Met Physics 2 Final HW questions

HW1: "Intensity is conserved". Verify this principle for the Sun's disk by deriving a formula for its average intensity as seen from a distance D from the Sun's center, given that its total radiant power output is P and its radius is Rs. For simplicity, assume that the Sun's intensity I is the same at all points on the visible disk (this is not strictly true). You will need to derive an exact expression for the solid angle subtended by the Sun's disk for arbitrary D > Rs. If done correctly, your final solution for I should not depend on D. (Hint: To find the solid angle subtended by the Sun, assume that it is directly overhead, so that the center coincides with the z-axis and the edge is defined by θ = θmax ((see Notes). Then integrate I cosθsinφ dθdφ with the appropriate limits.)

Answer F=P/4*pi*D^2 Set equation 2.58 equal to F (pi*I(Rs/D)^2) I=P/(2*pi*Rs)^2

HW2: If the intensity of radiation incident on a surface is uniform from all directions and denoted by the constant I, verify that the total flux is πI, as stated by (2.60). Note that this approximately describes the illumination of a horizontal surface under a heavily overcast sky. It also describes the relationship between the flux and intensity of radiation leaving a surface, if that surface is emitting radiation of uniform intensity in all directions.

Answer F=integral from 0 to 2pi and integral from 0 to pi/2 of I*sin(θ)cos(θ)dθdφ

HW3: Calculate the upward flux of reflected solar radiation at 12 μm, assuming an overhead sun (temperature 6000 K, solid angle Δω = 6.8 × 10^-5 sr) and a surface (flux) reflectivity rλ =0.1. Find the ratio of the above flux to that due to terrestrial emission at a temperature of 300 K.

Answer F_sun=B_λ*T_sun*Δω*rλ F_ter=ε*pi*B_λ*T_ter F_sun/F_ter

HW1: The broadband flux of solar radiation that reaches the top of the atmosphere is approximately 1370 Wm-2, when measured on a plane normal to the beam. Combine this with the solid angle you derived in a previous problem to compute the average radiant intensity of the sun's surface.

Answer I=F/Δω

HW3: Refer to the following figure. (on the answer side) Using radiative equilibriums at TOA and surface, derive the expression of Ts and Ta. Use globally averaged solar incoming flux as S. Assume surface emissivity ε=1.

Answer Look at the photo <3

HW3: Sometimes Planck's function B(T) may be expressed as a function of frequency ν or wavenumber instead of wavelength λ. Given that Bλ(T) dλ must equal Bν(T) dν when dν and dλ correspond to the same narrow interval of the spectrum, find the correct expression forBν as a function of ν only.

Answer Substitute in λ=c/ν for Bλ(T)=(2hc^2)/(λ^5*(e^(hc/kλT)-1))

HW2: The radius of sun is Rs and the distance from the sun to the Earth is Ds. When Ds varies, the radiation received at the Earth changes accordingly. Derive an expression for TOA solar flux, S, as a function of S0, Ds, and Ds. Show that ΔS/S0≈-2ΔDs/Ds . That is, a positive 1% change in Ds leads to a negative 2% change in S.

Answer Use ln on both sides of the equation S=pi*I(Rs/Ds)^2 to rewrite the equation, remember (pi*I*Rs) is a constant

HW4: At three different wavelengths λ1, λ2, and λ3, the profile of absorption coefficient due to a certain atmospheric constituent is given by βe(z) = k_n*ρ_0*e^[-z/H]where ρ_0 = 4 g/m3 is the density of the constituent at sea level, and H = 8 km is the scale height. The wavelength-dependent mass extinction coefficients k1, k2, and k3, respectively, are 0.05, 0.10, and 0.15 m2 kg-1. Find the altitudes z_n of the corresponding peaks of the absorption weighting functions W(z) for radiation incident at the top of the atmosphere with θ = 60°.

