IOM 2016, Теория — Задача 3. «Planet Cronus»

• 1) 353.6 Earth years

• 2) $\omega \approx (1{,}07 \pm 0{,}02)\cdot 10^{-8}\ \mathrm{s^{-1}}, \quad \frac{a}{g} \approx (4{,}56 \pm 0{,}05)\cdot 10^{-11}$

  The answer table prints the unit of ω as «c⁻¹» (Cyrillic «с» for seconds).

• 3) $P_{Sun} = \sigma T_1^4\left(\frac{r_1}{R_K}\right)^2\pi r^2$

• 4) $\frac{P_{Sun}}{P_K} = \left(\frac{T_1}{T_0}\right)^4\left(\frac{r_1}{2R}\right)^2 \approx (2{,}4 \pm 0{,}2)\cdot 10^{-4}$

• 5) $\Delta\tau_{rad} \approx \frac{\rho c r\lvert\Delta T\rvert}{3\sigma T_0^4}$ ≈ $24$ Cronus years

• 6) $T(t) \approx T_0\cdot 2^{-t/(4\tau_{1/2})}$

• 7) $\Delta\tau \approx \frac{4}{\ln 2}\frac{\lvert\Delta T\rvert}{T_0}\tau_{1/2}$ ≈ $350$ Cronus years

• 8) $\frac{dp_{\mathrm{v}}}{dT} \approx \frac{\kappa\mu}{RT^2}p_{\mathrm{v}}$

• 9) $p_{\mathrm{v}}(T) \approx p_{\mathrm{v}}(T_0)\cdot\exp\left[\frac{\kappa\mu}{R}\left(\frac{1}{T_0} - \frac{1}{T}\right)\right]$

• 10) $T_1 = \frac{T_0}{1 - \frac{RT_0}{\kappa\mu}\ln\varphi_0}$ ≈ $325.3$ K

• 11) $\tau_1 = \frac{4\tau_{1/2}}{\ln 2}\ln\left[1 - \frac{RT_0}{\kappa\mu}\ln\varphi_0\right] \approx -\frac{4RT_0\tau_{1/2}}{\kappa\mu\ln 2}\ln\varphi_0$ ≈ $415$ Cronus years

• 12) $h_{\max} = \frac{\varphi_0 p_{\mathrm{v}}(T_0)}{\rho_l g}$

• 13) $\frac{h_l}{h_{\max}} = 1 - \frac{1}{\varphi_0}\exp\left[\frac{\kappa\mu}{R}\left(\frac{1}{T_0} - \frac{1}{T}\right)\right] = 1 - \exp\left[\frac{\kappa\mu}{R}\left(\frac{1}{T_1} - \frac{1}{T}\right)\right]$

• 14) $\tau_2 = \frac{4\tau_{1/2}}{\ln 2}\ln\left[1 + \frac{RT_0}{\kappa\mu}\ln\left(\frac{4}{3\varphi_0}\right)\right] \approx \frac{4RT_0}{\kappa\mu\ln 2}\ln\left(\frac{4}{3\varphi_0}\right)\tau_{1/2}$ ≈ $950$ Cronus years

• 15) No, no.

TABLE OF ANSWERS:

*An allowed deviation $\Delta$ is indicated for any numerical answer: 1 point is awarded if a proposed numerical value falls within the range. If the deviation exceeds $\Delta$ and is less than $2\Delta$, the score is down to 0,5 points. No point is awarded for a numerical answer if the deviation is more than $2\Delta$.

Proposed Solution:

1. According to the third Kepler’s law one obtains that Cronus «year» is related to Earth year as $\dfrac{\tau_K}{\tau_E} = 50^{3/2} \approx 353{,}6$, i.e. $\tau_K \approx 353{,}6$ Earth years or $\tau_K \approx 1{,}115\cdot 10^{10}$ sec.

2. The duration of a «solar day» is due both to planet rotation around its axis and to its orbital motion around the Sun. That is, the period of Cronus rotation around its axis with respect to «distant stars» («stellar day») equals 1/19 «year», so $\omega = 19\dfrac{2\pi}{\tau_K} \approx 1{,}071\cdot 10^{-8}$ sec$^{-1}$. The absolute value of free fall acceleration near the Cronus surface $g = \dfrac{GM}{r^2} = \dfrac{4\pi}{3}G\rho r$ and the absolute value of centripetal acceleration at the equator $a = \omega^2 r$, therefore $\dfrac{a}{g} = \dfrac{3\omega^2}{4\pi G\rho} \approx 4{,}56\cdot 10^{-11}$. This means that the planet rotation has almost no influence on dynamical processes near the surface, in particular, one can consider the planet atmosphere to be spherically symmetric.

