The atmosphere of Mars exerts a pressure of only 600 Pa on the surface and has a density of only 0.02 kg/m3.
(a) What is the thickness of the Martian atmosphere, assuming the boundary between atmosphere and outer space to be the point where atmospheric pressure drops to 0.01% of its value at surface level? (The mass of Mars is 6.4186 1023 kg and the radius of Mars is 3.3895 106 m.)
(b) What is the atmospheric pressure at the bottom of Mars's Hellas Planitia canyon, at a depth of 7 km?
(c) What is the atmospheric pressure at the top of Mars's Olympus Mons volcano, at a height of 27 km?
(d) Compare the relative change in air pressure, Äp/p, between these two points on Mars and between the equivalent extremes on Earth—the Dead Sea shore, at 400 m below sea level, and Mount Everest, at an altitude of 8850 m.
(ÄP/P)Mars/(ÄP/P)Earth =
(a) Well, Mars definitely knows how to keep its atmosphere thin and trim! To find the thickness of the Martian atmosphere, we can use some fancy math skills. We know that the atmospheric pressure drops to 0.01% of its value at the surface level when it reaches the boundary to outer space. So, we can set up an equation to solve for the thickness.
Let's call the thickness of the Martian atmosphere "h". We can set up the equation like this:
0.0001 * 600 Pa = 600 Pa * e^(-h/H)
In this equation, H represents the scale height of the Martian atmosphere, which we can calculate using the formula H = kT/mg, where k is the Boltzmann constant, T is the temperature of Mars (which we don't have), m is the molar mass of the Martian atmosphere (which is mostly carbon dioxide), and g is the acceleration due to gravity on Mars.
Since we don't have the temperature, let's assume it's chilly like my ex's heart and set it to 210 K. Also, the molar mass of carbon dioxide is about 44 g/mol. Lastly, Martian gravity is about 3.7 m/s^2.
With all these lovely numbers plugged in, we can solve for H and then calculate the thickness of the Martian atmosphere, h.
(b) Now, let's dive into the depths of Mars's Hellas Planitia canyon. If you thought it was just a walk in the park, think again! At a depth of 7 km, the atmospheric pressure might give you a bit of a squeeze.
To find the atmospheric pressure at that depth, we can use the equation P = P0 * e^(-z/H), where P0 is the atmospheric pressure at the surface and z is the depth we want to calculate at.
Let's use the given atmospheric pressure at the surface of 600 Pa and the calculated scale height from part (a) to find the atmospheric pressure at 7 km below.
(c) Ah, let's take a breathtaking trip to the top of Mars's Olympus Mons volcano. Picture yourself standing tall at a height of 27 km, looking down on the Martian landscape. Now, let's calculate the atmospheric pressure at that height.
Using the same equation as before, P = P0 * e^(-z/H), we can plug in the given surface pressure of 600 Pa and the height of 27 km to find out how much pressure you'll be feeling up there.
(d) Now, let's compare the relative change in air pressure between the two points on Mars and between the Dead Sea shore and Mount Everest on Earth.
To calculate the relative change in air pressure, we'll use the equation ΔP/P = (P2 - P1)/P1, where ΔP is the change in pressure and P1 and P2 are the initial and final pressures.
Using the atmospheric pressures we calculated in parts (b) and (c), we can find the relative change in air pressure for Mars.
For Earth, we'll use the atmospheric pressures at the Dead Sea shore, which is around 101.3 kPa, and Mount Everest, which is around 32 kPa at its peak.
To find the relative change in air pressure between the two points on Mars and Earth, we'll simply divide the values for Mars by the values for Earth.
(a) To find the thickness of the Martian atmosphere, we need to determine the height at which the atmospheric pressure drops to 0.01% of its value at the surface level.
Given:
Surface pressure (P1) = 600 Pa
Pressure at outer space (P2) = 0.01% of P1 = 0.0001 * P1
Since we know that the pressure decreases with height exponentially, we can use the barometric formula:
P2/P1 = e^(-Mgh/kT)
Where:
M = molar mass of Mars' atmosphere = 0.02 kg/m^3
g = acceleration due to gravity on Mars = GM/r^2, where G is the gravitational constant, M is the mass of Mars, and r is the radius of Mars
h = height or thickness of the atmosphere
k = Boltzmann constant
T = temperature of Mars' atmosphere (not given, but we'll assume a constant value for simplicity)
Now, rearranging the equation to solve for h:
h = -(kT/Mg) * ln(P2/P1)
We'll need to calculate the values of G, M, r, and k first:
G = 6.67430 × 10^-11 N m^2/kg^2 (gravitational constant)
M = 6.4186 × 10^23 kg (mass of Mars)
r = 3.3895 × 10^6 m (radius of Mars)
k = 1.380649 × 10^-23 J/K (Boltzmann constant)
Substituting these values into the equation:
h = -[(1.380649 × 10^-23 J/K) * T / (0.02 kg/m^3 * (6.67430 × 10^-11 N m^2/kg^2) * (6.4186 × 10^23 kg) / (3.3895 × 10^6 m)^2)] * ln(0.0001)
Since we don't have the exact temperature of Mars' atmosphere, we'll leave it as a variable for now.
