Answer:
Explanation:
first we calculate the escape velocity using [tex]V_{esc} =\sqrt{\frac{2GM}{R} }[/tex]
Thermal velocity of an atom [tex]V_{therm} =\sqrt{\frac{2KT}{M} }[/tex]
Given [tex]G= 6.67*10^{-11} m^{3}/kgS^{2}\\\\Mx_{titan} = 1.35*10^{23}kg\\[/tex]
Resistance from the centr of the titan is:
R= R(titan) + height of exosphere = 2575 + 1400=3.975*10^6m
escape vel = [tex]V_{esc} =\sqrt{\frac{2GM}{R} }\\\\V_{esc} =\sqrt{\frac{2*6.67*10^{-11}*1.35*10^{23}}{3.975*10^{6}} }=2.129*10^{3}m/s[/tex]
Thermal velocity of an atom [tex]V_{therm} =\sqrt{\frac{2KT}{M} }\\\\V_{therm} =\sqrt{\frac{2*1.38*10^{-23}*200}{1.67*10^{-27}} }=1.8*10^{3}m/s[/tex]
