irfpy.util.physics

Collection of physical formulae

Code author: Yoshifumi Futaana

A simple command line calculator for physical formulae.

See also

• irfpy.util.unitq: the unit handling

• irfpy.util.constant: the constant with and without unit

Supported formula

• temperature_2_pressure (vth, vthp, vthe)

• pressure_2_temperature

• dflux_2_vdf (j2f)

• vdf_2_dflux (f2j)

• kinetic_energy

• energy_2_velocity

• temperature_2_thermal_velocity

• gyrofrequency (wg, wgp, wge)

• gyrotime

• gyroradius (rg, rgp, rge)

• plasmafrequency (wp, wpp, wpe)

• plasmafrequency_ordinary (fp, fpp, fpe)

• inertial_length

• skin_depth

• alfven_speed

• ion_acoustic_speed

• debye_length

• escape_speed

• mean_free_path

• mean_free_path_kin

• mean_free_path_vis

• orbital_semimajor_axis

• orbital_period

Each functions has (in most of the cases) the associated functions with _u behiind. This is for the unitful-version of each method using irfpy.util.unitq` module.

Example

For example, you can get the kinetic energy of a mass of 5 kg particle with 3 m/s object as 22.5 J.

>>> import irfpy.util.physics as f
>>> print(f.kinetic_energy(5, 3))
22.5

Most of them has two variants.

• MKSA (or other pre-defined) unit system

• User-defined unit system at runtime using irfpy.util.unitq system.

Usually, the second system has usually a suffix of _u.

For example, if you want to know the energy (in Joule) for a mass of 4 kg moving with 15 m/s.

>>> print(f.kinetic_energy(4, 15))   # Kinetic energy for 4kg mass point with 15m/s speed. 450J.
450.0

But the same works as follows.

>>> from irfpy.util import unitq as u
>>> print(f.kinetic_energy(u.nit('4 kg'), u.nit('15 m/s')).rescale(u.J))
450.0 J

Even, you can use another unit system.

>>> print(f.kinetic_energy(u.nit('4000 g'), u.nit('0.015 km/s')).rescale(u.J))
450.0 J

The :mod:irfpy.util.unitq module contains the definitions of various units and the constant.

>>> from irfpy.util.unitq import k
>>> print(f.kinetic_energy(k.proton_mass, u.speed_of_light))
0.5 m_p*c**2

Also note that the :mod:irfpy.util.constant module also contains the definitions of various constants.

irfpy.util.physics.temperature_2_pressure(kelvin, per_cm3)

Temperature to pressure

The temperature in Kelvin under the density per_cm3 are converted to the pressure in nPa.

$$p[\mathrm{nPa}]\times {10}^{-9}=(n{[\mathrm{cm}}^{-3}]\times {10}^6)kT$$

where k is the Boltzman constant.

>>> loschmidt = 2.6868e19    # 2.69e19 /cm3 -- Loschmidt's constant  
>>> temp_ideal = 273.15      # 0 degreeC
>>> pressure_at_surface = temperature_2_pressure(273.15, 2.6868e19)  # This should be the ground pressure, 1 bar ~= 1e5 Pa ~= 1e14 nPa
>>> print('{:.3e}'.format(pressure_at_surface))
1.013e+14

Also, unitful version.

>>> loschmidt = 2.6868e19 / (u.cm ** 3)
>>> temp_ideal = 273.15 * u.K
>>> pressure_at_surface = temperature_2_pressure_u(temp_ideal, loschmidt)
>>> print('{:.3e}'.format(float(pressure_at_surface.r(u.nPa))))
1.013e+14

irfpy.util.physics.temperature_2_pressure_u(*args)

Unitful version of temperature_2_pressure()

irfpy.util.physics.pressure_2_temperature(nPa, per_cm3)

Pressure to temperature.

See also temperature_2_pressure().