Answer τ(z)/μ=1 τ(z)=k_n*ρ_0*H*e^[-z/H] z_peak=H*ln(k_n*ρ_0*H/μ)

HW1: The moon is at a mean distance Dm = 3.84×10^5 km from the earth; the Sun is at a mean distance Ds = 1.496 × 10^8 km. The radius of the moon is Rm = 1.74 × 10^3 km; the radius of the sun is Rs = 6.96 × 10^5 km. (a) Compute the angular diameter (in degrees) subtended by the sun and the moon. (b) Compute the solid angle subtended by the sun and the moon. (c) Which appears larger from Earth, and by what percentage do the two solid angles differ? (d) If the above values were constant, would it be possible to explain the occurrence of total solar eclipses?

Answers A: 2*θ=2*arcsin(Rs/Ds) B: Δω=2pi(1-cosθ) C: Sun/Moon from B D: If the values are constant, we cannot explain total solar eclipses

HW1: (a) Only radiation with wavelengths smaller than 0.2424 μm is capable of dissociating molecular oxygen into atomic oxygen, according to the reaction: O2 + photon → O + O. Based on this information, how much energy is apparently required to break the molecular bond of a single molecule of O2? (b) A small light source emits 1W of radiation uniformly in all directions. The wavelength of the light is 0.5 μm. (b-a) How many photons per second are emitted by the light source? (b-b) If the light source were on the moon and were viewed by a telescope on Earth having a 20 cm diameter circular aperture, how many photons per second would the telescope collect? Ignore atmospheric attenuation. Assume a distance D = 3.84 × 10^5 km between the moon and the Earth.

Answers A: E=h*ν B-A: N=F*λ/h*c B-B: Nnew=Fnew*λ/h*c, Fnew=F/pi*r^2*4piD^2

HW2: The total radiation flux incident on a surface due to wavelengths between 0.3 μm and 1.0 μm is 200 W m-2. 1. (10 points) (a) What is the average spectral flux within this interval? Give your answer in units of W m-2 μm-1. (b) If the spectral flux is constant with wavelength, then what is the total flux contributed by wavelengths just between 0.4 μm and 0.5 μm? (c) What is the total flux (in Wm-2) contributed by radiation of exactly 0.5 μm wavelength?

Answers A: Equation 2.47 Spectral flux=total radiation flux incident/(λ2-λ1) B: Equation 2.47 Flux(0.4-0.5)=spectral flux*Δλ C: 0 because if the range=0 then power range=0

HW4: A certain cloud layer has geometric thickness H = 0.1 km and liquid water path L = 0.01 kg m-2. Taking Q_e≈2 and the solar zenith angle θ = 60°, compute the direct transmittance t_dir for (a) N = 100 cm-3 (characteristic of clean maritime environments), and (b) N = 1000 cm-3 (characteristic of continental environments). (Density of liquid water ρ_l=1 g/cm3)

Answers A: Equation 7.76, don't plug in N at first wait until the end B: Do the same with new N

HW1: (a) Within a certain material, an EM wave with λ= 1 μm is attenuated to 10% of its original intensity after propagating 10 cm. Determine the imaginary part of the index of refraction ni. (b) For red light (λ= 0.64 μm), ni in pure water is approximately 1.3×10-8; for blue light (λ= 0.48 μm), ni =1.0× 10^-9. The deep end of a typical home swimming pool is approximately 2.5 m deep. Compute the fraction of each wavelength that survives the two-way trip to the bottom of the pool and back, when illuminated (and viewed) from directly above. In light of your findings (and in view of the appearance of most swimming pools as seen from the air), comment on the common assumption that water is "colorless."

Answers A: F=F0*e^(-β_a*x), ni=(λ/4pix)*ln(F0/F) B: β_a=4pi*ni/λ, find fraction with e^(-2*β_a*2.5)

HW3: A satellite viewing a surface location under cloud-free conditions measures a 12 micron radiance of 6.2 W m-2μm-1sr-1. (a) Compute the brightness temperature T_B. (b) Compute the actual temperature, assuming that the atmosphere is completely transparent, and that the surface in question is known to have an emissivity of 0.9 at this wavelength. (c) Is the ratio of the brightness temperature to the actual temperature equal to the emissivity?

