Exoplanet Orbits

(Also see Part I.)

The star 51 Pegasi is a bit weird. When you plot its brightness over time, it seems to periodically become slightly more red and then slightly more blue. Why would it ever do that?

Well, one reason why a source of light might become redder (redshift) or bluer (blueshift) is due to the Doppler effect. In other words, when it gets redder, it's moving away from us, and when it gets bluer, it's approaching. In fact, we can directly turn redshift into radial velocity differences using pretty simple math, and obtain this diagram.

M_* = 1.06M_{\odot} \\
P = 4.23\ days

For reference:

M_{\odot} = 1.99 \times 10^{30}\ kg \\
M_J = 1.898 \times 10^{27}\ kg \\
G = 6.67 \times 10^{-11} Nm^2kg^{-2}

We would like to find the mass and semi-major axis of the planet 51 Pegasi b:

M_b = ? \\
a = ?

Kepler's third law relates the orbital period, semi-major axis and the gravitational parameter $\mu$:

P = \sqrt{ \frac{a^3 \times 4\pi^2}{\mu} } \\
\mu = G(M_b + M_*)

But $M_b$ is thought to be negligible, especially given the uncertainty in estimating stellar mass so it's common to short-hand $\mu$ as just:

\mu = G \times M_* = 1.06 \times 1.99 \times 10^{30} kg \times 6.67 \times 10^{-11} Nm^2kg^{-2} \\
\mu = 1.4067 \times 10^{20} Nm^2kg^{-1} = 1.4067 \times 10^{20} \times m^3s^{-2}

This lets us solve third law for the semi-major axis:

P = \sqrt{ \frac{a^3 \times 4\pi^2}{\mu} } \\
P^2 = \frac{a^3 \times 4\pi^2}{\mu} \\
\frac{P^2}{a^3} = \frac{4\pi^2}{\mu} \\
\frac{a^3}{P^2} = \frac{\mu}{4\pi^2} \\
a^3 = P^2 \times \frac{\mu}{4\pi^2} \\
a^3 = (4.23 \times 86400s)^2 \times \frac{1.4067 \times 10^{20} \times m^3s^{-2}}{4\pi^2} \\
a^3 = 1.336 \times 10^{11} s^2 \times 3.564 \times 10^{18}m^3s^{-2} \\
a^3 = 4.761 \times 10^{29} m^3 \\
a = \sqrt[3]{ 4.761 \times 10^{29} m^3 } \\
a = 7.81 \times 10^{9} m = 0.0522 AU \\
a = 0.052 AU\ (2\ s.f.)

We now know the values of $M_*$, $P$ and $a$ and want to calculate $M_b$. The radial velocity curve of 51 Pegasi shows a clean sinusoidal variation with semi-amplitude $K \approx 55\ ms^{-1}$. (Phase-folded radial velocity data from Mayor & Queloz (1995). Credit: The Planetary Society.)

We can use the radial velocity semi-amplitude to estimate the semi-major axis of the star's orbit around the barycenter, and then relate that value to the masses of both bodies.

Assuming negligible eccentricity ($e \approx 0.01$), the orbital velocity $|v|$ is approximately constant and equivalent to the observed wobble. This lets us relate it to the distance from barycenter $r$:

|v| = \frac{2\pi \times r}{P} \\
r = \frac{|v| \times P}{2\pi}

Since the orbit is nearly circular and the data is corrected for the star's systemic radial velocity, the semi-amplitude $K = 55\ ms^{-1}$ represents the maximum wobble velocity directly.

We obtain the distance $r$ of the star from the barycenter:

r = \frac{55 ms^{-1} \times 4.23 \times 86400s}{2\pi} \\
r = 3.199 \times 10^{6} m

And finally the mass of the planet $M_b$:

M_* \times r = M_b \times a \\
M_b = \frac{M_* \times r}{a} \\
M_b = \frac{1.06 \times 1.99 \times 10^{30} kg \times 3.199 \times 10^{6} m}{7.81 \times 10^{9} m} \\
M_b = 8.64 \times 10^{26} kg \\
M_b = \frac{8.64 \times 10^{26}}{1.898 \times 10^{27}} M_J \\
M_b = 0.46 M_J\ (2\ s.f.)

Additional orbital elements: eccentricity and inclination

The semi-major axis $a$ and the gravitational parameter $\mu$ don't fully characterize the orbit. There are two additional pairs of parameters we don't know:

1. The eccentricity $\varepsilon$ and the longitude of periapsis $\varpi$. These two values define how far the orbit deviates from the perfectly circular, and how the ellipse is oriented in the orbital plane.
2. The inclination $I$ and longitude of the ascending node $\Omega$, which define the orientation of the orbit's plane with respect to the plane which contains our line of sight.

The effects of orbital eccentricity would deform the radial velocity sine wave. For 51 Pegasi b, the curve is remarkably sinusoidal, confirming the near-zero eccentricity ($e \approx 0.01$). For more eccentric orbits, the correction can be derived as follows – the magnitude of orbital velocity $|v|$ is given by:

|v| = \sqrt{ \mu \times (\frac{2}{r} - \frac{1}{a}) }

The term $r$ is the linear distance from the barycenter. It can be obtained from eccentricity, semi-major axis and the current longitudinal position called true anomaly $\theta$. True anomaly can in turn be obtained from phase $t - t_0$, to which it's related through Kepler's equations:

\cos{\theta} = \frac{\cos{E} - \varepsilon}{1 - \varepsilon \cos{E}} \\
M = E - \varepsilon \sin{E}

Where $M$ is the mean anomaly, a fictitious angle which varies linearly with time. The above is probably a transcendental equation, so finding $\varepsilon$ must be done numerically. For example:

1. Guess a value of $\varepsilon$.
2. Plot a series of $|v|$ values based on the estimate.
3. If the deformation is too great, guess lower. If the deformation is not enough, guess higher.

I can think of no similar trickery to constrain $I$ and $\varpi$. We are only able to say $0 < \sin{I} < 1$. The lowest mass will be calculated if $\sin{I} = 1$ and the further it deviates from that value, the greater the mass must be to account for the observed $|v|$.

Conclusion

As we can't know the orbital inclination, the answers can only be stated as the lower bound:

M_b > 0.46\ M_J\ (2\ s.f.) \\
a_b > 0.052\ AU\ (2\ s.f.)

So this is a gas giant with about half the mass of Jupiter, orbiting seven times closer to its star than Mercury does to the Sun. This is called a hot jupiter and some people got a Nobel Prize for proving such a planet can exist.