>>> p = 1e5 * 1e9      # 1 bar as nPa
>>> loschmidt = 2.6868e19   # in /cm3
>>> temperature = pressure_2_temperature(p, loschmidt)
>>> print('{:.3f}'.format(temperature))   # Should be around 0degC
269.576

Unitful version also exists

>>> p = 1 * u.bar      # 1 bar
>>> loschmidt = 2.6868e19 / u.cm **3
>>> temperature = pressure_2_temperature_u(p, loschmidt)
>>> temperature = float(temperature.rescale(u.K))
>>> print('{:.3f}'.format(temperature))
269.576

irfpy.util.physics.pressure_2_temperature_u(*args)

Unitful version of temperature_2_pressure()

irfpy.util.physics.dflux_2_vdf(j, mass, energy)

Differnetial flux to velocity distribution function

This is MKSA version. For plasma-unit version (but still unitless), you may refer to dflux_2_vdf_pl(). For unitq version, you may refere to “math:dflux_2_vdf_u.

The relation is

$$f=\frac{m^2}{2E}j$$

Let us consider the differential flux of j=1e-5 /cm2 eV sr s. The particle is proton (for simplicity mass is assumed to be 1 amu), and the energy is 500 eV.

>>> j = 1e+5   # /cm2 eV sr s
>>> m = 1      # amu
>>> e = 500    # eV
>>> print('%.2e' % dflux_2_vdf_pl(j, m, e))
1.07e-10

Now, we get the velocity distribution function of 1.1e-10 s^3/m^6.

Obviously, we also call via MKSA unit version.

>>> jmksa = j * 1e4 / k.qe
>>> print('%.2e' % jmksa)   # in /m2 J sr s
6.24e+27

>>> mmksa = m * k.amu   # kg
>>> emksa = e * k.qe   # J

>>> print('%.2e' % dflux_2_vdf(jmksa, mmksa, emksa))
1.07e-10

For interactive use, unitq module will be a significant help.

>>> j_u = 1e5 / u.cm**2 / u.eV / u.sr / u.s
>>> m_u = 1 * u.amu
>>> e_u = 500 * u.eV
>>> vdf_u = dflux_2_vdf_u(j_u, m_u, e_u)
>>> u.pprint(vdf_u, fmt='%.2e')
1.07e-10 s**3/m**6

irfpy.util.physics.j2f(j, mass, energy)

Alias to dflux_2_vdf()

irfpy.util.physics.dflux_2_vdf_pl(j, mass, energy)

Differnetial flux with plasma-frequently-used units

This provides the alias but different unit system to dflux_2_vdf().

The input unit is j as [/cm2 eV sr s], mass for [amu], and energy for [eV].

The returned vdf is [s**3 / m**6].

irfpy.util.physics.j2f_pl(j, mass, energy)

Alias to dflux_2_vdf_pl()

irfpy.util.physics.vdf_2_dflux(f, mass, velocity)

J from f. (MKSA)

$$J=\frac{v^2}{m}f$$

>>> f = 1.07e-10    # [s^3 / m^6]
>>> j = vdf_2_dflux(f, 1.66e-27, 300e3)
>>> print('%.2e' % j)    # [/m^2 J sr s]
5.80e+27

The dflux should be converted back to the origical vdf.

>>> f2 = dflux_2_vdf(j, 1.66e-27, kinetic_energy(1.66e-27, 300e3))
>>> print('%.2e' % f2)
1.07e-10

The unitful version:

>>> f_u = 1.07e-10 * u.s ** 3 / u.m ** 6
>>> j_u = vdf_2_dflux_u(f_u, 1 * u.amu, 300 * u.km/u.s)
>>> u.pprint(j_u, fmt='%.2e')
5.80e+27 1/(m**2*s*sr*J)

>>> j_u.units = u.cu_diffflux
>>> u.pprint(j_u, fmt='%.2e')
9.29e+04 (1/cm**2 /sr /eV / s)

irfpy.util.physics.f2j(f, mass, velocity)

J from f. (MKSA)

$$J=\frac{v^2}{m}f$$

>>> f = 1.07e-10    # [s^3 / m^6]
>>> j = vdf_2_dflux(f, 1.66e-27, 300e3)
>>> print('%.2e' % j)    # [/m^2 J sr s]
5.80e+27

The dflux should be converted back to the origical vdf.

