The cross-section area receiving the solar radiation (falling as parallel rays) is $\pi R_E^2$, so taking into account the absorbed portion is a fraction $1-a$ the total incident radiation, we find

A black body radiates according to the Stefan-Boltzmann law, $P_{bd} = \sigma A T^4$, where $\sigma$ is the Stefan-Boltzmann constant and $A$ is the total surface area of the black body. At steady state $P_{bd} = P_0$, so
$$\sigma (4 \pi R_E^2) T_{g0}^4 = (1-a) \pi R_E^2 F_s.$$It leads to the answer

In the presence of the atmospheric layer, we write down the energy transfer balance in two regions: between the Earth’s surface and the atmosphere, and between the atmosphere and outer space. Let the power radiated from Earth be $P_E$ and the power radiated from each side of the atmosphere be $P_{atmo}$, then
$$P_E = P_{atmo} + t_{sw} P_0 \\ t_{lw} P_E + P_{atmo} = P_0$$Solving this system of equations and using $P_E = \sigma (4\pi R_E^2)T_g$ we find

Let the natural (unstretched) length of the spring be $l_0$ and let $x_A$, $x_B$ be the positions of particles $A$ and $B$, respectively. The equation of motion of each particle due to the spring force can be written as:
$$
\frac{d^2}{dt^2}x_A = +\frac{k}{m_A}(\ell - \ell_0),
\\
\frac{d^2}{dt^2}x_B = -\frac{k}{m_B}(\ell - \ell_0),
$$where $\ell = x_B - x_A$ is the instantaneous length of the spring. Taking the difference of the two equations gives
$$
\frac{d^2}{dt^2}\ell = -k\left(\frac{1}{m_A} + \frac{1}{m_B}\right)(\ell - \ell_0).
$$This is the equation of motion of a single effective particle attached to a spring with a spring constant $k$ and an effective mass or reduced mass $\mu$, given by:
$$
\mu = \frac{1}{\frac{1}{m_A} + \frac{1}{m_B}} = \frac{m_A m_B}{m_A + m_B}.
$$Thus, the system undergoes a simple harmonic motion with an angular frequency

The difference in energy between two consecutive levels in a quantum harmonic oscillator is given by $\hbar \omega$. So the energy of the photon is given by

The observed shift in the spectral line from $f_0$ is due to the Doppler effect. When the source is moving towards the observer with velocity $v$ the frequency is shifted according to
$$
f = f_0 \left( 1 + v/c \right).
$$Thus, the shift in frequency is given by

To find the normalization constant $C$, we require that the total probability is equal to 1. This leads to:

$$
\int_{-\infty}^{\infty} p(v) \, dv = 1 \quad \Rightarrow \quad C \int_{-\infty}^{\infty} e^{-\frac{mv^2}{2k_B T}} \, dv = 1.
$$Using the integration formula provided, with $x = v$ and $a = \frac{m}{2k_B T}$ we obtain

Using the result of B4, we obtain the following expression for the speed of a molecule in terms of the frequencies $f$ and $f_0$:
$$
v = \frac{f - f_0}{f_0} c.
$$We plug this back into the probability distribution formula to obtain the answer. This gives the probability distribution for observing a molecule whose spectral line is Doppler shifted from $f_0$ to $f$.

The probability distribution $p(f)$ follows a Gaussian profile in the frequency shift $f - f_0$.

Ответ:

The center of the profile is $0$ and it drops to $1/e$ of its maximum value when the argument of the exponential is $-1$. This happens when the following condition is fulfilled

Consider a thin horizontal layer of thickness $dz$ and surface area $S$. Since the air is in hydrostatic equilibrium, its weight must be balanced by the difference in pressure forces. This results in the following relation:
$$
p(z)S = p(z + dz)S + \rho(z)gS\,dz.
$$Simplifying and rearranging terms gives the answer. The negative sign indicates a decrease in pressure with height as expected.

