μ 2 r Substituting at at aro /v20 1-a (7) where % is the ratio of thrust to gravity, a small number for us. Since at=AV, Equation(7)reads (2 In a more rigorous analysis the factor 2 4= 1. 19 turns out to be actually 0.79 ∧-sa 1-0.79v (more exact) 16.522, Space P pessan Lecture 3 Prof. Manuel martinez Page 3 of 916.522, Space Propulsion Lecture 3 Prof. Manuel Martinez-Sanchez Page 3 of 9 2 e e v = 2 r µ , or 2 e 0 co e 1 v r - = 0. 2v r ⎛ ⎞ ⎜ ⎟ ⎝ ⎠ Substituting, 2 0 2 2 2 co 6 co co co ar 2 1 at at v 1 10 2v v at 1 v ⎛ ⎞ ⎜ ⎟ ⎛⎞ ⎛⎞ ⎝ ⎠ ⎜⎟ ⎜⎟ − + −− = ⎝⎠ ⎝⎠ ⎛ ⎞ ⎜ ⎟ − ⎝ ⎠ or ( )2 2 2 0 co 6 co co 4 ar v at = 1- v at 1 - v ⎛ ⎞ ⎜ ⎟ ⎛ ⎞ ⎝ ⎠ ⎜ ⎟ ⎝ ⎠ ( ) 1 4 1 4 1 0 4 2 co co 2 0 at 2a 2ar 1- = = = 2 v v r ⎛ ⎞ ⎛ ⎞ ⎜ ⎟ ⎜ ⎟ ⎜ ⎟ ν µ ⎝ ⎠ ⎜ ⎟ ⎜ ⎟ ⎝ ⎠ (7) where 2 0 a = r ν µ (8) is the ratio of thrust to gravity, a small number for us. Since at= V∆ , Equation (7) reads ( )1 4 V v 1- 2 esc. co ⎡ ⎤ ∆≅ ν ⎢ ⎥ ⎣ ⎦ (9) In a more rigorous analysis, the factor 21/4 = 1.19 turns out to be actually 0.79: 1 4 V = v 1 - 0.79 esc. co ⎡ ⎤ ∆ ⎢ ⎥ ⎣ ⎦ ν (10) (more exact)