Answers A: T_B=(hc/k_B*λ)*(ln(1+2hc^2/λ^5*I))^-1 B: Use T equation from A except it is now λ^5*I/ε C: T_B/T

HW3: Repeat the previous problem, only for a wavelength of 1 cm and an intensity of 2.103 × 10^-10 W m-2μm-1sr-1. How does your answer to part (c) change?

Answers A: T_B=(hc/k_B*λ)*(ln(1+2hc^2/λ^5*I))^-1 B: Use T equation from A except it is now λ^5*I/ε C: T_B/T

HW3: Consider a spherical object near the top of the atmosphere (S0=1370 W/m2), with radius r = 10 cm and heat capacity C of 1 × 10^4 J/K. (a) Find an expression for the heating rate dT/dt as a function of T, when T ≠TE, TE is radiative equilibrium temperature. (b) Assume that after the temperature reaches equilibrium, the incident solar radiation is suddenly shut off - for example, the object passes into the shadow of a planet. Use your solution to the previous part to find the temperature T of the object as a function of time t. Ignore the effects of finite thermal conductivity. (c) From your solution to (b), determine the time that it takes for the object to cool to 100 K, 10 K, and 1 K.

Answers A: Use equation C*dT/dt=F_net to solve for dT/dt when F_net=F_sun-F_emi, F_sun=p*r^2*S_0 F_emi=4pi*r^2*σ*T^4 B: Set S_0=0, move all T's onto the left side, integrate from T_E to T on the left and 0 to t on the right C: Use equation from B and plug in T

HW2: Consider a cloud that, when viewed from a point on the surface, occupies the portion of the sky defined by π/4<θ<π/2 and 0<φ<π/8. (a) What is the solid angle subtended by the cloud? (b) What percentage of the sky is covered by this cloud?

Answers A: Δω=integral from 0 to pi/8 and integral from pi/4 to pi/2 of sin(θ) B: Δω/2pi

HW4: A particular plane parallel cloud has liquid water density ρw = 0.1 g m-3 and thickness DZ = 100 m. At a certain wavelength, the mass extinction coefficient of the cloud droplets is ke,w = 150 m^2/kg, and the single scatter albedo is ωw = 1.0. However, the air in which the droplets are suspended is itself absorbing at this wavelength, having volume absorption coefficient βa,v = 10 km-1 and ωv = 0. (a) Compute the combined βe, βa, βs, and ω for the mixture. (b) Compute the total optical thickness τ of the cloud layer. (c) If radiation with intensity Iλ,top is incident on the top of the cloud from a zenith angle θ = 60o, compute the directly transmitted intensity Iλ,bot.

Answers A: βe=ke*ρw+βa,v βa=βa,v βs=ke*ρw+ωw ω=βs/βe B: τ=βe*ΔZ C: Iλ,bot=Iλ,top*e^(-τ/μ)

HW2: Compute the flux from an overhead spherical sun, as seen from a planet in an orbit of radius D, given that the sun has radius Rs and a uniform intensity I. Make no assumptions about the size of D relative to Rs. Use two different methods for your calculation: (a) Method 1: Integrate the intensity over the solid angle subtended by the sun, with the usual cosine-weighting relative to the local vertical. (b) Method 2: Compute the flux density emerging from the surface of the sun, translate that into a total power emitted by the sun, and then distribute that power over the surface of a sphere of radius D.Do your two solutions agree?

Answers A: θ_o=arcsin(Rs/D), F1=2pi*I*integral from 0 to θ_0 of cos(θ)sin(θ)dθ B: Sun's total energy=pi*I*4pi*Rs^2 Set Sun's total energy equal to 4pi*D^2*F2 and solve for F2

HW2: Compute the daily average (W/24, using 𝑊 = ∫ from t sunrise to 𝑡𝑠𝑢𝑛𝑠𝑒𝑡 𝑆_𝑜*𝑐𝑜𝑠𝜃_𝑠(𝑡)𝑑𝑡) top-of-the-atmosphere insolation [W m-2] for the following two cases: (a) the North Pole at the time of the Northern Hemisphere summer solstice (sun shines 24 hours at 90-23=67degree zenith angle) (b) the equator at the time of the equinox [θs=π(t-12)/12, while sun shines from t=6 to 18]. Assume that the solar flux normal to the beam is a constant 1370 Wm−2, and note that the North Pole is inclined 23° toward the Sun at the time of the solstice.

