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.