Question

) Titan’s Evolving Atmosphere: Titan’s exopshere lies nearly 1400 km above its surface. What is the escape velocity from this altitude? What is the thermal speed of a hydrogen atom at the exospheric temperature of about 200 K?

Answer 1

Answer:

Explanation:

first we calculate the escape velocity using $V_{esc} =\sqrt{\frac{2GM}{R} }$

Thermal velocity of an atom  $V_{therm} =\sqrt{\frac{2KT}{M} }$

Given $G= 6.67*10^{-11} m^{3}/kgS^{2}\\\\Mx_{titan} = 1.35*10^{23}kg$

Resistance from the centr of the titan is:

R= R(titan) + height of exosphere = 2575 + 1400=3.975*10^6m

escape vel = $V_{esc} =\sqrt{\frac{2GM}{R} }\\\\V_{esc} =\sqrt{\frac{2*6.67*10^{-11}*1.35*10^{23}}{3.975*10^{6}} }=2.129*10^{3}m/s$

Thermal velocity of an atom  $V_{therm} =\sqrt{\frac{2KT}{M} }\\\\V_{therm} =\sqrt{\frac{2*1.38*10^{-23}*200}{1.67*10^{-27}} }=1.8*10^{3}m/s$