Remark: the accuracy of numerical values for $\omega$ and $a/g$ in the table of answers is chosen to be 2% in order to distinguish the correct quantities from incorrect but numerically close ones which follow if one does not take into account the difference between the period of planet rotation and the solar day, i.e. when using $\omega = 20\dfrac{2\pi}{\tau_K}$ (the ensuing discrepancy is about 5%).

3, 4) If the radioactive «heater» is switched off, the only heating the planet receives is due to thermal radiation of the Sun, the heating power is $P_{Sun} = \sigma T_1^4\left(\dfrac{r_1}{R_K}\right)^2\pi r^2$; the heat loss is due to thermal radiation by the planet to outer space, its power is $P_K = \sigma T_0^4\cdot 4\pi r^2$ (the Sun and Cronus are treated as black bodies, so the Stephan-Boltzmann law applies). The ratio of these two quantities is $\dfrac{P_{Sun}}{P_K} = \left(\dfrac{T_1}{T_0}\right)^4\left(\dfrac{r_1}{2R_K}\right)^2 \approx 2{,}4\cdot 10^{-4}$.

5. Planet temperature decreases by a small amount $\Delta T = -4$ K during a time $\Delta\tau_{rad}$ determined from the equation:

(Here $C$ is the planet heat capacity). The second term in square brackets was found to be $\sim 10^{-4}$ and is negligible compared to 1. This term estimates the contribution of solar radiation to the planet thermal balance. Thus solar radiation has almost no influence on temperature dynamics around $T_0$ and can be neglected. Notice that this is another argument in favour of the assumption of spherical shape of the atmosphere (thermal radiation of the Sun violates this symmetry). Since $C \approx c\cdot M = \dfrac{4}{3}\pi\rho c r^3$ one obtains:

This time significantly exceeds the time of relaxation to «quasiequilibrium» altitude profile of the atmosphere and is significantly less that the half-life of radioactive «fuel» of the heating layer.

6, 7) The above results show that the problem of planet cooling can be considered to be spherically symmetric and the cooling process to be «quasiequilibrium» (i.e. it is reasonable to assume that at any time $t$ a heat coming from the planet heating layer to the surface is approximately balanced by the heat $P(t) \approx \sigma T^4(t)$ radiated by the surface to outer space). On the other hand $P(t)$ is directly proportional to the number of radioactive decays per second, so $P(t) = P(0)\cdot 2^{-t/\tau_{1/2}}$. Therefore $T(t) \approx T_0\cdot 2^{-t/(4\tau_{1/2})}$. Then the temperature drops by $\Delta T = -4$ K for a time determined from the equation $\dfrac{\lvert\Delta T\rvert}{T_0} \approx 1 - 2^{-\Delta\tau/4\tau_{1/2}} \Rightarrow \Delta\tau \approx \dfrac{4}{\ln 2}\dfrac{\lvert\Delta T\rvert}{T_0}\tau_{1/2} \approx 350$ Cronus years. Obviously the inequality $\Delta\tau_{rad} \ll \Delta\tau$ holds, which validates the assumption that thermal radiation «tunes» the planet temperature in accord with the power of radioactive decay.

Remark: If a contester has not justified spherical symmetry of the problem and quasistationary nature of the process (by comparing $\Delta\tau_{rad}$ and $\Delta\tau$) the maximum score cannot be awarded for items 6 and 7 even if the contester has obtained the correct dependence $T(t)$!

8, 9) The cycle considered is the Carnot cycle with the heater temperature $T + dT$ and the cooler temperature $T$. The cycle efficiency $\eta = \dfrac{dT}{T}$. The work done per cycle equals its area, i.e. $A = dp_{\mathrm{v}}(V_v - V_l) \approx dp_{\mathrm{v}}V_v$ (here $dp_{\mathrm{v}}$ is the change of pressure of saturated vapour when temperature increases from $T$ to $T + dT$ and a volume of liquid water $V_l$ is negligible compared to a volume of vapour $V_v$). The heat taken from the heater is the condensation heat of vapour $Q = \kappa m$, so for the cycle efficiency one can write:

(here the ideal gas law $\rho = \dfrac{\mu p_{\mathrm{v}}}{RT}$ is used for the vapour). By integrating this equation one obtains: $p_{\mathrm{v}}(T) \approx p_{\mathrm{v}}(T_0)\cdot\exp\left[\dfrac{\kappa\mu}{R}\left(\dfrac{1}{T_0} - \dfrac{1}{T}\right)\right]$.

Remark: Actually this part of the solution is a derivation of the Clausius–Clapeyron relation. If a contester refers to this relation right away, neglects specific volume of water, and uses the ideal gas law to evaluate specific volume of vapour, item 8) should be awarded 2 points out of 3, while item 9) should be awarded the maximum score.