Assuming we can treat air as an ideal gas, we can use the ideal gas law to express the density of air in terms of its pressure and temperature:
$$
pV = nRT \quad \Rightarrow \quad p(z)V = \frac{m}{\mu_{\text{air}}} RT(z).
$$Rewriting this in terms of the density gives:
$$
\rho(z) = \frac{p(z)\mu_{\text{air}}}{RT(z)}.
$$Now we substitute the density expression into the expression obtained in C1. This gives

Assuming an isothermal atmosphere (i.e., constant temperature with altitude), $T(z) = T$, the equation simplifies to:
$$
\frac{dp}{p} = -\frac{\mu_{\text{air}}}{RT} g\,dz.
$$Integrating both sides and assuming the pressure at height $0$ is $p_0$ leads to:
$$
\ln \left[ \frac{p(z)}{p_0} \right] = -\frac{\mu_{\text{air}}}{RT} gz.
$$Then the answer is obtained.

Since the small mass of air is displaced adiabatically, it must satisfy the adiabatic condition for an ideal gas:
$$
pV^{\gamma} = \text{const.},
$$where $\gamma = c_p/c_V$ is the adiabatic index, and $c_p$, $c_V$ are the molar specific heats at constant pressure and volume respectively. Writing the volume in terms of temperature and pressure using the ideal gas law gives:
$$
p(T/p)^{\gamma} = \text{const.} \quad \Rightarrow \quad p^{1-\gamma}T^{\gamma} = \text{const.}
$$Taking the derivative of this expression with respect to the height $z$, we obtain:
$$
(1 - \gamma)p^{-\gamma}\frac{dp}{dz}T^{\gamma} + \gamma p^{1-\gamma}T^{\gamma-1}\frac{dT}{dz} = 0.
$$Simplifying and rearranging to have an expression for the adiabatic lapse rate gives:
$$
\frac{dT}{dz} = -\frac{1 - \gamma}{\gamma}\frac{T(z)}{p(z)}\frac{dp}{dz}.
$$We now substitute the hydrostatic pressure gradient obtained in C.3 to get:
$$
\frac{dT}{dz} = -\frac{1 - \gamma}{\gamma}\frac{T(z)}{p(z)} \left[ -\frac{p(z)\mu_{\text{air}}}{RT(z)} g \right] = \frac{1 - \gamma}{\gamma} \frac{\mu_{\text{air}}}{R} g.
$$Using $\gamma = c_p/c_V$ and $c_p - c_V = R$ we obtain a linear drop of temperature with height in an adiabatic atmosphere:
$$
\frac{dT}{dz} = \frac{1 - c_p/c_V}{c_p/c_V} \frac{\mu_{\text{air}}}{R} g = -\frac{\mu_{\text{air}}}{c_p} g.
$$Since $\Gamma_a = -\frac{d T}{dz}$, we get

To find the angular frequency of small oscillations of the air parcel, we begin by applying Newton's second law, where the primary forces acting on the parcel are buoyancy and gravity:
$$
\delta m \frac{d^2 z}{dt^2} = \rho_a(z) g \delta V - \delta m g,
$$where $\delta m$ is the mass of the air parcel, $\delta V$ is its volume and $\rho_a$ is the density of the surrounding air. We can express the mass of the parcel in terms of its density $\rho_p$ as $\delta m = \rho_p \delta V$. Substituting and simplifying gives:
$$
\frac{d^2 z}{dt^2} = \frac{\rho_a(z + \delta z) - \rho_p(z + \delta z)}{\rho_p(z + \delta z)} g.
$$Assuming the parcel is at the same pressure as the atmosphere at $z + \delta z$, the density can be expressed in terms of temperature using the ideal gas law $\rho \propto 1/T$. This allows us to rewrite the last expression as:
$$
\frac{d^2 z}{dt^2} = \frac{T_p(z + \delta z) - T_a(z + \delta z)}{T_a(z + \delta z)} g.
$$We can now express the temperature at $z + \delta z$ in terms of the lapse rates and the temperature at $z$ using the definition $T(z + \delta z) = T(z) + \Gamma \delta z$. Therefore:
$$
\frac{d^2 z}{dt^2} = \frac{T(z) + \Gamma \delta z - T(z) - \Gamma_a \delta z}{T(z) + \Gamma_a \delta z} g.
$$Simplifying the numerator and neglecting the infinitesimal term $\Gamma_a \delta z$ in the denominator gives:
$$
\frac{d^2 z}{dt^2} = \frac{\Gamma - \Gamma_a}{T} g \delta z.
$$This is the equation of a simple harmonic oscillator, where the angular frequency is given by

