Question

V_ocean= volume of ocean

Docean  = volume of ocean

A) The total volume of Titan’s lakes and seas was recently estimated to be 70,000 km³ of liquid methane. If this volume was spread across all of Titan into a global ocean (ignore any surface topography for the remainder of this question), how deep would that global ocean be? The volume of a global ocean can be approximated by its depth multiple by the surface area of the moon (V_ocean = D_{ocean *} (4 * pi * R_Titan²)). In reality, Titan’s lakes and seas only take up ~1% of Titan’s surface area. Given this information, what is the average depth of a Titan sea if we assume that they all the lakes and seas have the same depth (Hint: use the formula for a global ocean but put in the surface area of Titan’s lakes and seas instead of the whole moon). Remember that Titan has a radius of 2575 km.

B) The total volume methane in Titan’s atmosphere can be estimated from its surface pressure. Titan’s troposphere (where most of the atmosphere’s mass is contained) is made up of 97.15% nitrogen and 2.85% methane by mass, with only trace amounts of other components (you heard in class that Titan’s atmosphere was 95% nitrogen and 5% methane, which is the composition by volume). Titan’s surface pressure is 1.5 bars or 150,000 Pa (a pascal is defined as 1 Pa = 1 kg m^-1 s^-2), so the partial pressure due to methane is: Pressure_CH4 =150,000 * 0.0285 = 4275 Pa. Surface pressure is a measure of force per unit area. Accordingly, the average surface pressure from methane can be set equivalent to the gravitation force exerted by all of the methane in Titan’s atmosphere (M_CH4 * g_Titan) divided by the moon’s surface area (4 * pi * R²)

Pressure_CH4 = M_CH4 * g_Titan / (4 * pi * R²)

Where M_CH4 is the mass of methane in Titan’s atmosphere. Use the above equation to determine the total mass of methane in Titan’s atmosphere.

C) In order to compare the amount of methane in the atmosphere to the amount in the observed in the lakes and seas, let’s determine the depth of a global ocean of methane with the mass from part (b). The density of liquid methane is r_CH4=450 kg/m³. Remembering that density is defined as mass over volume (r=M/V), determine the volume of liquid methane that would have the mass you calculated in part (b). As explained in (a), the volume of a global ocean can be approximated by its depth multiple by the surface area of the moon (V_ocean = D_{ocean *} (4 * pi * R_Titan²). Calculate D_ocean for the mass determined in part (b). How does is compare to the global ocean depth you calculated in part (a)?

Answer 1