10. It is important to note that since the total mass of water above the planet surface remains constant, the pressure exerted on the planet surface by water does not vary either: $p_0 = \dfrac{m_{\mathrm{H_2O}}g}{4\pi r^2}$. Before condensation it completely due to the pressure of vapour in a surface layer of atmosphere. According to the problem $p_0 = \varphi_0 p_{\mathrm{v}}(T_0)$. The condensation at the Cronus surface begins when this pressure equates the pressure of saturated vapour at a new temperature $T_1$ such that

whence $T_1 = \dfrac{T_0}{1 - \dfrac{RT_0}{\kappa\mu}\ln\varphi_0} \approx (325{,}3 \pm 0{,}4)$ K. Notice that before the onset of condensation the planet cooled by less than 5 K, so the above calculation of the time required to reduce the temperature by $\Delta T = -4$ K is quite realistic.

11. According to the above law of temperature variation, the condensation begins at

12. After the condensation begins the pressure $p_0 = \dfrac{m_{\mathrm{H_2O}}g}{4\pi r^2}$ is a sum of the pressure of liquid water layer of a thickness $h_l$ and the pressure of water vapour above its surface which is equal to the pressure of saturated vapour at a current temperature, i.e.

($\rho_l$ is the density of liquid water). The maximum ocean depth corresponds is reached when all available water has condensed, i.e. $h_{\max} = \dfrac{m_{\mathrm{H_2O}}}{4\pi r^2\rho_l} = \dfrac{\varphi_0 p_{\mathrm{v}}(T_0)}{\rho_l g} \Rightarrow \rho_l g = \dfrac{\varphi_0 p_{\mathrm{v}}(T_0)}{h_{\max}}$.

13. Therefore ocean depth at a temperature $T < T_1$ is determined from the equation $\varphi_0\left(1 - \dfrac{h_l}{h_{\max}}\right) = \exp\left[\dfrac{\kappa\mu}{R}\left(\dfrac{1}{T_0} - \dfrac{1}{T}\right)\right]$, whence

14. Ocean depth becomes equal to a quarter of the maximum at $T_2 = \dfrac{T_0}{1 + \dfrac{RT_0}{\kappa\mu}\ln\left(\dfrac{4}{3\varphi_0}\right)}$.

Therefore

15. Condensation of vapour releases heat. This process, which takes place evenly over the planet surface, is quite «fast», i.e. the heat release «follows» the condensation rate which, in turn, is determined by the rate of planet cooling. Of course, the release of vaporisation heat slows down the planet cooling to some extent and this influence must be analysed. Using the known pressure of saturated water vapour at $T_3 \equiv 373$ K one estimates $p_{\mathrm{v}}(T_0) \approx p_{\mathrm{v}}(T_3)\cdot\exp\left[\dfrac{\kappa\mu}{R}\left(\dfrac{1}{T_3} - \dfrac{1}{T_0}\right)\right] \approx 17$ kPa, so the total vaporisation heat released is $Q_c = \kappa m_{\mathrm{H_2O}} = \dfrac{\kappa\varphi_0 p_{\mathrm{v}}(T_0)\cdot 4\pi r^2}{g} = \dfrac{3\kappa\varphi_0 p_{\mathrm{v}}(T_0)r}{G\rho} \approx 8\cdot 10^{23}$ J. This heat is released for hundreds of Cronus years on the surface of a planet which material is a good heat conductor. Taking into account the heat capacity of the planet $C \approx \dfrac{4\pi}{3}\rho c r^3 \approx 1{,}4\cdot 10^{27}$ J/K one can see that its warming due to this process is very small, $\delta T_c \approx 5\cdot 10^{-4}$ K, even if the total amount of the released heat is absorbed. Obviously the planet cooling and the ensuing condensation of vapour is mostly due to reduction of radioactive heating. The temperature decreases approximately according to exponential law. Therefore the rates of cooling and condensation slow down almost at the same pace. Since the condensation lasts no less than cooling before it starts ($\tau_2 > 2\tau_1$), and a temperature change for the considered processes is about $\Delta T \approx 5$ K a decrease in the rate of cooling because of the release of vaporisation heat does not exceed $\dfrac{\delta T_c}{\Delta T} \approx 10^{-4}$. Therefore the influence of condensation on the cooling is negligible both at an accuracy of 5% and 0,5%.

Remark: Let us recapitulate: the essential part of the analysis of thermal dynamics of Cronus atmosphere is based on understanding of existence of several time scales:

• a scale of weak «fast» effects (like inhomogeneity of solar radiation and release of vaporisation heat) which is about one «day» or less; such effects have almost no influence on the dynamics because they are small and because they get averaged over a time scale of the order of Cronus year and more.
• a scale of «middle rate» relaxation processes (like restoration of the equilibrium temperature profile and the surface temperature) lasting from a Cronus year to decades;
• a scale of «slow» variation of external factors (variation of the heating power) taking hundreds and thousands Cronus years.