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Chapter 12: Problem 7

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
The gauge pressure at the bottom of a half kilometre deep freshwater ocean on Mars would be \(1855000 \, \mathrm{Pa}\). The depth you would need to go in the Earth's ocean to experience the same gauge pressure would be approximately \(189 \, \mathrm{m}\).

Step by step solution

01

Gauge Pressure on Mars

First, calculate the pressure at the bottom of such an ocean on Mars. This can be done using the formula for pressure at a certain depth, which is P = 蟻gh. Here, 蟻 is the fluid density, g is the gravitational acceleration, and h is the fluid depth. Given: 蟻 (density of freshwater) = \( 1000 \, \mathrm{kg/m}^3 \), g (gravity on Mars) = \( 3.71 \, \mathrm{m/s}^2 \), and h (depth of the ocean on Mars) = \( 0.500 \, \mathrm{km} = 500 \, \mathrm{m} \). Substitute these values into the pressure equation: P = 蟻gh = \( 1000 \, \mathrm{kg/m}^3 \) * \( 3.71 \, \mathrm{m/s}^2 \) * \( 500 \, \mathrm{m} \) = \( 1855000 \, \mathrm{Pa} \).
02

Gauge Pressure on Earth

Next, use the calculated pressure value to find the depth needed to experience the same gauge pressure on Earth. We can use the same equation used earlier, P = 蟻gh, and solve it for h: h = P / (蟻g). Now, 蟻 is still the density of freshwater, which is \( 1000 \, \mathrm{kg/m}^3 \) but g now refers to the gravity on Earth, which is \( 9.81 \, \mathrm{m/s}^2 \). The pressure P is the one that was calculated in Step 1. Substitute these values into the equation: h = P / (蟻g) = \( 1855000 \, \mathrm{Pa} \) / [\( 1000 \, \mathrm{kg/m}^3 \) * \( 9.81 \, \mathrm{m/s}^2 \)] = \( 189 \, \mathrm{m} \).

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Key Concepts

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

Fluid Density
When we talk about fluid density, we're referring to the mass per unit volume of the fluid. In scientific terms, it's expressed as kilograms per cubic meter (\( \frac{kg}{m^3} \)). It's a crucial factor in calculating various physical properties, such as buoyancy and, importantly for our purposes, pressure.

For any fluid at rest, such as the water in an ocean or lake, the density remains fairly constant with depth if we assume it鈥檚 incompressible, like freshwater in our example. Bear in mind that fluid density can vary with temperature and salinity, but in the case of this exercise, we are considering the density of freshwater to be a constant 1000 \( \frac{kg}{m^3} \) 鈥� a standard assumption in many physics problems involving water.
Gravitational Acceleration
Gravitational acceleration is the acceleration of an object due to the gravitational force of a massive body, like a planet or moon. On Earth, we usually take this as 9.81 meters per second squared (\( 9.81 \frac{m}{s^2} \)). This value, however, isn't universal; it changes depending on where you are in the universe. For instance, Mars has a gravitational acceleration of 3.71 \( \frac{m}{s^2} \), which is only about 38% of Earth's.

This difference has a noticeable effect on the pressure experienced at the same depth on different celestial bodies. Because gravitational force plays a pivotal role, the same volume of water would weigh less on Mars than on Earth, resulting in a lower pressure at the bottom of the Martian ocean compared to an ocean on Earth at the same depth.
Pressure-Depth Relationship
The pressure-depth relationship is a fundamental concept in fluid mechanics indicating that pressure increases with depth within a fluid. This relationship is described by the equation \( P = \rho gh \) where \( P \) represents the fluid pressure at a certain depth, \( \rho \) is the fluid density, \( g \) is the gravitational acceleration, and \( h \) is the depth of the fluid.

The relationship tells us that the deeper you go, the higher the pressure, due to the weight of the fluid above pushing down. It鈥檚 important to note that this equation assumes uniform gravitational acceleration and fluid density - a fair approximation for many practical calculations.

Applying this relationship to our Mars scenario shows that the gauge pressure at the bottom of a 0.500 km deep ocean would be significantly less compared to what you'd feel at the same depth in Earth's oceans because of the lesser gravitational acceleration.

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Most popular questions from this chapter

What gauge pressure is required in the city water mains for a stream from a fire hose connected to the mains to reach a vertical height of \(15.0 \mathrm{~m} ?\) (Assume that the mains have a much larger diameter than the fire hose.

A cubical block of density \(\rho_{\mathrm{B}}\) and with sides of length \(L\) floats in a liquid of greater density \(\rho_{\mathrm{L}}\). (a) What fraction of the block's volume is above the surface of the liquid? (b) The liquid is denser than water (density \(\rho_{\mathrm{W}}\) ) and does not mix with it. If water is poured on the surface of that liquid, how deep must the water layer be so that the water surface just rises to the top of the block? Express your answer in terms of \(L, \rho_{\mathrm{B}}, \rho_{\mathrm{L}},\) and \(\rho_{\mathrm{W}}\). (c) Find the depth of the water layer in part (b) if the liquid is mercury, the block is made of iron, and \(L=10.0 \mathrm{~cm}\).

You are doing experiments from a research ship in the Atlantic Ocean. On a day when the atmospheric pressure at the surface of the water is \(1.03 \times 10^{5} \mathrm{~Pa}\), at what depth below the surface of the water is the absolute pressure (a) twice the pressure at the surface and (b) four times the pressure at the surface?

A uniform lead sphere and a uniform aluminum sphere have the same mass. What is the ratio of the radius of the aluminum sphere to the radius of the lead sphere?

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? (Ignore the small variation of pressure over the surface of the window.)
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