EuPhO 2017, Теория — Задача 2. Disk in gas

• $a_0 \approx \frac{S\rho k_B}{mM}\sqrt{T_0T_1},\qquad v_{\max} \approx v_0 = \sqrt{\frac{k_BT_0}{m}}$

  Order-of-magnitude estimates, correct up to a numerical coefficient of order one.

The initial pressure on the thermal insulating layer is $P_0 = nk_BT_0$, where $n$ is number density of the gas. It originates from multiplying the flux $j_0 \propto v_{x0}$ and momentum that one molecule transfers $p_0 = 2mv_{x0}$ (elastic collision), where $v_{x0}$ is the normal component of molecule’s velocity, and taking the average $(\overline{2v_{x0}^2} \propto T_0)$. When applying the same idea to the surface with good thermal contact, we find out that the flux remains the same, although the momentum increases:

where $v_{x1}$ is the normal velocity component of the molecule flying away from the disk. Thus for pressure $P_1$:

which is correct to some numerical coefficient of the order of one.

The net force acting on the disk:

and then the initial acceleration:

Since $P_1 \gg P_0$, the disk will accelerate until its speed becomes of the order of average gas molecules speed. After the velocity $v$ of the disc becomes on the order of $v_0 = \sqrt{kT_0/m}$, the flux of molecules reaching the backside $j(v)$ decays faster than exponentially due to the nature of the molecular velocity distribution in the ideal gas (for example, $j(2v_0) \approx 10^{-3}j_0$ and $j(3v_0) \approx 10^{-6}j_0$). That leads to a proportional decrease in a propelling pressure $P_1$. In order to compensate for an initial bias $\sqrt{T_1/T_0} \approx 30$, it will take around a factor of one on the velocity of the disk. Therefore the maximum velocity of the disk:

Here we assumed that the disk will not cool close to $T_0$ before it reaches the maximum velocity. Let us show it. The acceleration time is approximately:

Since the power of heat removal $P_{\mathrm{th}}$ is maximal at the beginning (at zero velocity), we can upper-bound estimate the time for the disk to cool as $t_c = Q/P_{\mathrm{th}}$, where $Q$ is the total heat of the disk. The initial thermal power of heat removal can be estimated as:

and the total heat $Q \approx Nk_BT_1$. Given $M \approx Nm$, we obtain:

Finally, $t_a/t_c \approx \sqrt{T_0/T_1} \ll 1$, and indeed disk will not cool significantly before it reaches the velocity about $v_0$.

Grading scheme. Indented lines show partial points for partially correct solutions

Initial acceleration (5 pts)

• $P_0 = nkT = j_0 \times \Delta p_0$, $\Delta p_0 = 2mv_{x0}$ … 2 pts

  • $P \propto kT$ … 1 pts
• $\Delta p_1 = mv_{x1}$, $P_1 \approx nk\sqrt{T_0T_1}$ … 2 pts

  • $P_1 = nkT_1$, no points for $a_0$ further … 1 pts
  • only $\Delta p_1 = mv_{x1}$ … 1 pts
• Answer for $a_0$ … 1 pts

  • if some slight mistake … 0.5 pts
• Special rule

  • if $\langle v_{x0}\rangle = \sqrt{3k_BT/m}$ (student doesn’t understand the difference between velocity of the molecule and the component of the velocity) … −0.5 pts

Maximal velocity (4 pts)

• $P_1$ and $P_0$ depend on the velocity of the disk, $P_1$ drops significantly if $v \approx v_0$, thus $v_{\max} \approx v_0$ … 4 pts

  • $P_0' \approx \rho v^2$, but $P_1$ stays the same, thus $v_{\max} \approx v_0\sqrt{T_1/T_0}$ … 2 pts
  • Student understands that at least some pressure depends on the velocity of the disk … 1 pts
  • The velocity is maximal when disk cools to $T_0$ 0 pts

Justification of slow cooling (1 pts)

• Estimation of times $t_a$ and $t_c$ given … 1 pts

  • Student explicitly writes that $T_1' \approx T_1$ but doesn’t prove it … 0.5 pts