Worked example 2: why total solar eclipses happen

Use NASA's data to compare the angular sizes of the Sun (diameter $1.391 \times 10^{6}\,\mathrm{km}$ , distance $1.496 \times 10^{8}\,\mathrm{km}$ ) and the Moon (diameter 3475 km, distance 384 400 km).

\theta_{\text{Sun}} = \frac{1.391 \times 10^{6}}{1.496 \times 10^{8}} = 9.30 \times 10^{-3}\,\mathrm{rad} = 0.533^\circ, \qquad \theta_{\text{Moon}} = \frac{3475}{384\,400} = 9.04 \times 10^{-3}\,\mathrm{rad} = 0.518^\circ

The Sun is about 400 times wider than the Moon but also about 400 times farther away, so they look almost exactly the same size. That coincidence is why the Moon can just cover the Sun in a total eclipse (Lesson 2).

Measuring the distance to a star: parallax

Hold a finger at arm's length and look at it with one eye, then the other: it seems to jump against the background. That shift is parallax. As the Earth goes round the Sun, a nearby star seems to shift in the same way against much more distant stars. The parallax angle $p$ is half the total shift seen over six months: the angle at the star subtended by 1 AU (Figure 1.2). Since $p$ is tiny, $p = \frac{1\,\mathrm{AU}}{d}$ in radians, and with $p$ in arcseconds and $d$ in parsecs this becomes simply

d = \frac{1}{p} \qquad (d \text{ in pc}, \; p \text{ in arcseconds})

That is how the parsec got its name: the distance of a star with a parallax of one arcsecond. No star is that close; even the nearest has a parallax under 1″, which is why parallax was first measured only in 1838.

Figure 1.2: Parallax: a nearby star seen from opposite sides of the Earth's orbit. The parallax angle p is subtended at the star by 1 AU.

Worked example 3: the nearest star

Proxima Centauri is 4.25 light-years away (NASA). Express this distance in kilometres, in parsecs and in AU, and find its parallax.

d = 4.25 \times 9.46 \times 10^{15}\,\mathrm{m} = 4.02 \times 10^{16}\,\mathrm{m} = 4.02 \times 10^{13}\,\mathrm{km}, \qquad d = \frac{4.25}{3.26} = 1.30\,\mathrm{pc} = 2.7 \times 10^{5}\,\mathrm{AU}

p = \frac{1}{d} = \frac{1}{1.30} = 0.77''

Even the nearest star shows a shift of less than one arcsecond, about the angle a coin 2 cm wide covers when seen from 5 km away.

Worked example 4: building the scale model

In the opening scale model the Sun (diameter $1.391 \times 10^{6}\,\mathrm{km}$ ) is a football 22 cm across. Find the scale factor and the model sizes of the Earth and its distance from the Sun.

k = \frac{0.22\,\mathrm{m}}{1.391 \times 10^{9}\,\mathrm{m}} = 1.58 \times 10^{-10}

D_{\text{Earth}} = 1.2756 \times 10^{7} \times 1.58 \times 10^{-10} = 2.0 \times 10^{-3}\,\mathrm{m}, \qquad d_{\text{Earth}} = 1.496 \times 10^{11} \times 1.58 \times 10^{-10} = 24\,\mathrm{m}

On this scale Neptune is 714 m from the football, and Proxima Centauri 6400 km.

The Earth and the Moon photographed from Mars orbit by the HiRISE camera on NASA's Mars Reconnaissance Orbiter on 20 November 2016. Notice how small the Moon is beside the Earth (its diameter is about a quarter of Earth's); the Moon has been brightened in this composite so that it shows up. Image: NASA/JPL-Caltech/University of Arizona, public domain, via Wikimedia Commons.

Warning

Common mistakes. (1) Treating the light-year as a unit of time: it is a distance. (2) Using degrees in $\theta = \frac{D}{d}$ : the formula needs radians (multiply degrees by $\frac{\pi}{180}$ ). (3) Using $d = \frac{1}{p}$ with $p$ in degrees or radians: it works only for $p$ in arcseconds and $d$ in parsecs. (4) Forgetting that the parallax angle is subtended by 1 AU, half of the shift between opposite sides of the orbit.

Tip

Three rungs of the ladder. Inside the solar system use AU (8.3 light-minutes each); between the stars use light-years or parsecs (1 pc = 3.26 ly). To change units, divide distance by $c$ for light-time, and use $d = \frac{1}{p}$ for parallax.

To Scale: The Solar System (Wylie Overstreet and Alex Gorosh). A group of friends build a scale model of the solar system on a dry lake bed in Nevada, with the Earth the size of a marble. Watch how far apart the planets have to be, just like the football model in this lesson.