>>> f2 = dflux_2_vdf(j, 1.66e-27, kinetic_energy(1.66e-27, 300e3))
>>> print('%.2e' % f2)
1.07e-10

The unitful version:

>>> f_u = 1.07e-10 * u.s ** 3 / u.m ** 6
>>> j_u = vdf_2_dflux_u(f_u, 1 * u.amu, 300 * u.km/u.s)
>>> u.pprint(j_u, fmt='%.2e')
5.80e+27 1/(m**2*s*sr*J)

>>> j_u.units = u.cu_diffflux
>>> u.pprint(j_u, fmt='%.2e')
9.29e+04 (1/cm**2 /sr /eV / s)

irfpy.util.physics.kinetic_energy(mass, velocity)

Return the kinetic energy.

$$KE=\frac{1}{2}m{v}^2$$

>>> print(kinetic_energy(4, 15))   # Kinetic energy for 4kg mass point with 15m/s speed. 450J.
450.0
>>> print(kinetic_energy(u.nit('4 kg'), u.nit('15 m/s')).rescale(u.J))
450.0 J

irfpy.util.physics.kinetic_energy_u(*args)

Unitful version of kinetic_energy()

irfpy.util.physics.kinetic_energy_proton(velocity_in_km_s)

Return the proton energy.

Parameters:

velocity_in_m_s – Velocity in km/s.

Returns:

The kinetic energy in eV.

Simple formulation: 1 eV energy proton is about 13.8 km/s

>>> print('{:.3f}'.format(kinetic_energy_proton(13.8)))
0.994

Solar wind (4e5 m/s) proton has energy of 835 eV. The call below is the unitful version of the formulation.

>>> energy = kinetic_energy_proton_u(400000 * u.m / u.s)
>>> print('{:.3f}'.format(float(energy.nr(u.keV))))
0.835

irfpy.util.physics.kinetic_energy_electron(velocity_in_km_s)

Return the electron energy

Parameters:

velocity_in_km_s – Velocity in km/s

Returns:

The kinetic energy in eV

Simple formulation: 1 eV energy electron is about 600 km/s

>>> print('{:.3f}'.format(kinetic_energy_electron(600)))
1.023

or one can use unitful version.

>>> energy = kinetic_energy_electron_u(600000 * u.m / u.s)
>>> print('{:.3e}'.format(float(energy.nr(u.J))))
1.640e-19

irfpy.util.physics.energy_2_velocity(mass, energy)

Return the velocity from the kinetic energy

Return the velocity from the kinetic energy. This is the inversion function of kinetic_energy().

>>> print(energy_2_velocity(4, 450.))   # 4kg mass with 450 J => v=15 m/s
15.0

You may also use unitful version.

>>> print(energy_2_velocity_u(u.nit('4 kg'), u.nit('450 J')))
15.0 m/s

irfpy.util.physics.energy_2_velocity_proton(energy_eV)

Return the velocity (in km/s) of the proton with the given energy (eV)

Parameters:

energy_eV – Energy of proton in eV.

Returns:

Speed of proton in km/s.

>>> print('{:.2f}'.format(energy_2_velocity_proton(1)))    # 1 eV proton has 13.8 km/s
13.84

or use unit-version.

>>> speed = energy_2_velocity_proton_u(1 * u.eV)
>>> u.pprint(speed)
13.841 km/s

irfpy.util.physics.energy_2_velocity_electron(energy_eV)

Return the velocity (in km/s) of the electron with the given energy (eV)

Parameters:

energy_eV – Energy of electron in eV.

Returns:

Speed of electron in km/s.

>>> print('{:.2f}'.format(energy_2_velocity_electron(1)))    # 1 eV proton has 593 km/s
593.10

or use unit-version.

>>> speed = energy_2_velocity_electron_u(1 * u.eV)
>>> u.pprint(speed)
593.097 km/s

irfpy.util.physics.temperature_2_thermal_velocity(mass, temperature)

Return the thermal velocity from temperature

$$v_{th}=\sqrt{\frac{kT}{m}}$$

Thermal velocity is calculated from the temperature.

Roughly the proton with 1eV temperature has 9.8 km/s thermal velocity.

>>> print('%.3f' % temperature_2_thermal_velocity(1.67e-27, 11600))
9792.924

Unitful version also supported.

