An introduction to exoplanets
An introduction to exoplanets

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An introduction to exoplanets

4.3  The radial velocity see-saw gives mass

If the star is bright enough and the planet is massive enough, which is indeed the case for many of the transiting planets discovered from the ground, then astronomers can use the radial velocity (RV) method to measure the mass of the planet. You learned about the RV method in Week 3. In a nutshell: as the planet orbits the star, the star wobbles, creating a Doppler shift in the light from the star. This produces tiny changes in the wavelengths of the fingerprints in the spectrum of the star. These tiny changes are measurable and can be used to work out the mass of the planet.

With both the radius and the mass of a planet we can work out a really useful quantity – its density. This gives us a good idea what the planet is actually made of, as you learned in Week 2. You’ll look at this idea in more detail later this week.

Activity 2  How the radial velocity reveals the planet mass

Timing: Allow about 15 minutes

This interactive application is set up to simulate the radial velocity curve for one of the hottest of the known hot Jupiters, a planet called WASP-12 b. Its orbital period is just 26 hours! The star WASP-12 has a mass of 1.35 MSun, and this value for the star’s mass doesn’t change. The planet transits and so the inclination is close to 90°.

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Interactive feature not available in single page view (see it in standard view).

WASP-12 b completes an orbit in just over a day – 1.09 days to be precise, measured from the time between the transit dips in the light curve.

1. What is the value of 1.09 days expressed in units of years?

Answer

You need to work out what fraction of a year is equal to 1.09 days. There are 365.25 days in a year – the 0.25 comes from the leap day inserted every four years. So the number required is 1.09 ÷ 365.25 = 0.003 rounded to 3 decimal places.

If you know the mass of the star and the orbital period, you can work out the distance between the star and the planet. This comes from Kepler’s Third Law, which was mentioned in Week 3. It is probably the most important equation in astronomy, and the interactive application you are using is calculating its consequences.

Using the slider, adjust the value of aplanet until the orbital period is 0.003 Earth years. You can also use the arrow keys on the keyboard to amend the values.

2. What value of aplanet is needed to make the orbital period equal to 0.003 Earth years?

Answer

The application gives a small range of possible values centred on about 0.025 AU.

The green graph shows how the measured radial velocity for WASP-12 would behave if you made a continuous series of measurements. As you saw in Week 3, for an orbital inclination of exactly 90° you see the star move exactly towards you and exactly away from you once per orbit. By comparing the maximum and minimum observed values of the radial velocity, you can work out how fast the star is moving because of the pull of the orbiting planet.

The star moves faster if the planet has more mass, as you will see if you move the slider to adjust the planet mass. Precise measurements of the star’s radial velocity give you precise measurements of the mass of the orbiting exoplanet. This is true even though you can’t see the planet itself.

Observations of WASP-12 show it has an orbital speed of 226 m/s, measured from the radial velocity curve which is derived from the size of the wavelength changes in the spectra. Use the slider to adjust the value of Mplanet until the stellar orbital speed is 226 m/s.

3. What value of Mplanet is needed to make the stellar orbital speed 226 m/s?

Answer

Mplanet = 1.46 MJ

(Your answer may be slightly different from this – you should get this value when aplanet= 0.025 AU.)

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