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Oceans on Mars. Scientists have found evidence that Mars may once have had an ocean 0.500 \(\mathrm{km}\) deep. The acceleration due to gravity on Mars is 3.71 \(\mathrm{m} / \mathrm{s}^{2}\) , (a) What would be the gauge pressure at the bottom of such an ocean, assuming it was freshwater? (b) To what depth would you need to go in the earth's ocean to experience the same gauge pressure?

Short Answer

Expert verified
(a) 1,855,000 Pa (b) 189.1 m

Step by step solution

01

Understand Gauge Pressure

Gauge pressure is the pressure relative to atmospheric pressure. The formula to calculate gauge pressure \( P_g \) at a certain depth in a fluid is \( P_g = \rho gh \), where \( \rho \) is the density of the fluid, \( g \) is the gravitational acceleration, and \( h \) is the depth of the fluid.
02

Calculate Gauge Pressure on Mars

For Mars, we need to calculate the gauge pressure at a depth of 0.500 km using the Martian gravity and assuming freshwater (density, \( \rho = 1000 \, \mathrm{kg/m^3} \)). Convert the depth to meters, giving 500 m. Applying the formula: \[ P_g = \rho gh = 1000 \, \mathrm{kg/m^3} \times 3.71 \, \mathrm{m/s^2} \times 500 \, \mathrm{m} \] \[ P_g = 1,855,000 \, \mathrm{Pa} \]
03

Calculate the Equivalent Depth on Earth

To find the depth on Earth with the same gauge pressure, use the earth's gravitational constant (9.81 \, \mathrm{m/s^2}). Set the gauge pressure equal to the calculated Martian pressure and solve for depth \( h \): \[ 1000 \, \mathrm{kg/m^3} \times 9.81 \, \mathrm{m/s^2} \times h = 1,855,000 \, \mathrm{Pa} \] Solving for \( h \): \[ h = \frac{1,855,000}{1000 \times 9.81} \] \[ h = 189.1 \, \mathrm{m} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Fluid Mechanics
Fluid mechanics is the study of fluids and the forces acting on them. It is a fundamental branch of physics that deals with how liquids and gases behave in various conditions. One of the crucial concepts in fluid mechanics is pressure. Pressure is defined as the force exerted by a fluid per unit area. Understanding pressure is vital for analyzing fluid behavior in different scenarios, including planetary oceans.

A fluid can exert pressure due to its weight and density, and this pressure can vary with depth. In the context of our exercise, fluid mechanics helps us understand how a hypothetical Martian ocean would exert pressure at its bottom. This is known as gauge pressure, which is the pressure within the fluid minus atmospheric pressure. The understanding of this concept allows scientists to make informed predictions and calculations regarding ancient Martian oceans.
Mars Gravity
Mars gravity plays a significant role in determining the pressure exerted by fluids on the Martian surface. Compared to Earth, Mars has a weaker gravitational pull because it is smaller and less massive. The acceleration due to gravity on Mars is approximately 3.71 m/s².

This lower gravity affects how pressure is calculated in Martian conditions. With a weaker gravitational field, the pressure at a given depth would be less than on Earth. In our exercise, this is a key factor when determining the gauge pressure at the bottom of the hypothetical Martian ocean. It's these very differences in gravitational pull that allow us to explore how fluid behavior changes across planets.

Understanding Martian gravity is crucial for various scientific explorations as it affects everything from human activity on Mars to the planet's ability to sustain liquid water.
Ocean Depth
Ocean depth directly influences the pressure exerted at the ocean floor. The deeper you go in a fluid, the more pressure there is. This is because more fluid is above you, contributing to the force exerted downward.

In our exercise, the concept of ocean depth helps us understand how an ocean as deep as 0.500 km on Mars would behave in terms of pressure. Depth is a critical factor in fluid pressure equations and plays an integral role in oceanography studies on Earth and theoretical ocean studies on Mars.

Understanding the concept of depth not only helps in calculating pressure but also in analyzing potential environments that could harbor life or preserve signs of past life. Simply put, ocean depth is a fundamental parameter in the study of any large body of water, whether on Earth, Mars, or any other celestial body.
Pressure Calculation
Calculating pressure, especially gauge pressure, is a straightforward process once the formula is understood. The equation for calculating gauge pressure at a depth of a fluid is given by:
  • \( P_g = \rho gh \)
where:
  • \( P_g \) is the gauge pressure,
  • \( \rho \) is the density of the fluid (for water, typically taken as 1000 kg/m³),
  • \( g \) is the gravitational acceleration,
  • \( h \) is the depth of the fluid in meters.
The gauge pressure is essentially the weight of the fluid at a certain depth, excluding any atmospheric pressure.

In the standard exercise, pressure calculations were made for an ocean on Mars, using the density of freshwater and the Martian gravity constant to find the pressure at a specific depth. Then, knowing this pressure, a similar depth calculation was performed for Earth's ocean using Earth's gravity constant. Such calculations are essential for understanding fluid behavior under different gravitational conditions.

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Most popular questions from this chapter

A piece of wood is 0.600 \(\mathrm{m}\) long, 0.250 \(\mathrm{m}\) wide, and 0.080 \(\mathrm{m}\) thick. Its density is 600 \(\mathrm{kg} / \mathrm{m}^{3}\) . What volume of lead must be fastened underneath it to sink the wood in calm water so that its top is just even with the water level? What is the mass of this volume of lead?

A tall cylinder with a cross-sectional area 12.0 \(\mathrm{cm}^{2}\) is partially filled with mercury; the surface of the mercury is 5.00 \(\mathrm{cm}\) above the bottom of the cylinder. Water is slowly poured in on top of the mercury, and the two fluids don't mix. What volume of water must be added to double the gauge pressure at the bottom of the cylinder?

You are designing a diving bell to withstand the pressure of seawater at a depth of 250 \(\mathrm{m}\) . (a) What is the gauge pressure at this depth? (You can ignore changes in the density of the water with depth. \((b)\) At this depth, what is the net force due to the water outside and the air inside the bell on a circular glass window 30.0 \(\mathrm{cm}\) in diameter if the pressure inside the diving bell equals the pressure at the surface of the water? (You can ignore the small variation of pressure over the surface of the window.)

A golf course sprinkler system discharges water from a horizontal pipe at the rate of 7200 \(\mathrm{cm}^{3} / \mathrm{s}\) . At one point in the pipe, where the radius is 4.00 \(\mathrm{cm}\) , the water's absolute pressure is \(2.40 \times 10^{5} \mathrm{Pa}\) . At a second point in the pipe, the water passes through a constriction where the radius is \(2.00 \mathrm{cm} .\) What is the water's absolute pressure as it flows through this constriction?

A rock with mass \(m=3.00 \mathrm{kg}\) is suspended from the roof of an elevator by a light cord. The rock is totally immersed in a bucket of water that sits on the floor of the elevator, but the rock doesn't touch the bottom or sides of the bucket. (a) When the elevator is at rest, the tension in the cord is 21.0 \(\mathrm{N}\) . Calculate the volume of the rock. ( b) Derive an expression for the tension in the cord when the elevator is accelerating upward with an acceleration of magnitude a. Calculate the tension when \(a=2.50 \mathrm{m} / \mathrm{s}^{2}\) upward. (c) Derive an expression for the tension in the cord when the clevator is accelerating downwand with an acceleration of magnitude \(a\) . Calculate the tension when \(a=2.50 \mathrm{m} / \mathrm{s}^{2}\) downward. (d) What is the tension when the elevator is in free fall with a downward acceleration equal to \(g ?\)

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