The size of everything
Tonight's idea
Nothing in space is the size you think it is, and nothing is as close as the pictures suggest. The best tool for seeing how big and how far is a ratio.
First, how far: measure distance in light-time. Light is the fastest thing there is, about 300,000 kilometres every second. Light from the Moon takes 1.3 seconds to reach us. Light from the Sun takes 8.3 minutes, so if the Sun vanished you would carry on seeing it for eight minutes. Light from the next nearest star, Proxima Centauri, takes 4.2 years. The light arriving from it now set off when you were about seven. A light-year is simply the distance light covers in a year, about \(9.46 \times 10^{12}\) km. It is a distance, not a time, even though it has "year" in the name.
Numbers that size mean nothing to a human brain, so astronomers shrink everything by the same factor and build a scale model. Suppose the Sun is a basketball, 24 cm across. The real Sun is \(1.39 \times 10^6\) km across, so our model shrinks everything by the ratio \(24 \text{ cm} : 1{,}390{,}000 \text{ km}\). Apply that same ratio to the Earth, which is 12,742 km across, and you get about 2 mm: a peppercorn. Apply it to the Earth–Sun distance of \(1.496 \times 10^8\) km and the peppercorn sits 26 metres from the basketball. Jupiter is a large marble 134 m away. Neptune is a small pea over three quarters of a kilometre off.
And Proxima Centauri? It is a small red star, about one seventh of the Sun's width, so in our model it is a ping-pong ball about 4 cm across, not another basketball. Keep the same ratio for the distance and it lands roughly 6,900 km away: put the Sun in New York and the nearest star is in Berlin. Space is mostly space.
Here is the deeper connection. In 1619 Johannes Kepler found that a planet's year \(T\) and its distance from the Sun \(a\) obey one rule: \(T^2 = a^3\), when \(T\) is measured in Earth years and \(a\) in Earth–Sun distances (an astronomical unit, AU). Far planets have long years, and the rule says exactly how long. You can check it tonight with a table and a pencil.
Watch one: Mars by Kepler's rule
Mars orbits at 1.524 AU. Kepler predicts \(T^2 = 1.524^3\). Compute: \(1.524^2 = 2.323\), and \(2.323 \times 1.524 = 3.540\). So \(T = \sqrt{3.540}\). Squeeze it: \(1.8^2 = 3.24\) is too small, \(1.9^2 = 3.61\) is too big, \(1.88^2 = 3.534\) is very close. Mars's year should be about 1.88 Earth years. This squeezing between two guesses is how you take a square root without a button; you will use it again in the stretch.
The measured value is 1.881 years. A rule written down four centuries ago, from naked-eye observations, predicts the Martian year to three decimal places.
Your turn
- How many kilometres are in a light-second? A light-minute? Use 300,000 km/s. Then check: the Sun is \(1.496 \times 10^8\) km away. How many light-minutes is that?
- In the basketball model, the Moon is 384,400 km from Earth and 3,474 km across. How far is the Moon from the peppercorn, and how big is it? (Every real kilometre becomes \(24 \div 1{,}390{,}000\) cm in the model, about 0.0000173 cm. Multiply each real distance by that.)
- Complete this table. Compute \(T^2\) and \(a^3\) for each planet and compare.
Planet \(T\) (years) \(a\) (AU) \(T^2\) \(a^3\) Mercury 0.241 0.387 Venus 0.615 0.723 Earth 1 1 Mars 1.881 1.524 3.538 3.540 Jupiter 11.86 5.203 Saturn 29.46 9.537 - A comet takes 8 years to orbit the Sun. Use Kepler's rule to find its average distance in AU. (Find a number whose cube is 64.)
Stretch: predicting and signalling
The asteroid belt sits about 2.8 AU from the Sun. Predict the length of a year for an asteroid there. You will need the square root of \(2.8^3\); squeeze it between two whole numbers first, then between two numbers with one decimal place.
Second problem. The Voyager 1 probe is more than 20 light-hours away (truly: it left in 1977 and has been coasting outward ever since). Engineers send it a command. How long before they could possibly hear back that it worked? If a mistake is in the command, how long is the earliest the fix could arrive?
Answers
- A light-second is 300,000 km; a light-minute is 18,000,000 km (\(1.8 \times 10^7\)). The Sun: \(1.496 \times 10^8 / 1.8 \times 10^7 \approx 8.3\) light-minutes.
- Moon distance: \(384{,}400 \times 24 / 1{,}390{,}000 \approx 6.6\) cm from the peppercorn. Moon size: \(3{,}474 \times 24 / 1{,}390{,}000 \approx 0.06\) cm, a grain of fine sand. The whole Earth–Moon system fits in your palm; the Sun is 26 m away.
- Mercury: 0.058 and 0.058. Venus: 0.378 and 0.378. Earth: 1 and 1. Jupiter: 140.7 and 140.9. Saturn: 867.9 and 867.4. They agree to within rounding every time.
- \(T^2 = 64 = a^3\), so \(a = 4\) AU.
Stretch. \(2.8^3 = 21.952\). \(4^2 = 16\) and \(5^2 = 25\), so \(T\) is between 4 and 5. \(4.6^2 = 21.16\), \(4.7^2 = 22.09\), so \(T \approx 4.7\) years (more precisely 4.69). The dwarf planet Ceres, at 2.77 AU, takes 4.6 years. For Voyager: the command takes 20 hours to arrive and the reply takes 20 hours to return, so 40 hours minimum before confirmation. A fix sent the moment the mistake is noticed also takes 20 hours to arrive, so engineers check such commands very carefully before pressing send.