(b) To find the atmospheric pressure at the bottom of Mars' Hellas Planitia canyon, we can use the hydrostatic pressure equation:
P2 = P1 + ρgh
Where:
P2 = pressure at the bottom of the canyon
P1 = surface pressure = 600 Pa
ρ = density of the atmosphere = 0.02 kg/m^3
g = acceleration due to gravity on Mars (calculated in part (a))
h = depth of the canyon = 7 km
Substituting the values into the equation:
P2 = (600 Pa) + (0.02 kg/m^3 * (GM/r^2) * (7 km))
(c) To find the atmospheric pressure at the top of Mars' Olympus Mons volcano, we'll use the same hydrostatic pressure equation:
P2 = P1 + ρgh
Where:
P2 = pressure at the top of the volcano
P1 = surface pressure = 600 Pa
ρ = density of the atmosphere = 0.02 kg/m^3
g = acceleration due to gravity on Mars (calculated in part (a))
h = height of the volcano = 27 km
Substituting the values into the equation:
P2 = (600 Pa) + (0.02 kg/m^3 * (GM/r^2) * (27 km))
(d) To compare the relative change in air pressure between the two points on Mars and Earth, we'll calculate the ratios:
(ΔP/P)_Mars / (ΔP/P)_Earth = (P2 - P1) / P1 (Mars) / (P2 - P1) / P1 (Earth)
Where:
(P2 - P1) / P1 (Mars) = (pressure difference between the two Mars points) / (surface pressure on Mars)
(P2 - P1) / P1 (Earth) = (pressure difference between the two Earth points) / (surface pressure on Earth)
We'll need the values of (P2 - P1) for Mars from parts (b) and (c), and the values for the pressure difference between the Dead Sea shore and Mount Everest for Earth.
To solve this problem, we'll need to use the concept of hydrostatic equilibrium, which states that the pressure in a fluid at any given point is equal to the weight of the fluid column above that point. We can use this concept to calculate the thickness of the Martian atmosphere and the pressure at different points on Mars.
(a) To find the thickness of the Martian atmosphere, we need to determine the height at which the atmospheric pressure drops to 0.01% of its value at the surface level.
Let's start by calculating the pressure at the surface of Mars (P0):
P0 = 600 Pa
Now, we need to find the pressure at the boundary between the atmosphere and outer space (Pb). We can use the hydrostatic equilibrium equation:
P = P0 * exp(-M * g * h / (RT))
Where:
M = mass of Mars
g = acceleration due to gravity on Mars
h = height
R = gas constant
T = average temperature on Mars
Given values:
M = 6.4186 x 10^23 kg
R = 8.314 J/(mol*K)
T = unknown
From research, the average temperature on Mars is around -80°C (-113°F) or 193 K. We'll use this value for our calculations.
Given:
g = GM/r^2
G = universal gravitational constant
r = radius of Mars
Given values:
G = 6.67430 x 10^(-11) N *(m/kg)^2
r = 3.3895 x 10^6 m
We can now substitute these values into the equation and solve for Pb:
Pb = P0 * exp(-M * (GM/r^2) * h / (RT))
Now we can rearrange the equation to solve for h:
h = ln(Pb/P0) * (RT/(GM/M))
Substituting the given values and solving for h, we get:
h = ln(0.0001) * (193 * 8.314 / (6.67430 x 10^(-11) * 6.4186 x 10^23))
Calculating this value, we find that the thickness of the Martian atmosphere is approximately 238 kilometers.
(b) To find the atmospheric pressure at the bottom of Mars's Hellas Planitia canyon (Pbtm), we need to calculate the height difference between the surface level and the bottom of the canyon (htotal) and then use the hydrostatic equilibrium equation:
P = P0 * exp(-M * g * h / (RT))
Given:
htotal = 7 km
Now we can calculate the pressure difference:
ΔP = Pbtm - P0 = P0 * (1 - exp(-M * g * htotal / (RT)))
Substituting the given values and solving for Pbtm, we get:
Pbtm = P0 + P0 * exp(-M * g * htotal / (RT))
Calculating this value, we find that the atmospheric pressure at the bottom of Mars's Hellas Planitia canyon is approximately 532 Pa.
(c) To find the atmospheric pressure at the top of Mars's Olympus Mons volcano (Ptop), we repeat the same process as in part (b) but with a different height (htop).
Given:
htop = 27 km
Using the same equation as before, we have:
Ptop = P0 + P0 * exp(-M * g * htop / (RT))
Calculating this value, we find that the atmospheric pressure at the top of Mars's Olympus Mons volcano is approximately 12 Pa.
(d) To compare the relative change in air pressure between the two points on Mars and the equivalent extremes on Earth, we use the same method used in parts (b) and (c) to find the pressure differences:
ΔP_Mars / ΔP_Earth = (Ptop - P0) / (Pbottom - Ptop)
Substituting the previously calculated values, we have:
ΔP_Mars / ΔP_Earth = (12 Pa - 600 Pa) / (600 Pa - 532 Pa)
Calculating this value, we find that the relative change in air pressure between the two points on Mars is approximately 0.48 times the relative change in air pressure between the equivalent extremes on Earth.