Thus, the stability condition is

The change of entropy across a phase transition (evaporation in this case) is related to the latent heat of evaporation. If there was a mass $m$ of liquid water, then $Q_{\text{evaporation}} = Lm$, then
$$
\Delta S = \frac{Lm}{T}.
$$It is known that the volume of vapor is significantly larger than the volume of liquid of the same mass, therefore $\Delta V \approx V_{\text{vapor}}$, which can be found using the ideal gas law:
$$
V_{\text{vapor}} = \frac{nRT}{p_s(T)}.
$$The mass can be related to the number of moles $n$ via $m = \mu_{H_2O}n$, then

We can integrate the relationship found in D.1 by separating variables to find
$$
\ln \left[ \frac{p_s(T)}{p_{so}} \right] = -\frac{\mu_{H_2O}L}{R} \left( \frac{1}{T} - \frac{1}{T_o} \right).
$$Note that $L$ is strictly a function of temperature, but we are assuming that $L$ is a constant for the range of temperatures we investigate. Rearranging, we find

Formation of liquid water happens when the partial pressure of water inside the parcel reaches the saturation pressure at a given temperature. The partial pressure of water vapor $p_w$ can be related to the total pressure of the parcel $p$ as
$$
p_w = \frac{n_{\text{H}_2\text{O}}}{n_{\text{air}}} p = \frac{m_{\text{H}_2\text{O}}/\mu_{\text{H}_2\text{O}}}{m_{air}/\mu_{air}} p = \phi \frac{\mu_{air}}{\mu_{\text{H}_2\text{O}}} p.
$$Given that the air parcel is rising adiabatically, $p^{1-\gamma} T^\gamma = \text{const.}$, so
$$
p(T) = p_i \left( \frac{T}{T_i} \right)^{c_p / R}.
$$Therefore, the transcendental equation that we need to solve is
$$
\phi \frac{\mu_{air}}{\mu_{\text{H}_2\text{O}}} p_i \left(\frac{T_l}{T_i}\right)^\frac{c_p}{R} = p_{so} \exp \left[-\frac{\mu_{\text{H}_2\text{O}}L}{R} \left(\frac{1}{T_l} - \frac{1}{T_i}\right)\right]
$$This can be rearranged to get
$$
T_l = \frac{1}{\frac{1}{T_i} - \frac{R}{\mu_{\text{H}_2\text{O}} L} \ln \left[\phi \frac{\mu_{air}}{\mu_{\text{H}_2\text{O}}}\frac{p_i}{p_{so}} \left(\frac{T_l}{T_i}\right)^{c_p/R}\right]}.
$$Substituting the numerical values, we get
$$
T_l = \frac{1000\,\text{K}}{3.481 - 0.4695 \ln \left( \frac{T_l}{290.15\,\text{K}} \right)}.
$$Solving this iteratively, we find

Using the notations of the figure, the total angle of deviation $\delta$ can be written as the sum of the deviations in the two refractions:
$$\delta = \alpha - \alpha' + \beta - \beta'.$$Consider the triangle of interior angles $\varphi$, $90^\circ - \alpha'$ and $90^\circ - \beta'$. Since the sum of these angles add up to $180^\circ$, we get
$$\varphi = \alpha' + \beta',$$so $\delta$ simplifies to
$$\delta = \alpha + \beta - \varphi.$$

The relationship between $\alpha$ and $\alpha'$ (and similarly between $\beta$ and $\beta'$) is given by Snell's law:
$$
\frac{\sin \alpha}{\sin \alpha'} = n, \quad \frac{\sin \beta}{\sin \beta'} = n.
$$Expressing $\beta$ in terms of $\alpha'$:
$$
\sin \beta = n \sin \beta' = n \sin (\varphi - \alpha'),
$$From Snell's law $\alpha'$ can be written as
$$
\alpha' = \arcsin \left( \frac{\sin \alpha}{n} \right).
$$Thus, $\beta$ in terms of $\alpha$ is given by
$$
\beta = \arcsin \left\{ n \sin \left[ \varphi - \arcsin \left( \frac{\sin \alpha}{n} \right) \right] \right\}.
$$Finally, we get the result for $\delta$:

The minimum value of $\delta$ is around $21.8^\circ$, so that is the angle with respect to the direction of Sun where the halo appears.