HW 5 Q1 (Danny LeBrun)

All of it was done on paper so you'll have to look at these picture uploads:


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HW5 Q1 (William Matzko)

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1.  For radial velocities, we have the relation k = 28.4 (m/s) p^-(1/3)(M_p*sin(i)) (M_s)^(-2/3), where M_p is the planet mass in Jupiter masses, and M_s is the mass of the star in solar masses. For astrometry, we have  α  = (M_p/M_s)(a)(d)^(-1), where a is the semimajor axis in AU, and d is the distance to the system in parsecs. Just to make things easier to look at, we can make some simplifying assumptions. We are already told to assume sin(i) = 1 in all instances and M_s is the mass of the Sun. We might as well further assume that our target is one Jupiter mass. Further, we know P is proportional to a ^(3/2). The two above equations then reduce to α = 0.001 (a) (d)^-1 k = 28.4 (m/s) (a)^(-1/2) We are given that astrometry is sensitive to  α = 10 micro arcseconds and above, and radial velocities are sensitive to k = 1 m/s and above. Note that only the astrometry method depends on d. We must now find when the astrometry method becomes more sensitive than the rad...
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HW5 Q2, Q3, Q5 (William Matzko)

2. We know the phase amplitude is  ε = A*g*(R/r)^2, where A is the geometric albedo, R is the radius of the planet, r = a(1-e^2)/(1+ecos(f)), and g = 10^(-0.4*m). Here, m is just  m = 0.09( α/100) + 2.39( α/100)^2 - 0.65( α/100), where  α is in degrees (the 100's in the denominator are likewise in degrees). We are working with a Venus analog, so we can assume, in this part, that e = 0, so that r = a. The phase amplitude will be maximum when  α = 0. So, our maximum phase amplitude is then  ε = (0.67) (1)(6050km/108E6km)^2, where I have plugged in the geometric albedo of Venus, the radius of Venus, and Venus' semimajor axis. This yeilds  ε = 2E-9.  Now, we are to assume i = 40 degrees and w = 270 degrees. The effect of w is to orient the orbit such that we have a maximum phase angle at f = 0 . We know the relation between  α and these orbital elements: cos( α) = -sin(i)sin(w+f). Plugging in the numbers and solving for  α yields  α = 50 d...
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HW 5 Q6 (William Matzko)

6. (a)Usually, the temperature on a planet decreases with altitude. However, in inversion layers, this is not the case. In an inversion layer, temperature increases with altitude. Such a thing is important in the atmosphere of exoplanets because it can give insight into what that planet's atmosphere could be composed of. These inversion layers are usually composed of molecules that absorb UV radiation, so finding these inversion layers on exoplanets is desirable if we want to try and find life on them. (b) We know that P = P_0 exp(-z/H), where P is the pressure, P_0 is the pressure at the surface level, z is the altitude above surface level and H is the scale height. We are given that, on Earth z = 8.8 km and H = 8 km. This corresponds to a pressure of ~0.333*P_0. On a modified planet, to find the z at which we have the same pressure, we first must compute the scale height. We can express the scale height equation as H = k*T/(u*m*g), where k is the Boltzmann constant, T is the te...
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