Math Detours · Grade 10 · Connections · 2

Kepler's laws and measuring the universe

T² ∝ a³, parallax, standard candles, and a cosmic distance ladder built out of logarithms.

Name ____________________Date ____________

Tonight's idea

Every distance in astronomy is inferred, and the inferences stack: each rung of the ladder is calibrated by the one below.

Rung 1: the Solar System, with Kepler. In 1619 Kepler found that for every planet the square of the orbital period \(T\) is proportional to the cube of the orbit's size \(a\). Here \(a\), the semi-major axis, is half the long diameter of the ellipse: nearest and farthest distances from the Sun add to \(2a\). In years and astronomical units (1 AU is the Earth–Sun distance) the constant is 1:

\(T^2 = a^3\). Mars: \(a = 1.524\) AU, \(T = 1.881\) yr. \(1.524^3 = 3.540\), \(1.881^2 = 3.538\).

Newton later showed why (tonight's stretch). The law turns a period, easy to measure, into a distance. The AU itself comes from radar bounced off Venus: 149.6 million km.

Rung 2: nearby stars, with parallax. Seen from opposite ends of Earth's orbit, six months apart, a nearby star shifts against the far ones. Half that shift is its parallax \(p\). A star with \(p = 1\) arcsecond (an arcsecond is 1/3600 of a degree) is 1 parsec away, 3.26 light-years, and in general \(d = 1/p\) parsecs. Proxima Centauri has \(p = 0.768''\), so \(d = 1.30\) pc \(= 4.25\) light-years.

Sun Earth, January Earth, July nearby star p d = 1/p
Half the star's apparent shift over six months is \(p\).

Rung 3: everything else, with standard candles. A source of luminosity \(L\) at distance \(d\) has brightness \(b = L/(4\pi d^2)\). If you know \(L\) (the object is a known type, calibrated on nearby examples by parallax), then measuring \(b\) gives \(d\). Cepheid variables and type Ia supernovae are the famous candles.

Watch one: Jupiter, Ceres, and a twin star

Jupiter: \(a = 5.203\) AU, \(T = 11.86\) yr. \(5.203^3 = 140.85\) and \(11.86^2 = 140.66\). Ceres orbits at \(a = 2.77\) AU, so predict \(T = a^{3/2} = 2.77^{1.5} = 4.61\) yr; observed, 4.60 yr.

Two stars are identical twins. Star A is 10 pc away by parallax; star B appears 25 times fainter. Brightness falls as \(1/d^2\), so B is \(\sqrt{25} = 5\) times farther: 50 pc.

Your turn

  1. Check Kepler's third law for Venus (\(a = 0.723\) AU, \(T = 0.615\) yr) and Saturn (\(a = 9.54\) AU, \(T = 29.46\) yr). Give \(a^3\) and \(T^2\) to four significant figures.
  2. An asteroid orbits the Sun every 8 years. How far from the Sun is it, on average? Halley's comet has period 75.3 years and comes within 0.586 AU of the Sun. Find its semi-major axis \(a\), then its farthest distance from the Sun. Which planet does that take it beyond?
  3. A star has parallax \(0.1''\). How far away is it in parsecs and in light-years? Barnard's Star has \(p = 0.547''\): its distance?
  4. A type Ia supernova at 20 million light-years calibrates the candle. Another is seen 10,000 times fainter. How far away is it? And one seen a million times fainter?

Stretch: where Kepler's law comes from, and weighing the Sun

For a circular orbit gravity is the centripetal force (the inward force that keeps something on a circle): \(\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}\), with \(v = 2\pi r/T\). Show that \(T^2 = \dfrac{4\pi^2}{GM} r^3\). Then weigh the Sun from Earth's orbit: \(r = 1.496 \times 10^{11}\) m, \(T = 3.156 \times 10^{7}\) s, \(G = 6.674 \times 10^{-11}\) in SI units.

Answers

  1. Venus: \(a^3 = 0.3779\), \(T^2 = 0.3782\). Saturn: \(a^3 = 868.3\), \(T^2 = 867.9\). Both agree to better than one part in a thousand.
  2. \(a = T^{2/3} = 8^{2/3} = 4\) AU exactly. Halley: \(a = 75.3^{2/3} = 17.83\) AU; farthest distance \(2 \times 17.83 - 0.586 = 35.1\) AU, beyond Neptune (30.1 AU).
  3. \(p = 0.1''\): 10 pc \(= 32.6\) ly. Barnard's Star: \(1/0.547 = 1.83\) pc \(= 5.96\) ly.
  4. 10,000 times fainter means \(\sqrt{10000} = 100\) times farther: 2 billion light-years. A million times fainter: 1,000 times farther, 20 billion light-years.

Stretch. Cancel \(m\) and one \(r\): \(\frac{GM}{r} = v^2 = \frac{4\pi^2 r^2}{T^2}\), so \(T^2 = \frac{4\pi^2}{GM} r^3\): \(T^2 \propto r^3\) with a constant depending only on the central mass. Then \(M = \dfrac{4\pi^2 r^3}{G T^2} = \dfrac{4\pi^2 (1.496\times10^{11})^3}{6.674\times10^{-11} \times (3.156\times10^7)^2} = 1.99 \times 10^{30}\) kg. The same trick on Io's orbit weighs Jupiter, which is how planets with moons were first weighed.

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