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Solar System dynamics and exoplanets

Learning objectives

  • Outcome

    Apply Kepler's third law T2=4π2GMa3T^2 = \dfrac{4\pi^2}{GM}a^3 to compute an orbital period from a semi-major axis and vice versa, and to estimate the mass of the central body.

  • Outcome

    Compute a planet's minimum mass msin⁡im\sin i from the radial-velocity amplitude KK and period PP using the mass function.

  • Outcome

    Determine a planet's radius from the transit depth δ=(Rp/R∗)2\delta = (R_p/R_*)^2, and from transit plus radial velocities together determine its mean density and infer the planet's type.

Understand the idea

Text

The motion of planets, their moons and exoplanets is governed by universal gravitation: F=GMmr2F = G\dfrac{Mm}{r^2}. Kepler's laws follow from this law, and for a circular orbit around a body of mass MM the period is related to the semi-major axis: T2=4π2GM a3.T^2 = \frac{4\pi^2}{GM}\,a^3. Note that the period depends on aa and MM but not on the planet's own mass, so the observed TT and aa yield the mass of the central body. Real systems are not ideal: mutual perturbations, resonances, tides and migration cause small deviations from Keplerian motion, and it is precisely these deviations that let us measure planetary masses in the Solar System. Exoplanets are found mainly by indirect methods: the Doppler shift of stellar lines (radial velocities), the dimming of light during a transit, and also microlensing and direct imaging. The radial-velocity method gives the projected mass msin⁡im\sin i, a transit gives the radius ratio δ=(Rp/R∗)2\delta = (R_p/R_*)^2, and the two together give the planet's density and hence clues to its composition. The methods are sensitive to different planets: hot giants are easier to find than small cold ones, so the known exoplanet sample is biased.

Worked examples

Direct

Problem

A planet moves in a circular orbit around a star of mass M=1 M⊙M = 1\,M_\odot with semi-major axis a=1.5a = 1.5 AU. Find the orbital period TT in years and in days.

Solution

Write Kepler's third law in the convenient form with aa in AU and mass in solar masses: T=a3M/M⊙ years.T = \sqrt{\frac{a^3}{M/M_\odot}}\ \text{years}. Substitute a=1.5a = 1.5 and M/M⊙=1M/M_\odot = 1: T=(1.5)3=3.375≈1.84 years.T = \sqrt{(1.5)^3} = \sqrt{3.375} \approx 1.84\ \text{years}. In days: 1.84×365≈6721.84 \times 365 \approx 672 days. Check: for the Earth a=1a = 1 AU and M=1 M⊙M = 1\,M_\odot give exactly T=1T = 1 year, so the form of the formula is correct.

Application

Problem

For a star of mass M=1 M⊙M = 1\,M_\odot, spectral lines shift periodically with amplitude K=12K = 12 m/s and period P=4P = 4 days. Estimate the planet's minimum mass msin⁡im\sin i.

Solution

Start from the mass function: m3sin⁡3i(M+m)2=PK32πG.\frac{m^3\sin^3 i}{(M+m)^2} = \frac{P K^3}{2\pi G}. Since m≪Mm \ll M, take M+m≈MM + m \approx M and obtain msin⁡i≈(PK3M22πG)1/3.m\sin i \approx \left(\frac{P K^3 M^2}{2\pi G}\right)^{1/3}. Convert the period: P=4×86400=3.456×105P = 4 \times 86400 = 3.456\times10^5 s. Evaluate the product in the numerator: K3=123=1728K^3 = 12^3 = 1728, so PK3=3.456×105×1728=5.97×108P K^3 = 3.456\times10^5 \times 1728 = 5.97\times10^8; the squared mass is M2=(1.989×1030)2=3.96×1060M^2 = (1.989\times10^{30})^2 = 3.96\times10^{60}; the product is 2.36×10692.36\times10^{69}. The denominator is 2πG=4.19×10−102\pi G = 4.19\times10^{-10}. The fraction equals 5.63×10785.63\times10^{78}, and the cube root gives msin⁡i≈1.8×1026 kg≈30 M⊕≈0.09 MJup.m\sin i \approx 1.8\times10^{26}\ \text{kg} \approx 30\,M_\oplus \approx 0.09\,M_\text{Jup}. This is a Neptune-type planet. The value is a lower bound: for an inclination i<90∘i < 90^\circ the true mass is larger.

Transfer

Problem

A planet transiting the disk of a star with radius R∗=1 R⊙R_* = 1\,R_\odot dims the brightness by δ=1%\delta = 1\%. Radial velocities at a nearly edge-on inclination (i≈90∘i \approx 90^\circ) give a mass m≈0.9 MJupm \approx 0.9\,M_\text{Jup}. Find the planet's radius and mean density, and decide whether it is rocky or gaseous.

