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\section{ABEP plasma investigation}

So let's make some calculations for the ABEP plasms in VLEO conditions.
First, for calculating the the time necesary for the EED function to be formed.
The momentum relaxation time can be estimated as the inverse of the elastic collision frequency \(\nu_m\)
\begin{equation}
    \tau_m = \frac{1}{\nu_m} = \frac{1}{n_n \sigma_m v_e},
\end{equation}
where \(n_n\) is the neutral gas density, \(\sigma_m\) is a momentum transfer cross section (depends on the type of gas and energy), and \(v_e\) is the average electron velocity (which depends on electron temperature). The establishment time for the electron distribution function in a gas discharge is typically 2-3 times the momentum relaxation time \(\tau_m\) \cite{5}.

For ABEP plasma in VLEO conditions typical temperature and pressure are: \(T_g\) = 200K, \(p\) = 1 Pa.
For the neutral gas density one can obtain:
\begin{equation}
    n_n = \frac{1}{k_b T_g} \approx 4 \cdot 10^{20}\ m^{-3}.
\end{equation}
For the mean electron velocity one have:
\begin{equation}
    v_e = \sqrt{\frac{8 k_b T_e}{\pi m_e}} \approx 6 \cdot 10^6\ \frac{m}{s},
\end{equation}
where \(k T_e\) is the electron energy (assumed to be 10 eV in this calculation), \(m_e\) is the electron mass.

In an \(N_2\) discharge, the momentum transfer cross section \(\sigma_m\) is a key parameter that characterizes the scattering of electrons by nitrogen molecules. The value of \(\sigma_m\) depends on the energy of the electrons, as the momentum transfer cross section is energy-dependent. Typical values for the momentum transfer cross section are obtained from experimental data or cross-section databases. For our particular case, \(\sigma_m \approx 10^{-20} \ m^2 \)  is a good approximation \cite{4}.

Finally, for the momentum relaxation time:
\begin{equation}
    \tau_m \approx 5 \cdot 10^{-8} s,
\end{equation}
It takes 10\ -\ 100 electron - neutral collisions to form the EEDF, so for the time formation:
\begin{equation}
    \tau_e \approx (10-100) \cdot 1.5 \cdot 10^{-7} s \approx 10^{-6} - 10^{-5}\ s.
\end{equation}

Let's now calculate the residence time of air in the ABEP conditions. Paper \cite{6} assumes an orbital speed \( V_o \) of 8 km/s and a maximum chamber length \( L_c \) is about 1 m.
For this conditions, the residence time:
\begin{equation}
    \tau_r = \frac{L_c}{V_o} \approx \cdot 10^{-4}\ s.
\end{equation}

One can conclude from this results that the time necessary to EEDF to form is smaller than the residence time of a gas in an experimental chamber, and so it's possible to some extent to take snapshots and analyze the EEDF throughout the dynamically developing experiment in ABEP.

Following conditions from paper \cite{6}, let's obtain the EEDF for \(N_2\) plasma medium.
For a rough reduced field approximation one can obtain:
\begin{equation}
    \frac{E}{n_n} = \frac{U}{\lambda n_n} = \frac{U}{v_e \tau_m n_n} \approx 10^{-19}\ V \cdot m^2 = 100\ Td.
\end{equation}

Here, \(U\) is a voltage of the discharge, \(lambda\) is the electron mean free path, and \(n_n\) is the still neutral gas density. The unit Td for the reduced electric field depends with SI units as:
\( 1\ Td = 10^{-21}\ V \cdot m^2 \).
For the ionization degree let's consider the case showed on fig. 3(f) from \cite{6}:
\begin{equation}
    \gamma = \frac{n_e}{n_n} = \frac{10^{16}}{3 \cdot 10^{20}} \approx 3 \cdot 10^{-5}.
\end{equation}

Overall, for our particular conditions we can conclude that the EEDF of \(N_2\) plasma will be formed due to the electron-neutral collisions and, in smaller part, by the electron-electron collisions.
\pagebreak

\begin{figure} [ht]
    \centering
    \includegraphics[width=0.6\linewidth]{figures/fig6.png}
    \caption{\label{fig:figure6} Snapshot of EEDF in \(N_2\) discharge for VLEO ABEP conditions.}
\end{figure}

As we can see on figure \ref{fig:figure6}, the inelastic processes (in this case - vibrational excitation) at low degrees of ionization can have an exceptionally strong influence on the function \( f(\epsilon) \), creating "dips" in the distribution at energy levels corresponding to atomic excitation and altering its "tail." Under  equilibrium conditions, of course, the loss of fast electrons due to excitation and ionization, in accordance with the principle of detailed balance, is compensated by reverse processes. However, in the context of our strictly nonequilibrium problem, this does not occur. It is also important to note that the reverse processes of molecular de-excitation are thresholdless, meaning that electrons of any energy can participate, which further disrupts the equilibrium.

\pagebreak

\begin{figure} [h!]
    \centering
    \includegraphics[width=0.6\linewidth]{figures/fig7.png}
    \caption{\label{fig:figure7} Different vibrational temperatures of N2 plasma at E/N = 50 Td.}
\end{figure}

On figure \ref{fig:figure7} we can see the influence of the vibrational temperature on the form of EEDF. Here, the higher \(T_v\), the less number of electrons are able to excite the next vibrational levels of \(N_2\). At the same time, the tail of distribution functions tends to grow with the increase of \(T_v\). It can be explained easily: the higher \(T_v\), the more high-energy electrons exists in the plasma (due to the de-excitation processes). In addition, one can discover the visible line break near 15.6 eV - it is the ionization threshold for a nitrogen molecule. The reason for this visibility is the same - at higher vibrational temperatures there would be more electrons capable of ionizing the gas around.