>>> print('%.3f' % temperature_2_thermal_velocity_u(k.mp_u,
...        1 * u.eVT).nr(u.nit('km/s')))
9.787

irfpy.util.physics.temperature_2_thermal_velocity_u(*args)

Unitful version of temperature_2_thermal_velocity().

irfpy.util.physics.vth(mass, temperature)

Return the thermal velocity from temperature

$$v_{th}=\sqrt{\frac{kT}{m}}$$

Thermal velocity is calculated from the temperature.

Roughly the proton with 1eV temperature has 9.8 km/s thermal velocity.

>>> print('%.3f' % temperature_2_thermal_velocity(1.67e-27, 11600))
9792.924

Unitful version also supported.

>>> print('%.3f' % temperature_2_thermal_velocity_u(k.mp_u,
...        1 * u.eVT).nr(u.nit('km/s')))
9.787

irfpy.util.physics.gyrofrequency(mass, charge, magnetic_flux_density)

Gyro frequency, ${\omega }_g$.

$${\omega }_g=\frac{qB}{m}$$

>>> omega = gyrofrequency_u(k.mp_u, k.e_u, 10 * u.nT)
>>> u.pprint(omega, fmt='%.4f')
0.9579 rad/s

irfpy.util.physics.gyrofrequency_u(mass, charge, magnetic_flux_density)

Gyro frequency, ${\omega }_g$.

$${\omega }_g=\frac{qB}{m}$$

>>> omega = gyrofrequency_u(k.mp_u, k.e_u, 10 * u.nT)
>>> u.pprint(omega, fmt='%.4f')
0.9579 rad/s

irfpy.util.physics.wg(mass, charge, magnetic_flux_density)

Gyro frequency, ${\omega }_g$.

$${\omega }_g=\frac{qB}{m}$$

>>> omega = gyrofrequency_u(k.mp_u, k.e_u, 10 * u.nT)
>>> u.pprint(omega, fmt='%.4f')
0.9579 rad/s

irfpy.util.physics.wg_u(mass, charge, magnetic_flux_density)

Gyro frequency, ${\omega }_g$.

$${\omega }_g=\frac{qB}{m}$$

>>> omega = gyrofrequency_u(k.mp_u, k.e_u, 10 * u.nT)
>>> u.pprint(omega, fmt='%.4f')
0.9579 rad/s

irfpy.util.physics.wgp(mass, charge, magnetic_flux_density)

Gyro frequency, ${\omega }_g$.

$${\omega }_g=\frac{qB}{m}$$

>>> omega = gyrofrequency_u(k.mp_u, k.e_u, 10 * u.nT)
>>> u.pprint(omega, fmt='%.4f')
0.9579 rad/s

irfpy.util.physics.wge(mass, charge, magnetic_flux_density)

Gyro frequency, ${\omega }_g$.

$${\omega }_g=\frac{qB}{m}$$

>>> omega = gyrofrequency_u(k.mp_u, k.e_u, 10 * u.nT)
>>> u.pprint(omega, fmt='%.4f')
0.9579 rad/s

irfpy.util.physics.gyrotime(mass, charge, magnetic_flux_density)

Gyro time

Return the gyration time. $2\pi /{\omega }_g$.

irfpy.util.physics.gyroradius(mass, charge, bfield, vel)

Gyroradius.

$$r_g=\frac{v}{{\omega }_g}=\frac{mv}{qB}$$

irfpy.util.physics.rg(mass, charge, bfield, vel)

Gyroradius.

$$r_g=\frac{v}{{\omega }_g}=\frac{mv}{qB}$$

irfpy.util.physics.rgp(mass, charge, bfield, vel)

Gyroradius.

$$r_g=\frac{v}{{\omega }_g}=\frac{mv}{qB}$$

irfpy.util.physics.rgp_u(mass, charge, bfield, vel)

Gyroradius.

$$r_g=\frac{v}{{\omega }_g}=\frac{mv}{qB}$$

irfpy.util.physics.rge(bfield, vel)

Gyroradius.

$$r_g=\frac{v}{{\omega }_g}=\frac{mv}{qB}$$

irfpy.util.physics.plasmafrequency(mass, charge, density)

Plasma frequency, ${\omega }_p$

$${\omega }_p=\sqrt{\frac{n{e}^2}{m{ϵ}_0}}$$

Return the plasma frequency.

