Where Ei is roughly 2-3 times the actual ionization energy Ei to include the radiative losses due to excitation by electron impact, followed by prompt photon emission Notice that, since ne ion dx - e this can also be written as 5 2 5. Solving for Derivatives We combine here for clarity, the main equations: dx m.v. ee -mv (26) d(noTe) = -ene-m t dx (27) d E (28) It is just a matter of algebra to solve for each of the gradients separately( including the potential gradient -E==). The results are (30m) vn)-V 2E +5kT =5mevevevi+vion kTe +mv(v (29) 5 dx 3meneveve-ne vion ) m,(2 2V.-v n) 2E 5kT (30) 3v。 3kTe-mve k dx=.3m v,mv?-myion3kTe(2v, -V) mv2-kT。2E+5kT miVe mv eEx=5mevevemv? +mvionI5kTe(2v, -vn) V2 2E +5kT (32) Here vo=-.(generally negative) 16.522, Space P pessan Lecture 18 Prof. Manuel martinez Page 9 of 2016.522, Space Propulsion Lecture 18 Prof. Manuel Martinez-Sanchez Page 9 of 20 Where ' Ei is roughly 2-3 times the actual ionization energy Eito include the radiative losses due to excitation by electron impact, followed by prompt photon emission. Notice that, since e e ion d n = dx Γ ν this can also be written as ' e e i ex d 5 kT + E = -e E dx 2 ⎡ ⎤ ⎛ ⎞ ⎢ ⎥ Γ Γ ⎜ ⎟ ⎣ ⎦ ⎝ ⎠ (24) 5. Solving for Derivatives We combine here for clarity, the main equations: e i x ion e d d d = =- = n dx dx dx Γ Γ Γ ν (25) ( ) i i i x i ion i n dv m v = eE - m v v - v dx (26) ( ) e e ex e ee d n kT = -en E - m dx Γ ν (27) ' e e e x e ion i d 5 kT = -e E - n E dx 2 ⎛ ⎞ ΓΓν ⎜ ⎟ ⎝ ⎠ (28) It is just a matter of algebra to solve for each of the gradients, separately (including the potential gradient x -E = x ∂φ ∂ ). The results are ( ) ' 2 i i i e e i i e e e i ion e i i i n e 5 55 dv v 2E + 5kT kT - m v = m v v + kT +m v v - v - 3 dx 3 3 v 3 ⎛ ⎞ ⎡ ⎤ ⎜ ⎟ ν ν ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ (29) ( ) ' 2 e ie e i i e e e e e ion i i n e 5 5 dn 2E + 5kT kT - m v = m n v - n m 2v - v - 3 dx 3 3v ⎛ ⎞ ⎡ ⎤ ⎜ ⎟ ν ν ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ (30) ( ) 2 ' 2 2 e ii e i e e i i e e e i i i ion e i n i e 5 22 dT m v - kT 2E + 5kT kT - m v k = - m v m v - m kT 2v - v - 3 dx 3 3 m v 3 ⎛ ⎞ ⎡ ⎤ ⎜ ⎟ ν ν ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ (31) ( ) 2 ' 2 2 i i e e i i x e e e i i i ion e i n e 55 5 v 2E + 5kT kT - m v eE = m v m v +m kT 2v - v - 3 3 3 v3 ⎛ ⎞ ⎡ ⎤ ⎜ ⎟ ν ν ⎢ ⎥ ⎝ ⎠ ⎣ ⎦ (32) Here e e ex e ν = ν = n Γ (generally negative)