Answer to Question #26449 in Astronomy | Astrophysics for Dave
Titan’s size (R = 2575 km) and density (ρ = 1.9 g cm -3) are much smaller than Earth’s (R = 6378 km and ρ = 5.5 g cm -3). Therefore, we would expect that its surface gravity gT is weaker than Earth’s g, so any material on Titan would weigh much less than it does on Earth (the weight of any mass m on any planet (or moon) is given by the product of its mass m and surface gravity g, i.e., W = mg). However, atmospheric pressure on Titan’s surface is greater than Earth’s, about 1.6 Earth atmospheres (pressure is the weight per square meter on the surface of the planet (or moon), i.e., P = W / A where A is the area)! This means that the amount of mass in Titan’s atmosphere above any square meter of Titan’s surface must be much greater than the amount of mass in the atmosphere of Earth above any square meter on Earth’s surface. Calculate the amount mass of atmosphere per square meter (we’ll call this σ = m/A) on Titan’s surface compared to that on Earth. (Hint: You’ll need to calculate Titan’s surface gravity relati
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