Solution

Find the radius from the transit depth: δ=(RpR∗)2⇒RpR∗=0.01=0.1.\delta = \left(\frac{R_p}{R_*}\right)^2 \Rightarrow \frac{R_p}{R_*} = \sqrt{0.01} = 0.1. Hence Rp=0.1 R⊙=0.1×6.96×108 m=6.96×107R_p = 0.1\,R_\odot = 0.1 \times 6.96\times10^8\ \text{m} = 6.96\times10^7 m, about 1111 Earth radii — the size of a gas giant. Volume: V=43πRp3=43π(6.96×107)3≈1.41×1024 m3.V = \frac{4}{3}\pi R_p^3 = \frac{4}{3}\pi (6.96\times10^7)^3 \approx 1.41\times10^{24}\ \text{m}^3. Convert the mass: m=0.9×1.898×1027=1.71×1027m = 0.9 \times 1.898\times10^{27} = 1.71\times10^{27} kg. Density: ρ=mV=1.71×10271.41×1024≈1.2×103 kg/m3=1.2 g/cm3.\rho = \frac{m}{V} = \frac{1.71\times10^{27}}{1.41\times10^{24}} \approx 1.2\times10^3\ \text{kg/m}^3 = 1.2\ \text{g/cm}^3. This is close to Jupiter's density (1.331.33 g/cm$^3$) and far below the density of rock (33–55 g/cm$^3$), so the planet is gaseous, not rocky.

Practice

Text

A simulation instead of night-time observing. A real transit needs a telescope, clear weather and hours of time, so we work with a model or archival data.

  1. Open an online simulator of transits and radial-velocity curves, or take archival data for some exoplanet from a public database and plot it in a spreadsheet.
  2. Set a star of 1 M⊙1\,M_\odot and a planet on a circular orbit; record the resulting PP and aa.
  3. Using M=4π2a3GT2M = \dfrac{4\pi^2 a^3}{G T^2}, recover the stellar mass from the "observations" and compare it with the value you set.
  4. From the transit depth find Rp/R∗R_p/R_*, from the amplitude KK find msin⁡im\sin i, compute the density and decide whether the planet is rocky or gaseous.
  5. By varying the planet's radius and period, note at which parameters the signal falls below the noise level — this illustrates observational selection.

Safety. Never look at the Sun without a special filter; carry out any observation of bright objects only through a certified solar filter mounted in front of the aperture.

Visual reference

Purpose

Show how one and the same orbit appears in two independent observational signals: the periodic shift of stellar lines and the periodic dimming of the star's brightness.

Representation

A schematic of two stacked graphs sharing a common time axis. The upper one is the radial-velocity curve vr(t)=Ksin⁡(2πt/P)v_r(t) = K\sin(2\pi t/P) with the amplitude KK and period PP labelled. The lower one is the light curve with dips of depth δ=(Rp/R∗)2\delta = (R_p/R_*)^2 at transit times; below one dip a small sketch shows the star and the planet crossing its disk. The dips coincide in time with the zero crossings of the sine curve.

Real-world examples

GPS navigation in a car

Scenario

A car's GPS receiver uses signals from satellites orbiting at an altitude of about 20 200 km20\,200\ \text{km}. Each satellite completes one orbit in about 11 h 58 min11\ \text{h}\ 58\ \text{min}, moving at roughly 3.9 km/s3.9\ \text{km/s}. The receiver measures signal travel times to within a few nanoseconds to fix the car's position to about 5 m5\ \text{m}.

Connection

The satellite's orbital radius and period are linked by Kepler's third law, T2∝a3T^2 \propto a^3, which follows from Newton's law of gravitation. The same laws allow us to calculate the orbits of exoplanets from their observed transit periods. The travel time of the signal also uses the speed of light, tying directly to distance measurement in astronomy.

Tides at the coast

Scenario

At a coastal town, the difference between high and low tide is normally about 2.0 m2.0\ \text{m}. During a full moon, when the Sun and Moon pull in line, the tidal range can increase to about 3.0 m3.0\ \text{m} (spring tide). The Moon is 384 400 km384\,400\ \text{km} away; its tidal effect on Earth's oceans is about twice that of the Sun, even though the Sun is much more massive, because tidal force falls off as 1/r31/r^3.

Connection

This exercises Newton's law of gravitation, F=Gm1m2r2F = G \frac{m_1 m_2}{r^2}, and the concept of tidal forces, which arise from the difference in gravitational acceleration across a body. The same differential gravity causes a star to wobble due to an orbiting exoplanet, a signal detected by the radial-velocity method.

Detecting an exoplanet transit

Scenario

A 13-year-old uses a 100 mm100\ \text{mm} telescope and a smartphone camera to record the brightness of the star HD 209458 over 44 hours. The star's brightness dips by about 1.5%1.5\% for 33 hours when the planet HD 209458 b passes in front. The planet orbits every 3.53.5 days.

Connection

The dip depth gives the ratio of planet radius to star radius. The transit duration and orbital period, combined with Kepler's third law, yield the semi-major axis and orbital inclination. This is the transit method, one of the main ways exoplanets are discovered and characterized.