Unitless version is MKSA; you may also use the alias version wp.

>>> print('{:.1f}'.format(plasmafrequency(1.67e-27, 1.60e-19, 10e6)))
4160.9
>>> print('{:.1f}'.format(wp(1.67e-27, 1.60e-19, 10e6)))
4160.9

irfpy.util.physics.wp(mass, charge, density)

Plasma frequency, ${\omega }_p$

$${\omega }_p=\sqrt{\frac{n{e}^2}{m{ϵ}_0}}$$

Return the plasma frequency.

Unitless version is MKSA; you may also use the alias version wp.

>>> print('{:.1f}'.format(plasmafrequency(1.67e-27, 1.60e-19, 10e6)))
4160.9
>>> print('{:.1f}'.format(wp(1.67e-27, 1.60e-19, 10e6)))
4160.9

irfpy.util.physics.wpp(density)

Plasma frequency, ${\omega }_p$

$${\omega }_p=\sqrt{\frac{n{e}^2}{m{ϵ}_0}}$$

Return the plasma frequency.

Unitless version is MKSA; you may also use the alias version wp.

>>> print('{:.1f}'.format(plasmafrequency(1.67e-27, 1.60e-19, 10e6)))
4160.9
>>> print('{:.1f}'.format(wp(1.67e-27, 1.60e-19, 10e6)))
4160.9

irfpy.util.physics.wpe(density)

Plasma frequency, ${\omega }_p$

$${\omega }_p=\sqrt{\frac{n{e}^2}{m{ϵ}_0}}$$

Return the plasma frequency.

Unitless version is MKSA; you may also use the alias version wp.

>>> print('{:.1f}'.format(plasmafrequency(1.67e-27, 1.60e-19, 10e6)))
4160.9
>>> print('{:.1f}'.format(wp(1.67e-27, 1.60e-19, 10e6)))
4160.9

irfpy.util.physics.plasmafrequency_ordinary(mass, charge, density)

Plasma ordinary frequency, $f_p$

$$f_p=\frac{{\omega }_p}{2\pi }$$

For example, the electron ordinary plasma frequency is

$$f_{pe}\sim 8980\sqrt{n_e}\mathrm{Hz}$$

>>> n_e = 100 / (u.cm ** 3)
>>> f_pe =plasmafrequency_ordinary_u(k.me_u, k.qe_u, n_e).nr(u.kHz)
>>> print('{:.1f}'.format(float(f_pe)))
89.8

irfpy.util.physics.fp(mass, charge, density)

Plasma ordinary frequency, $f_p$

$$f_p=\frac{{\omega }_p}{2\pi }$$

For example, the electron ordinary plasma frequency is

$$f_{pe}\sim 8980\sqrt{n_e}\mathrm{Hz}$$

>>> n_e = 100 / (u.cm ** 3)
>>> f_pe =plasmafrequency_ordinary_u(k.me_u, k.qe_u, n_e).nr(u.kHz)
>>> print('{:.1f}'.format(float(f_pe)))
89.8

irfpy.util.physics.fpp(density)

Plasma ordinary frequency, $f_p$

$$f_p=\frac{{\omega }_p}{2\pi }$$

For example, the electron ordinary plasma frequency is

$$f_{pe}\sim 8980\sqrt{n_e}\mathrm{Hz}$$

>>> n_e = 100 / (u.cm ** 3)
>>> f_pe =plasmafrequency_ordinary_u(k.me_u, k.qe_u, n_e).nr(u.kHz)
>>> print('{:.1f}'.format(float(f_pe)))
89.8

irfpy.util.physics.fpe(density)

Plasma ordinary frequency, $f_p$

$$f_p=\frac{{\omega }_p}{2\pi }$$

For example, the electron ordinary plasma frequency is

$$f_{pe}\sim 8980\sqrt{n_e}\mathrm{Hz}$$

>>> n_e = 100 / (u.cm ** 3)
>>> f_pe =plasmafrequency_ordinary_u(k.me_u, k.qe_u, n_e).nr(u.kHz)
>>> print('{:.1f}'.format(float(f_pe)))
89.8

irfpy.util.physics.inertial_length(mass, charge, density)

