Saturn’s largest moon, Titan, is actually larger than the planet Mercury. It is also one of the few bodies in the solar system that has a dense atmosphere. Data from the Voyager 1 probe indicates that the atmospheric pressure on Titan’s surface is about 1.60 x 105 Pa, even though free-fall acceleration at Titan’s surface is only 1.36 m/s2. Assuming that Titan’s atmosphere has a uniform density of 2.32 kg/m3, at what height above the surface is Titan’s atmospheric pressure equal to 1.01 x 105 Pa?
To find the height above the surface where Titan's atmospheric pressure is equal to 1.01 x 10^5 Pa, we can use the relationship between pressure, density, and height in a fluid.
The pressure in a fluid can be calculated using the formula:
P = P_0 + ρgh
Where:
P is the total pressure at a certain height
P_0 is the reference pressure (in this case, the atmospheric pressure at the surface of Titan)
ρ is the density of the fluid (in this case, the atmosphere of Titan)
g is the acceleration due to gravity (1.36 m/s^2 on Titan)
h is the height above the surface of Titan
Let's now rearrange the equation to solve for h:
h = (P - P_0)/(ρg)
Now, we can substitute the given values into the equation:
P = 1.01 x 10^5 Pa
P_0 = 1.60 x 10^5 Pa
ρ = 2.32 kg/m^3
g = 1.36 m/s^2
Plugging in these values:
h = (1.01 x 10^5 - 1.60 x 10^5) / (2.32 * 1.36)
h = -5.95 x 10^4 / 3.15
h ≈ -18898 m
The negative sign indicates that the height is below the surface. Therefore, the atmospheric pressure on Titan is equal to 1.01 x 10^5 Pa at a height of approximately 18898 meters (or 18.9 km) below the surface.