Homework 2

1)
As the planet mass is ~13% that of the star, it matters considerably.
velocity:

r is a fraction of the semi-major axis inversely proportional to relative mass:

Plugging in the numbers, V is 98.75 m/s. A speed of almost 100 m/s seems quite feasible.


2)
The planet orbiting VB10 was discovered by looking for periodicities in astrometric data. It was refuted by the suggested parameters (and a large portion of the physically plausible parameters) being ruled out by improvements in the radial velocity technique. The RV refutation was helped by the astrometric orbital parameters indicating a near edge-on inclination.

3)
a.
For the astrometric method, planet mass (as a fraction of stellar mass) shows an inverse relationship with semi-major axis. For the radial velocity method, it increases with the square root of stellar mass and semi-major axis.

b.
The cross-over depends strongly on stellar mass and distance, ranging from ~1.5e11 m (1 AU) for the plausible near/low mass case to ~3e12 m (20 AU) for a more distant and likely overly high mass star. (The upper left portion of the second graph should probably be considered unphysical)

4)
 Starting with the Planck equation, a spherical star, and a spherical grain in thermal equilibrium (like the in-class derivation, but with variable opacity):

Evaluating the integral in Wolfram Alpha yields:

Substituting:

(The equation given has 2 instead of sqrt(2), but I think it was described as a typesetting error?)

5)
https://github.com/pdn4kd/freezing-tyrion/blob/master/HW2-5.py

Homework 2

Link to assigment 2

Assignment 2

Link to Assignment 2

HW1

1)
a.
Estimating the maximum angular separation of Earth and Jupiter from the Sun at a distance of 10     parsecs: Earth's Aphelion is 1.01673 AU, Jupiter's is 5.45492 AU.1 AU at 1 pc is 1", so scaling proportionally, these are (to excessive significant figures) 0.101673" and 0.545492" from the Sun. Earth is questionable with the nominal HST resolution of 0.05" (the star and planet would be adjacent), but Jupiter (some 20 pixels away) could be distinguished.
b.
Using the planet-to-star flux ratio approximation R_p²/a²,

Planet	Radius (m)	Semi-Major Axis (m)	Flux Ratio	Magnitudes Fainter
Earth	6.371e6	1.49598023e11	1.814e-9	21.9
Jupiter	6.9911e7	7.78299e11	8.069e-9	20.2

2)
The problems of defining a planet at the high mass end are how they become increasingly star-like. Newly formed brown dwarfs are notable for having effective temperatures (few thousand kelvin), compositions (hydrogen-rich), and densities very close to still-forming red dwarfs. Possible solutions include looking at the spectra for deuterium, helium-3, and lithium abundances (higher in brown dwarfs than stars, at least once the stars have had time to burn those elements/isotopes).

3)
a.
Using Kepler's 3rd law in its full Newtonian glory:
T² == 4*pi²/(G*(M1+M2))*a³, or T == 2*pi*a^1.5/sqrt(G(M1+M2)).
For the stars (a == 1200 AU, M1 == M2 == 0.9 M_sol), T == 9.78e11 s, or 31000 (Julian) years.
b.
For the planet, T == 1.004e7 s, or 0.318 (Julian years), so Tstar/Tplanet ~= 97400 orbits

4)
Using Kepler's 3rd law in simplified form:
T² == a³,  a == T^2/3 == (120/365.25)^2/3 == 0.47613 AU
Periastron == a(1-e)
e == 1 - Periastron/a == 0.36992 ~= 0.37

5)

 I would not expect the decreasing density vague power law guess (mass ~ radius^2.5) to hold for higher masses (topping out at perhaps 2x Jupiter's radius even with >10x Jupiter's mass), given predictions for brown dwarf sizes and the relatively small radii of low mass red dwarfs.

Homework 1

https://drive.google.com/drive/folders/0B23b_IF-Qh6TMmxzaGpub1ZTQXc

Homework 1

1a) If the distance (d) to the Sun from Earth is ~ 1 AU and the distance to the Sun from Jupiter is ~ 5.3 AU, then the maximum angular separation of both planets from a distance (D) of 10 parsecs is given by the equation,

θ = d/D.

Plugging in the correct values for Earth and Jupiter we get,

θ_Earth= 4.84813681 × 10^-7 radians = 0.1” (arcsecs)

and

θ_Jupiter= 2.56951251 × 10^-6 radians = 0.53” (arcsecs)

Considering the stated resolution for the Hubble Space Telescope is 0.05” (arcsecs), we can resolve the separation of either planet from the Sun at a distance of 10 parsecs away.

1b) If the radius (r) of Earth is 6.371 x 10^6 meters and the radius of Jupiter is 69.911 x 10^6 meters, then the planet-to-star flux ratio is approximately given by

flux ratio ≈ r^2 / d^2.

So plugging in the correct values for Earth and Jupiter we get,

Earth-Sun flux ratio ≈ 1.81774025 × 10^-9

and

Jupiter-Sun flux ratio ≈ 2.28383156 × 10^-7

2) Problems defining the planets on the high-mass end is, in part, due to the over-lap of high-mass planets and low-mass brown dwarfs. Low-end properties of mass and temperature for brown dwarfs have been observed to be ~13 Jupiter masses and ~300K-750K. Also, the high-end properties of mass and temperature for exoplanets have been observed to be ~28 Jupiter masses and ~7500K. From this, it is obvious that there is over-lap between the characteristics of brown dwarfs and exoplanets. Direct spectrometry of the objects could help clarify the line between these two. However, this is difficult considering, from 1a, the resolving power of the HST is just small enough to resolve a planet at 1 AU from its host star. Even though the HST has the angular resolution to resolve this planet perhaps, with a flux ratio (from 1b) as small as ~10^-9, the planet's reflected light would still be washed out by the light from its host star.

3a) The orbital period of a binary star system is given by

So plugging in the correct values for HD 80606 and HD 80607 and solving for T we get,

T ≈ 30,984 years

3b) The orbital period for a planet-star system is given by the same formula as above where M1 >> M2. So plugging in the correct values for HD 80606 and HD 80606b we get,

T ≈ 0.318 years

Meaning that HD 80606b makes ~97,433 orbits for every one binary orbit of
HD 80606 and HD 80607.

4) If the orbital period is known then we can solve for the length of the semi-major axis using,

Plugging in the orbital period of 120 days and using a M1 = 1 solar mass >> M2 we get,

a ≈ 0.48 AU

Then, knowing that the closest approach of the planet is 0.3 AU we can solve for the eccentricity using

Radius of periastron = a(1-e)

Plugging in the values for the radius of periastron and the semi-major axis we get that

e ≈ 0.375.

5)

Assuming that the relationship between mass and radius is a power law given by

m^n = r ; n is any real number

Then the average of n for all the planets is

n ≈ 0.162.

Therefore the power law relationship between mass and radius is

m^0.162 ≈ r.

Planets larger than Jupiter would follow this relationship until their densities become so high that electron degeneracy pressures would overcome the classical concepts of pressure and temperature as they relate to ideal gases.