Astrophysics PS2

Author

Will St. John + Kiki Murphy

Problem 1

Consider the extrasolar planet system known as TRAPPIST-1 (the system with the most Earth-size habitable zone planets!). Refer to https://arxiv.org/abs/1703.01424 for relevant orbital parameters. Rank the orbital angular momentum of each planet from highest to lowest. Before you do any calculations take a stab at the ranking using your intuition and the derived equation. Compare your guess to the final ranking. Any surprises?

Answer: For our intuition approach, looking at the angular momention equation

\[L = \mu\sqrt{GMa(1-e^2)}\]

allows us to make some simplifications given our system’s properties. For starters, the system mass \(M\) is approximately the same for all orbits due to the mass of the star in comparison to the exoplanets. Additionally, the constant \(G\) will not change for each system. If we consider the reduced mass \(mu\), we can make the following simplification for a system with a very large (\(m_1\)) and very small mass (\(m_2\)).

\[\mu = \frac{m_1 m_2}{m_1 + m_2} \approx \frac{m_1 m_2}{m_1} = m_2\]

Thus, the primary characteristics for determining the angular momentum, by intuition, is the mass of the exoplanet, the semi-major axis, and the eccentricity.

$$\(L \sim m_2 \sqrt{a(1-e^2)}\)

Using the values given in the paper, we predict the following.

  1. g (largest L)
  2. c
  3. b
  4. f
  5. e
  6. d (smallest L)

Checking our values with a calculation…

import astropy.units as u
import astropy.constants as c

star_m = 0.0802 * c.M_sun

def L(m, a, e):
    return ((star_m * m)/(star_m + m)) * (c.G * (star_m + m) * a * (1 - e**2)) ** 0.5

# m, a, e
trappist_b = [0.85 * c.M_earth, 11.11 * u.au * 10 ** -3, 0.081]
trappist_c = [1.38 * c.M_earth, 15.21 * u.au * 10 ** -3, 0.083]
trappist_d = [0.41 * c.M_earth, 21.44 * u.au * 10 ** -3, 0.070]
trappist_e = [0.62 * c.M_earth, 28.17 * u.au * 10 ** -3, 0.085]
trappist_f = [0.68 * c.M_earth, 37.1 * u.au * 10 ** -3, 0.063]
trappist_g = [1.34 * c.M_earth, 45.1 * u.au * 10 ** -3, 0.061]

lb = L(trappist_b[0], trappist_b[1], trappist_b[2]).to(u.kg * u.m ** 2 / u.s)
lc = L(trappist_c[0], trappist_c[1], trappist_c[2]).to(u.kg * u.m ** 2 / u.s)
ld = L(trappist_d[0], trappist_d[1], trappist_d[2]).to(u.kg * u.m ** 2 / u.s)
le = L(trappist_e[0], trappist_e[1], trappist_e[2]).to(u.kg * u.m ** 2 / u.s)
lf = L(trappist_f[0], trappist_f[1], trappist_f[2]).to(u.kg * u.m ** 2 / u.s)
lg = L(trappist_g[0], trappist_g[1], trappist_g[2]).to(u.kg * u.m ** 2 / u.s)

print(f"L trappist b: {lb:.2e}")
print(f"L trappist c: {lc:.2e}")
print(f"L trappist d: {ld:.2e}")
print(f"L trappist e: {le:.2e}")
print(f"L trappist f: {lf:.2e}")
print(f"L trappist g: {lg:.2e}")
L trappist b: 6.73e+38 m2 kg / s
L trappist c: 1.28e+39 m2 kg / s
L trappist d: 4.51e+38 m2 kg / s
L trappist e: 7.81e+38 m2 kg / s
L trappist f: 9.85e+38 m2 kg / s
L trappist g: 2.14e+39 m2 kg / s

we get agreement with the largest and smallest values,

  1. g (largest L)
  2. c
  3. f
  4. e
  5. b
  6. d (smallest L)

which is relatively expected for a simple intution guess with no calculations.

Problem 2

Comet C/2013 A1 (Siding Spring) was discovered in January 2013. Refer to https://arxiv.org/abs/2012.12172 for relevant orbital parameters. Calculate the following five orbital properties: orbital period, perihelion distance in AU, aphelion distance in AU, perihelion velocity in km/s, and aphelion velocity in km/s. You may ignore uncertainties.

Answer: The equations for orbital period, aphelion, perihelion, aphelion velocity, and perihelion velocity are shown below, respectively.

\[P^2 = \frac{4\pi^2}{G(m_1 + m_2)}a^3\] \[\text{perihelion} = a - ae\] \[\text{aphelion} = a + ae\] \[v_p^2 = \frac{GM}{a}\left(\frac{1+e}{1-e}\right)\] \[v_a^2 = \frac{GM}{a}\left(\frac{1-e}{1+e}\right)\]

Using the equations above, and the values for \(a\), and \(e\) in the paper provided (in addition to some constants), we can solve for the five orbital properties listed earlier.

import math
a = 2.590376582621090E4 * u.au
e = 0.9999458751571103

P = ((4 * math.pi ** 2) * a ** 3 / (c.G * c.M_sun)) ** 0.5
perihelion = a - a * e
aphelion = a + a * e
v_p = (c.G * c.M_sun * ((1 + e)/(1 - e)) / a) ** 0.5
v_a = (c.G * c.M_sun * ((1 - e)/(1 + e)) / a) ** 0.5

print(f"Orbital Period: {P.to(u.yr):.2e}")
print(f"Perihelion: {perihelion.to(u.au):.2e}")
print(f"Aphelion: {aphelion.to(u.au):.2e}")
print(f"Perihelion Velocity: {v_p.to(u.km / u.s):.2e}")
print(f"Aphelion Velocity: {v_a.to(u.km / u.s):.2e}")
Orbital Period: 4.17e+06 yr
Perihelion: 1.40e+00 AU
Aphelion: 5.18e+04 AU
Perihelion Velocity: 3.56e+01 km / s
Aphelion Velocity: 9.63e-04 km / s

Problem 3

A planet with a mass 33.3 times larger than that of the Earth orbits a 1.7 Solar mass star with an orbital period of 8.88 years. What are the semimajor axis of the orbit (in AU) and the reduced mass (in Solar masses)?

Answer: Using the equation for orbital period (see above), we can rearrange to solve for the semimajor axis.

\[a = \left(\frac{P^2 G(m_1 + m_2)}{4\pi^2}\right)^{1/3}\]

Additionally, we can solve for the reduced mass using the equation outlined in Problem 1. Using the equations above the values provided, we can write a simple script to solve for the semimajor and reduced mass

m1 = 33.3 * c.M_earth
m2 = 1.7 * c.M_sun
P = 8.88 * u.yr

a = (P ** 2 * c.G * (m1 + m2) / (4 * math.pi ** 2)) ** (1/3)
mu = ((m1 * m2)/(m1 + m2))

print(f"Semimajor axis: {a.to(u.au):.4e}")
print(f"Reduced mass: {mu.to(u.M_sun):.4e}")
Semimajor axis: 5.1179e+00 AU
Reduced mass: 1.0001e-04 solMass