On the thermal velocity of gas (or any other particles) So I wonder if the thermal velocity of a gas can be considered as a 'real' velocity of particles, or it is just totally quantum phenomenon.
To elaborate my question, the wikipedia(https://en.wikipedia.org/wiki/Thermal_velocity) says that the thermal velocity of Hydrogen at 20C is about 1754m/s. However, the sun and the whole solar system are revolving around the center of the milky way of 200km/s speed, which is far faster than the usual 'thermal velocity' of Hydrogen.
Then, I got to wonder if an observer at the center of the milky way would see the Earth as a super hot planet, since the atmosphere is also moving at that fast speed with the Earth.
 A: The thermal velocity you're referring to is the root mean square velocity. It arises using ideal gas law equations. The first equation is a calculation of the pressure exerted on the walls of a container of volume $V$ by a gas.
$$P = \frac{1}{3} \frac{nMv^2}{V}$$
Where $P$ is pressure, $n$ number of molecules, $M$ molar mass, $v$ average velocity and $V$ container volume. This is combined with the ideal gas law
$$PV = nRT$$
To give us
$$\frac{1}{3} \frac{nMv^2}{V}=nRT$$
When we recognise that kinetic energy is $\frac{1}{2}Mv^2$, we get the average velocity of the gas as
$$ v_{rms} =(\frac{3RT}{M})^{1/2}$$
Anyway the reason for deriving this is we can look at the assumptions which are made, and that's namely the ideal gas law. And this law assumes a reference frame at rest with respect to the particles. Even if you took a jar of hydrogen and moved it at $05.c$ relative to you, the particle velocity is relative to the jar.
EDIT

To clarify the point, after reading the comments.
P is defined in the reference frame where the container is at rest. You can arbitrarily shift reference frame to one where the container is moving. But when you calculate the force on the walls of the container, you will just have an extra bunch of initial velocities to account for. But the force per unit area exerted on the container will be the same.
To answer your question about temperature of a planet moving with velocity relative to us far greater than the average velocity of a gas, consider the atmosphere of a distant planet which is moving at (for arguments sake a constant velocity) of $v_p=5000km/s$ relative to us. The kinetic energy of the atmosphere in our reference frame is:
$KE_{atm} = \frac{1}{2} mv_p^2 + E_{thermal}$, i.e the kinetic energy of some massive body (atmosphere) moving at $5,000km/s$ and some additional thermal component which for now we assume nothing about.
We can shift our reference frame by hopping in a space ship and flying towards the planet away from the earth at a constant velocity of $3,000km/s$. Now the atmosphere of the distant planet has velocity relative to us of $v_r=2,000km/s$, which gives kinetic energy:
$KE_{atm} = \frac{1}{2} mv^2 + E_{thermal}$. This is a smaller value than before.
We can choose any reference frame we desire and increase or decrease the apparent kinetic energy of the planet's atmosphere as much as we like. However in every reference frame, there is some component of energy $E_{thermal}$ that we can't get rid of. This is the thermal energy. And the velocity due to this thermal energy, given by $ v_{rms} =(\frac{3RT}{M})^{1/2}$ is the "real" motion you referred to.
