Question

The atmosphere of Mars exerts a pressure of only 600\. Pa on the surface and has a density of only \(0.0200 \mathrm{~kg} / \mathrm{m}^{3}\). 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 \(0.0100 \%\) of its yalue at surface level? b) What is the atmospheric pressure at the bottom of Mars's Hellas Planitia canyon, at a depth of \(7.00 \mathrm{~km} ?\) c) What is the atmospheric pressure at the top of Mars's Olympus Mons volcano, at a height of \(27.0 \mathrm{~km} ?\) d) Compare the relative change in air pressure, \(\Delta p / p\), between these two points on Mars and between the equivalent extremes on Earth-the Dead Sea shore, at \(400 . \mathrm{m}\) below sea level, and Mount Everest, at an altitude of \(8850 \mathrm{~m}\).

Short answer

#a) Finding the thickness of Martian atmosphere# #tag_title#Step 3: Determine the constant α#tag_content#Using the surface pressure and surface density provided, we can solve for the constant α: α = \(\frac{p_0}{\rho_0 c}\). #tag_title#Step 4: Calculate the thickness of the Martian atmosphere#tag_content#Now that we have the constant α, we can plug it into the equation from Step 2 to determine the height \(h_A\): \(h_A = \frac{1}{\alpha} \ln \frac{p_0}{0.0001p_0}\). b) Calculating the atmospheric pressure at the bottom of Hellas Planitia #tag_title#Step 1: Calculate the height difference#tag_content#The depth of Hellas Planitia, \(h_B\), is approximately 7 km. Hence, the height difference is -7 km (since it is below the surface). #tag_title#Step 2: Calculate the atmospheric pressure at the bottom of Hellas Planitia#tag_content#Using the relationship we derived in a), we can determine the pressure at the bottom of Hellas Planitia: \(p(h_B) = p_0 \cdot \mathrm{exp}(- \alpha h_B)\). c) Calculating the atmospheric pressure at the top of Olympus Mons #tag_title#Step 1: Calculate the height difference#tag_content#The height of Olympus Mons, \(h_C\), is approximately 21 km. #tag_title#Step 2: Calculate the atmospheric pressure at the top of Olympus Mons#tag_content#Using the relationship we derived in a), we can determine the pressure at the top of Olympus Mons: \(p(h_C) = p_0 \cdot \mathrm{exp}(- \alpha h_C)\). d) Comparing relative change in air pressure on Mars and Earth #tag_title#Step 1: Calculate the relative change in air pressure for Mars#tag_content#From part a), we calculated the pressure at the bottom of Hellas Planitia and the top of Olympus Mons. To determine the relative change in air pressure on Mars, we can use the formula: \(\frac{p(h_C) - p(h_B)}{p_0}\). #tag_title#Step 2: Calculate the relative change in air pressure for Earth#tag_content#Using the known values for Earth's atmosphere, calculate the relative change in air pressure from the highest point (Mount Everest) to the lowest point (Mariana Trench). \(\frac{p_{Everest} - p_{Mariana}}{p_{0, Earth}}\). #tag_title#Step 3: Compare the relative changes in air pressure#tag_content#Calculate the ratio of the relative change in air pressure between Mars and Earth: \(\frac{(\frac{p(h_C) - p(h_B)}{p_0})}{(\frac{p_{Everest} - p_{Mariana}}{p_{0, Earth}})}\).

Step-by-step solution

Derive the relationship between pressure, height, and density

Since we have density of the Martian atmosphere, we can use mass conservation to establish a relationship between the pressure and height. Let \(p(h) = p_0 \cdot \mathrm{exp}(- \alpha h)\), where \(\rho(h) = \rho_0 \cdot \mathrm{exp}(- \alpha h)\) is the density at height \(h\), \(\rho_0\) is the surface density, and \(p_0\) is the surface pressure. By integrating the mass conservation equation, we have: \(p = \frac{1}{\alpha c} \rho\), where \(c = \frac{g}{k}\), and \(\alpha\) is a constant to be determined.

Calculate the thickness of Martian atmosphere

The thickness of the Martian atmosphere will be the height when the pressure becomes 0.0100% of its surface value. So, let \(h_A\) be the height, and we have \(p(h_A) = 0.0001p_0\). Using the relationship derived in step 1, we can solve for \(h_A\): \(h_A = \frac{1}{\alpha} \ln \frac{p_0}{0.0001p_0}.\)

Key concepts

Atmospheric Pressure

Atmospheric pressure is the force exerted per unit area by the weight of the air above that area in the Earth's atmosphere. On Mars, atmospheric pressure is significantly lower than on Earth due to the planet's thinner atmosphere. It’s the pressure exerted by Mars' atmosphere at a particular location and varies with altitude.

Understanding atmospheric pressure on Mars is essential for many reasons, including the design of spacecraft that need to land safely or for predicting weather conditions. It also influences the boiling point of liquids and can affect the health and performance of astronauts on the Martian surface. In the exercise, atmospheric pressure was calculated at different altitudes to understand how it changes with elevation.

Density

Density measures the mass per unit volume of a substance. In the context of planetary atmospheres, density refers to the amount of air molecules within a certain volume of space. Mars' atmospheric density, for instance, is just about 0.0200 kg/m³ at surface level. This is drastically less than Earth's, accounting for the thinner atmosphere and reduced gravitational pull.

As altitude increases, the density of an atmosphere generally decreases because there are fewer air molecules at higher elevations. This relationship is depicted through exponential functions in the provided solution and helps explain phenomena like why it’s hard to breathe on high mountains due to thin air – a result of lower density.

Altitude

Altitude is the height of an object or point in relation to sea level or ground level. On Mars, altitude variations can considerably affect atmospheric pressure and density. With increasing altitude, air pressure decreases – a characteristic evident when calculating the pressure at different elevations of Martian topography, such as the depths of Hellas Planitia and the heights of Olympus Mons.

Different altitudes on Mars experience different atmospheric conditions, which can affect exploration and potential colonization. Conditions at different altitudes vary just like on Earth, and understanding these differences is vital when planning missions or researching Martian climate patterns.

Mass Conservation

The law of mass conservation states that mass is neither created nor destroyed within a closed system. In atmospheric science, this principle suggests that the total amount of air above a particular area remains constant unless there is an inflow or outflow of mass.

Applying mass conservation to the Martian atmosphere allows scientists to derive exponential relationships between pressure, density, and altitude. As shown in the exercise solution, it is possible to integrate these relationships to predict how pressure will change across different altitudes, assuming a static atmosphere without significant inflows or outflows. The step-by-step solution uses mass conservation to establish a foundational equation that connects atmospheric pressure and density with altitude on Mars.