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Practice

Jupiter's mean distance from the Sun is $7.785 \times 10^{8}\,\mathrm{km}$ . Express it in AU, and find how long sunlight takes to reach Jupiter.
A star is 8.6 light-years away. Express its distance in parsecs and in kilometres.
A star has a parallax of 0.10″. Find its distance in parsecs and in light-years.
Voyager 1 took the Pale Blue Dot image from about $6 \times 10^{9}\,\mathrm{km}$ . Express this in AU, and find how long its radio signals took to reach Earth.
Jupiter (diameter 142 984 km) is 4.2 AU from the Earth at its closest. Find its angular diameter in arcseconds.
Starting from the definition of the parsec, show that 1 pc ≈ 206 265 AU.
In a scale model the Earth is a ball 1.0 cm across. How far from it should the Moon be, and how big and how far away should the Sun be?
A newspaper says: "The star is 50 light-years old." Evaluate this statement.

Show answersHide answers

$\frac{7.785 \times 10^{11}}{1.496 \times 10^{11}} = 5.20\,\mathrm{AU}$ ; $t = \frac{7.785 \times 10^{11}}{3.00 \times 10^{8}} = 2.6 \times 10^{3}\,\mathrm{s} = 43\,\mathrm{min}$ .
$\frac{8.6}{3.26} = 2.6\,\mathrm{pc}$ ; $8.6 \times 9.46 \times 10^{12}\,\mathrm{km} = 8.1 \times 10^{13}\,\mathrm{km}$ .
$d = \frac{1}{0.10} = 10\,\mathrm{pc} = 33\,\mathrm{ly}$ .
$\frac{6 \times 10^{12}}{1.496 \times 10^{11}} = 40\,\mathrm{AU}$ ; $t = \frac{6 \times 10^{12}}{3.0 \times 10^{8}} = 2 \times 10^{4}\,\mathrm{s} \approx 5.6\,\mathrm{h}$ .
$\theta = \frac{1.43 \times 10^{8}}{4.2 \times 1.496 \times 10^{11}} = 2.28 \times 10^{-4}\,\mathrm{rad} = 2.28 \times 10^{-4} \times 206\,265 = 47''$ .
At 1 pc, 1 AU subtends $1'' = \frac{1}{3600} \times \frac{\pi}{180} = 4.848 \times 10^{-6}\,\mathrm{rad}$ . So $d = \frac{1\,\mathrm{AU}}{4.848 \times 10^{-6}} = 206\,265\,\mathrm{AU}$ .
Scale: 1.0 cm for 12 756 km. Moon: $\frac{384\,400}{12\,756} = 30\,\mathrm{cm}$ away. Sun: $\frac{1.391 \times 10^{6}}{12\,756} = 109\,\mathrm{cm}$ across, $\frac{1.496 \times 10^{8}}{12\,756} = 1.2 \times 10^{4}\,\mathrm{cm}$ = 117 m away.
The statement confuses distance with age. A light-year is a distance; the star is 50 light-years away, so the light we see left it 50 years ago. The star itself is very likely millions or billions of years old.

Quick check

Key points

1 AU = $1.496 \times 10^{11}\,\mathrm{m}$ (Earth–Sun); 1 ly = $9.46 \times 10^{15}\,\mathrm{m}$ (a distance); 1 pc = $3.09 \times 10^{16}\,\mathrm{m} = 3.26\,\mathrm{ly} = 206\,265\,\mathrm{AU}$ .
Light-time $t = \frac{d}{c}$ : 1.3 s from the Moon, 8.3 min from the Sun, 4.25 years from Proxima Centauri.
Angular size $\theta \approx \frac{D}{d}$ (radians); the Sun and Moon both look about 0.5° across.
Parallax: $d\,(\mathrm{pc}) = \frac{1}{p\,('')}$ ; the nearest star has $p = {0.77}^{''}$ .

Sources and credits

Data: NASA, Planetary Fact Sheet and Sun Fact Sheet; NASA Science, Our Nearest Celestial Neighbor? An Exotic 3-Star System; NASA/JPL Photojournal, PIA23645: Pale Blue Dot Revisited.
Image: Earth and Its Moon, as Seen From Mars, NASA/JPL-Caltech/University of Arizona, Wikimedia Commons, public domain.
Image: Pale Blue Dot from Voyager 1, NASA/JPL-Caltech, Wikimedia Commons, public domain.
Video: To Scale: The Solar System, Wylie Overstreet and Alex Gorosh, embedded from YouTube.

Written by Ganzaa: the lesson text, derivations, diagrams, worked examples and practice exercises.

1A football in Amahoro Stadium

2Units for astronomical distances

3Angles in the sky: angular size

4Measuring the distance to a star: parallax

A football in Amahoro Stadium

Units for astronomical distances

Angles in the sky: angular size

Measuring the distance to a star: parallax