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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) 1855 kPa on Mars; (b) 189.1 m depth on Earth matches this pressure.

Step by step solution

01

Understand the problem

We are asked to calculate the gauge pressure at the bottom of a Martian ocean that is 0.500 km deep, with gravity on Mars being 3.71 m/s². Then, we need to find out the corresponding depth in Earth's ocean that gives the same gauge pressure.
02

Convert ocean depth from km to meters

The given depth of the Martian ocean is 0.500 km. Convert this to meters by multiplying by 1000:\( 0.500 \times 1000 = 500 \ \mathrm{m} \)
03

Calculate the gauge pressure on Mars

Gauge pressure is calculated using the formula \( P = \rho g h \) where \( \rho \) is the density of water (assume \( 1000 \ \mathrm{kg/m^3} \)), \( g \) is the acceleration due to gravity, and \( h \) is the depth.\[ P = 1000 \ \mathrm{kg/m^3} \times 3.71 \ \mathrm{m/s^2} \times 500 \ \mathrm{m} \]\[ P = 1855000 \ \mathrm{N/m^2} \text{ or } 1855 \ \mathrm{kPa} \]
04

Calculate the required depth on Earth

Use the calculated gauge pressure \( 1855 \ \mathrm{kPa} \) and Earth's gravitational acceleration \( 9.81 \ \mathrm{m/s^2} \) to find the depth \( h \) on Earth.\[ h = \frac{P}{\rho g} = \frac{1855 \ \times 10^3}{1000 \times 9.81} \]\[ h \approx 189.1 \ \mathrm{m} \]

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Key Concepts

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

Martian gravity
Mars, being a unique planet in our solar system, has gravity that is quite different from what we experience here on Earth. The gravitational acceleration on Mars is about 3.71 m/s². This is only about 38% of Earth's gravity, making it significantly weaker compared to the 9.81 m/s² on Earth.

Understanding gravity is crucial when dealing with physical equations and measurements, especially when calculating pressures under large bodies of liquid. Due to its lower gravity, the pressure exerted by a column of water (or any liquid) on Mars would be different from that on Earth, even if the liquid column had the same height and density.

With less gravitational force pulling down on an object or liquid, Martian gravity changes how fluids behave on the planet's surface. This results in different experiences of forces like buoyancy and pressure.
Freshwater ocean
While the idea of a freshwater ocean on Mars is speculative yet fascinating, it brings up important concepts about water's properties. Freshwater, on Earth and potentially Mars, has a density of approximately 1000 kg/m³. This value is crucial for calculating pressures in water bodies using equations that require fluid density.

On Mars, a hypothetical ocean composed of freshwater would provide a unique environment to study how waters might behave differently compared to the Earth's oceans. This raises interesting questions about potential ecosystems, the chemistry of Martian water, and the planet's ability to have supported life in its past.
  • Density affects pressure exerted by the water column.
  • Water's high density makes it an efficient medium for pressure transmission.
This simplifies calculations when using the pressure equation because the density remains the same as on Earth.
Pressure equation
The pressure equation forms the backbone of many problems dealing with fluid mechanics. It is essential for calculating the pressure exerted by a fluid at a given depth.

The formula for gauge pressure in a fluid is given by:
  • \( P = \rho g h \)
Where:
  • \( P \) is the gauge pressure.
  • \( \rho \) is the fluid density.
  • \( g \) is the gravitational acceleration.
  • \( h \) is the depth of the fluid column.
This formula helps us understand how pressure increases with depth due to the weight of the fluid above. It highlights the direct proportionality between pressure and both the depth and density, allowing us to solve problems involving pressure differences in fluids efficiently. Understanding this equation is key to solving the exercise described, as it directly links the conditions of different celestial environments through comparison.
Earth's gravitational acceleration
Earth's gravitational acceleration, typically averaged at 9.81 m/s², plays a crucial role in everyday phenomena we observe, like objects falling and fluid pressure.

On Earth, this constant is involved in determining how deep one must dive underwater to experience a specific pressure, aligning with the same principle used on Mars in this exercise.
  • Gravity affects weight and the downward force exerted by objects and fluids.
  • At higher gravitational acceleration, the same mass exerts more force, increasing pressure.
This helps us calculate and compare pressures across different environments. By applying Earth's gravity in the pressure equation, we can determine how deep we must venture in Earth's ocean to match the conditions on Mars. This demonstrates the versatility of applying physical laws across celestial settings, deepening our comprehension of atmospheres, pressures, and forces.

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Most popular questions from this chapter

When an open-faced boat has a mass of 5750 kg, including its cargo and passengers, it floats with the water just up to the top of its gunwales (sides) on a freshwater lake. (a) What is the volume of this boat? (b) The captain decides that it is too dangerous to float with his boat on the verge of sinking, so he decides to throw some cargo overboard so that 20\(\%\) of the boat's volume will be above water. How much mass should he throw out?

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 the 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 the side length is 10.0 \(\mathrm{cm} .\)

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 \(2.40 \times 10^{5}\) 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 barrel contains a \(0.120-\mathrm{m}\) layer of oil floating on water that is 0.250 \(\mathrm{m}\) deep. The density of the oil is 600 \(\mathrm{kg} / \mathrm{m}^{3} .\) (a) What is the gauge pressure at the oil-water interface? (b) What is the gauge pressure at the bottom of the barrel?

A barge is in a rectangular lock on a freshwater river. The lock is 60.0 \(\mathrm{m}\) long and 20.0 \(\mathrm{m}\) wide, and the steel doors on each end are closed. With the barge floating in the lock, a \(2.50 \times 10^{6} \mathrm{N}\) load of scrap metal is put onto the barge. The metal has density 9000 \(\mathrm{kg} / \mathrm{m}^{3} .\) (a) When the load of scrap metal, initially on the bank, is placed onto the barge, what vertical distance does the water in the lock rise? (b) The scrap metal is now pushed overboard into the water. Does the water level in the lock rise, fall, or remain the same? If it rises or falls, by what vertical distance does it change?

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