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What is the total mass of the Earth's atmosphere? (The \(\mathrm{ra}\) dius of the Earth is \(6.37 \times 10^{6} \mathrm{m},\) and atmospheric pressure at the surface is \(1.013 \times 10^{5} \mathrm{N} / \mathrm{m}^{2} .\) )

Short Answer

Expert verified
The total mass of Earth's atmosphere is approximately \(5.15 \times 10^{18}\) kg.

Step by step solution

01

Calculate the Earth's Surface Area

Firstly, calculate the surface area of the Earth using the provided radius. The formula for the surface area of a sphere is \( 4\pi r^{2} \). Substituting the radius of the Earth (\(6.37 \times 10^{6}\) m), we get \(4\pi(6.37 \times 10^{6})^{2}\).
02

Calculate the Atmosphere's Force

The total force exerted by the atmosphere is the atmospheric pressure at the surface multiplied by the surface area of the Earth. Substituting the known values (pressure is \(1.013 \times 10^{5}\) N/m² and the surface area calculated in the first step), the total force will be \(1.013 \times 10^{5}\) N/m² x \(4\pi(6.37 \times 10^{6})^{2}\) m².
03

Calculate the Total Mass of the Atmosphere

Lastly, calculate the total mass of the atmosphere. The force due to gravity is the atmospheric force divided by the acceleration due to gravity. Since, on Earth, the acceleration due to gravity is approximately \(9.8\) m/s², the total mass of the atmosphere will be the total force of the atmosphere divided by \(9.8\) m/s².

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Key Concepts

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

Atmospheric Pressure
Atmospheric pressure is a vital concept in understanding the mass of the Earth's atmosphere. It refers to the force exerted by the weight of air molecules on a unit area of the Earth's surface. This force is due to the air molecules colliding with the surface and exerts a pressure that we experience daily, although we may not be conscious of it.

Imagine the atmosphere as a vast ocean of air, with layers of gases pressing down towards the Earth under the influence of gravity. The atmospheric pressure at the Earth's surface is a standard reference point and is measured by a barometer. The average atmospheric pressure at sea level is approximately 101,300 newtons per square meter (\(1.013 \times 10^{5} \text{N/m}^2\)), which is also known as one atmosphere (\text{atm}).

This pressure is crucial for calculating the total force exerted by the atmosphere, as it helps us understand the relationship between force and area. Without knowing the atmospheric pressure, we would be unable to accurately determine how much mass is suspended above us in the form of gases.
Earth's Surface Area
The Earth's surface area is a fundamental quantity needed to calculate the mass of the Earth's atmosphere. Covering approximately 510 million square kilometers, the Earth's surface is spherical, and its size can be calculated using the formula for the surface area of a sphere, which is \( 4\pi r^{2} \), where \( r \) is the radius of the Earth.

To put this into context for our problem, with the radius of the Earth given as \(6.37 \times 10^{6} \) meters, the surface area can be determined by squaring the radius, and then multiplying by four times \(\pi\). This calculation provides us with the total area across which the atmospheric pressure acts, which is essential for calculating the total force exerted by the Earth's atmosphere.

Understanding the vastness of this area is crucial: it is not just a patch of land here or there; we're considering the entirety of Earth's surface. Multiplying the atmospheric pressure by this huge surface area gives us an insight into the scale of the force created by the Earth's atmosphere pressing down on the surface.
Gravitational Force
Gravitational force is the invisible 'hand' that holds the Earth's atmosphere in place. It is the force of attraction between two masses, such as the Earth and atmosphere. Everything with mass is pulled towards every other thing with mass by gravity, which is why the atmosphere doesn't just float off into space.

The strength of this force on the surface of the Earth is known as the acceleration due to gravity and is approximately \(9.8 \) m/s². With this, we can understand how the mass of the atmosphere can be inferred from the force it exerts: by dividing the total atmospheric force by Earth's gravitational acceleration.

It's a beautiful demonstration of how gravity not only keeps our feet on the ground but also maintains a layer of gases essential for life as we know it. So, when you consider the formula used in the textbook solution, remember that it represents more than numbers and units; it's a small window into the profound interaction between Earth's mass and the sea of air above us.

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Most popular questions from this chapter

The United States possesses the eight largest warships in the world-aircraft carriers of the Nimilz class-and is building two more. Suppose one of the ships bobs up to float \(11.0 \mathrm{cm}\) higher in the water when 50 fighters take off from it in 25 min, at a location where the free-fall acceleration is \(9.78 \mathrm{m} / \mathrm{s}^{2} .\) Bristling with bombs and missiles, the planes have average mass 29 000 kg. Find the horizontal area enclosed by the waterline of the S1-billion ship. By comparison, its flight deck has area 18 000 \(\mathrm{m}^{2}\). Below decks are passageways hundreds of meters long, so narrow that two large men cannot pass each other.

The small piston of a hydraulic lift has a cross-scrional area of \(3.00 \mathrm{cm}^{2},\) and its large piston has a cross-sectional area of \(\left.200 \mathrm{cm}^{2} \text { (Figure } 14.4\right) .\) What force must be applied to the small piston for the lift to raise a load of \(15.0 \mathrm{kN} ?\) (In service stations, this force is usually exerted by compressed air.)

Normal atmospheric pressure is \(1.013 \times 10^{5} \mathrm{Pa}\). The approach of a storm causes the height of a mercury barometer to drop by \(20.0 \mathrm{mm}\) from the normal height. What is the atmospheric pressure? (The density of mercury is \(\left.13.59 \mathrm{g} / \mathrm{cm}^{3} .\right)\)

(a) Calculate the absolute pressure at an ocean depth of \(1000 \mathrm{m} .\) Assume the density of seawater is \(1024 \mathrm{kg} / \mathrm{m}^{3}\) and that the air above exerts a pressure of \(101.3 \mathrm{kPa}\) (b) At this depth, what force must the frame around a circular submarine porthole having a diameter of \(30.0 \mathrm{cm}\) exert to counterbalance the force exerted by the water?

Find the order of magnitude of the density of the nucleus of an atom. What does this result suggest concerning the structure of matter? Model a nucleus as protons and neutrons closely packed together. Each has mass \(1.67 \times 10^{-27} \mathrm{kg}\) and radius on the order of \(10^{-15} \mathrm{m}\)

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