Answers A: θ_s=90°-degrees inclined W=S_0*cos(θ_s)*24 from equation Insolation=W/24 B: Use the W equation and substitute in the θ_s from part B, solve for W and insolation

HW3: a. Regardless of whether an emitter is a perfect blackbody or not, one may define the color temperature of the emitter in terms of its wavelength of maximum emission, according to Wien's law. The wavelength of maximum emission from the sun is approximately 0.475 μm. Determine its color temperature. b. Given the following information, use the Stefan-Boltzmann relationship to compute the effective emitting temperature of the Sun: Solar constant at top of Earth's atmosphere S0 =1370 W m-2; mean radius of Earth's orbit 1.496 × 10^8 km; radius of the Sun's photosphere 6.96 × 10^5 km. How does this value compare with the color temperature of the Sun derived in the previous problem?

Answers A: λmax*T_c (T_c is 6099K) B: 4pi*r_sun^2*σ*T_sun^4 (T_sun equation in photo)

HW4: A ground-based radiometer operating at λ = 0.45 μm is used to measure the solar intensity Iλ (0). For a solar zenith angle θ = 30°, Il(0) = 1.74 × 10^7 W m-2μm-1sr-1. For θ = 60°, Iλ(0) = 1.14 × 10^7 W m-2 μm-1sr-1. From this information, determine the top-of-the-atmosphere solar intensitySλ and the atmospheric optical thickness τλ.

Answers Iλ(0)=Sλ*e^(-τ/μ), take the ln of both sides then plug in the I(0) and degree values, set equal and solve

HW1: The atmospheric boundary layer is that region near the surface that is "well-mixed" by mechanical and/or convective turbulence originating at the surface. Its thickness may range from a few meters to several kilometers. Heat added by conduction from the surface is typically distributed quickly throughout the boundary layer. At a certain location in the tropics, the sunrises at 06 Local Solar Time (LST), is directly overhead at noon, and sets at 18 LST (=6 PM). Assume that, during the twelve hours that the sun is up, the net flux of solar energy absorbed by dry vegetation and immediately transferred to the overlying air is F(t) = F0 cos[π(t-12)/12], where t is the time of day (in hours), and F0 = 500 Wm-2. (a) Ignoring other heating and cooling terms, compute the total solar energy (in J/m2) added to the boundary layer over one 24-hour period. (b) If the boundary layer depth ΔZ = 1 km, its average air density is ρ_a = 1 kg/m3, and the specific heat capacity at constant pressure is c_p = 1004 J/(kg K), compute the temperature increase ΔT implied by your answer to (a). (c) If, instead, the boundary layer depth started out at sunrise only 10m deep and remained at that depth throughout the day, what would be the corresponding change of temperature? Why is it much more likely that the boundary layer would deepen quickly after sunrise?

Answers: A: ΔQ=(24/pi)*F0 B: ΔT=ΔQ/m*c_p where m is ρ_a*Δz C: Change is temp=100x the original, this causes the boundary layer to rise

HW4: A cloud layer has a vertical profile of βe that is quadratic in altitude z between cloud base z_base and cloud top z_top, with maximum β_e,m at the midpoint of the cloud. At the base and top of the cloud, βe = 0. (a) Find the quadratic equation that describes βe(z) within the cloud layer. (b) Find an expression for the total optical path τ measured vertically through the cloud layer. (c) Typical values for the above parameters for solar radiation incident on a thin cloud layer might be z_base = 1.0 km, z_top = 1.2 km, and β_e,m = 0.015 m-1. Compute the total optical path for this case. (d) Based on your answer to (c), compute the vertical transmittance t through the cloud.

Answers: Look at photo!