>>> d = inertial_length(1.67e-27, 1.60e-19, 10e6)
>>> print('{:.2f}'.format(d))
72049.89

irfpy.util.physics.skin_depth(density)

irfpy.util.physics.alfven_speed(mass, density, mag)

Return Alfven speed

$$v_a=\frac{B}{\sqrt{\mu 0nm}}$$

irfpy.util.physics.ion_acoustic_speed(mass, temp_i, temp_e, gamma_i, gamma_e)

Ion acoustic speed

$$v_c=\sqrt{\frac{{\gamma }_{e}k{T}_e+{\gamma }_{i}k{T}_i}{m_i}}$$

${\gamma }_e=1$ is frequently used.

irfpy.util.physics.debye_length(ne, temp_e, charge)

debye length.

$${\lambda }_D=\sqrt{\frac{{ϵ}_{0}k{T}_e}{n_e{q}_e^2}}$$

irfpy.util.physics.escape_speed(mass_kg, radius_km)

Return the escape speed in km/s.

$$v_{\mathrm{esc}}=\sqrt{\frac{2GM}{R}}$$

For the Earth:

>>> me = 5.972e24   # Mass of Earth in kg
>>> re = 6371.      # Radius of Earth in km
>>> esc_e = escape_speed(me, re)    # Escape speed of Earth
>>> print('{:.1f}'.format(esc_e))
11.2

For the Sun:

>>> ms = 1.989e30 * u.kg
>>> rs = 695700 * u.km
>>> esc_s = escape_speed_u(ms, rs)
>>> print('%.1f' % esc_s.nr(u.km / u.s))
617.8

irfpy.util.physics.mean_free_path(density_cm3, crosssection_cm2)

Return the mean free path in cm.

A simple theory (scattering theory) gives the mean free path of

$$\lambda =\frac{1}{n\sigma }$$

where $n$ is the number density and $\sigma$ is the crosssection.

This is based on the scattering theory.

Parameters:

• density_cm3 – Density in cm^-3

• crosssection_cm2 – Cross section in cm^-2

Returns:

Mean free path in the unit of cm.

>>> n = 2.5e19    # Approximate molecule number density at Earth surface.
>>> sigma = 1e-15   # Typical collision cross-section
>>> mfp = mean_free_path(n, sigma)
>>> print('{:.3e}'.format(mfp))
4.000e-05

Or derive with unit support.

>>> from irfpy.util import unitq as u
>>> n = 2.5e25 * u.nit('m^-3')    # 2.5 x 10^25 m^-3
>>> sigma = 1e-15 * u.nit('cm^2')  # 1 x 10^25 cm^2
>>> mfp = mean_free_path_u(n, sigma)
>>> mfp = mfp.r(u.m)    # The unit is rescaled to meter.
>>> print('{:.2e}'.format(mfp))     # The mean free path value as the unit of meter.
4.00e-07 m

irfpy.util.physics.mean_free_path_kin(density_cm3, crosssection_cm2)

Return the mean free path in cm.

Kinetic theory (including temperature) suggests a correction by a factor of $\sqrt{2}$ compared to mean_free_path().

$$\lambda =\frac{1}{\sqrt{2}n\sigma }$$

where $n$ is the number density and $\sigma$ is the crosssection.

Parameters:

• density_cm3 – Density in cm^-3

• crosssection_cm2 – Cross section in cm^-2

Returns:

Mean free path in the unit of cm.

>>> n = 2.5e19    # Approximate molecule number density at Earth surface.
>>> sigma = 1e-15   # Typical collision cross-section
>>> mfp = mean_free_path_kin(n, sigma)
>>> print('{:.2e}'.format(mfp))  # Returned value is in cm.
2.83e-05

Or derive with unit support.

