Given T, this relates n, to n,. A second relation is needed and very often it is a specification of the overall pressure P=(ne+n,+nn)kT =(2ne +n)kT Combining(3)and(5) =s( s(T)(n-2n2) Where nP kT is the total member density of all particles We then have n2+2sn。-Sn=0 Since s increases very rapidly with t, the limits of (6)are n sn→0 (Weak ionization) T→0 heT→2 (Full ionization) Once an electron population exists, an electric field E will drive a current density j through the plasma. To understand this quantitatively consider the momentum balance of a" typical"electron It sees an electrostatic force FE=-eE [t also sees a"frictional" force due to transfer of momentum each time it collides with some other particle(neutral or ion). Collisions with other electrons are not counted, because the momentum transfer is in that case internal to the electron species. The ions and neutrals are almost at rest compared to the fast-moving electrons and we define an effective collision as one in which the electrons directed momentum is fully given up. Suppose there are ve of these collisions per second (ve=collision frequency per electron). The electron loses momentum at a rate m, Vev, where V. =mean directed velocity of electrons, and so 16.522, Space P pessan Lecture 10 Prof. Manuel martinezGiven T, this relates ne to nn . A second relation is needed and very often it is a specification of the overall pressure P = (n + ni + n ) kT = (2n + n ) kT (5) e n e n Combining (3’) and (5), 2 ( ) ⎛ ⎜ P n = S T - 2ne ⎟ = S T ⎝ kT ⎞ e ( )(n - 2ne ) ⎠ P Where n= is the total member density of all particles. kT We then have 2 n + 2Sn - Sn = 0 e e S2 (6) n = -S + + Sn = n e 1 + 1 + n S T( ) Since S increases very rapidly with T, the limits of (6) are n ⎯⎯⎯⎯⎯→ Sn (→ 0) (Weak ionization) e T0 → T 0 → ⎯⎯⎯⎯⎯ n ne T→∞ → 2 (Full ionization) G Once an electron population exists, an electric field E will drive a current density G j through the plasma. To understand this quantitatively, consider the momentum balance of a “typical” electron. It sees an electrostatic force G G F = -eE E (7) It also sees a “frictional” force due to transfer of momentum each time it collides with some other particle (neutral or ion). Collisions with other electrons are not counted, because the momentum transfer is in that case internal to the electron species. The ions and neutrals are almost at rest compared to the fast-moving electrons, and we define an effective collision as one in which the electron’s directed momentum is fully given up. Suppose there are νe of these collisions per second ( νe =collision frequency per electron). The electron loses momentum at a rate JG JJG -m Ve ν , where Ve =mean directed velocity of electrons, and so e e G JG FFriction = -m Ve ν (8) e e On average, 16.522, Space Propulsion Lecture 10 Prof. Manuel Martinez-Sanchez Page 2 of 12