>>> from irfpy.util import unitq as u
>>> n = 2.5e25 * u.nit('m^-3')    # 2.5 x 10^25 m^-3
>>> sigma = 1e-15 * u.nit('cm^2')  # 1 x 10^25 cm^2
>>> mfp = mean_free_path_kin_u(n, sigma)
>>> mfp = mfp.r(u.m)    # The unit is rescaled to meter.
>>> print('{:.2e}'.format(mfp))     # The mean free path value as the unit of meter.
2.83e-07 m

irfpy.util.physics.mean_free_path_vis(mu_Pa_s, p_Pa, T_K, m_amu)

Based on kinetic theory, the m.f.p is calcuated from viscosity.

$$l=\frac{\mu }{p}\sqrt{\frac{\pi k_{B}T}{2m}}$$

Here, $\mu$ is the viscosity, p is the pressure, $k_B$ is the Boltzmann constant, T is the temperature, and m is the mass of the gas.

https://en.wikipedia.org/wiki/Mean_free_path#Kinetic_theory_of_gases

It is not easy to get the viscosity for specific gas. For example of the air at 20 degreeC, the viscosity is ~1.8x10^-5 Pa s.

Parameters:

• mu_Pa_s – Viscosity in the unit of Pa s. Just a note 1 Pa s = 10 P = 1000 cP.

• p_Pa – Pressure in the unit of Pa.

• T – Temperuter in Kelvin.

• m_amu – Mass in atomic mass unit.

Returns:

Mean free path in the unit of cm.

>>> mu = 1.8e-5    # Pa s
>>> p = 101100   # Pascal
>>> T = 20 + 273.15   # Kelvin
>>> m = 28    # amu

>>> l = mean_free_path_vis(mu, p, T, m)
>>> print('{:.2e}'.format(l))
6.58e-06

Unitful version:

>>> mu = 0.018 * u.nit('centipoise')    # Air viscosity
>>> p = 1013 * u.nit('hPa')    # Air pressure
>>> T = (20 + 273.15) * u.nit('K')    # Unit does not support the conversion of degC <-> degK !!
>>> m = 28 * u.amu

>>> l = mean_free_path_vis_u(mu, p, T, m)    # Calculate the mean free path
>>> l = l.r(u.angstrom)    # Rescale to the unit of Ångström
>>> print('{:.2e}'.format(l))
6.57e+02 angstrom

irfpy.util.physics.orbital_semimajor_axis(period, mass)

Calculate the semi-major axis from the period of satellite.

Parameters:

• period – Period in [s]

• mass – Mass of the central body in [kg]

Returns:

Semi-jmajor axis (the radius for a circular orbit) in [km]

Using the Kepler’s law:

$$\frac{T^2}{a^3}=\frac{4{\pi }^2}{GM}$$

# Lunar orbiter with period of 7000s. The semi-major axis is ~1825 km (100 km alt). >>> from irfpy.util import planets as pla >>> a = orbital_semimajor_axis(7000, pla.moon[‘mass’]) >>> print(f’{a:.1f}’) 1825.8

# Moon around the earth. 27.5 day period give the distance of ~60 Rm (384000 km) >>> a_u = orbital_semimajor_axis_u(27.4 * u.day, … pla.earth[‘mass’] * u.kg) >>> a = a_u.nr(u.km) # Represented by days in float. >>> print(f’{a:.1f}’) 383946.1

irfpy.util.physics.orbital_period(a, m)

Calculate the orbital period.

Parameters:

• a – Semi-jmajor axis (the radius for a circular orbit) in [km]

• m – Mass of the central body in [kg]

Returns:

Period in [s]

Using the Kepler’s law:

$$\frac{T^2}{a^3}=\frac{4{\pi }^2}{GM}$$

# Lunar orbiter at 100 km. It is about 2 hours (~7050s) >>> from irfpy.util import planets as pla >>> p = orbital_period(1834, pla.moon[‘mass’]) >>> print(f’{p:.1f}’) 7046.9

# Moon around the earth. ~27 days >>> p_u = orbital_period_u(pla.moon[‘semi-major-axis’] * u.km, … pla.earth[‘mass’] * u.kg) >>> p = p_u.nr(u.day) # Represented by days in float. >>> print(f’{p:.1f}’) 27.4