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Chapter 19: Problem 53

To get a feeling for the emptiness of the universe, compare its density \(\left(4 \times 10^{-28} \mathrm{kg} / \mathrm{m}^{3}\right)\) with that of Earth's atmosphere at sea level \(\left(1.2 \mathrm{kg} / \mathrm{m}^{3}\right) .\) How much denser is Earth's atmosphere? Write this ratio using standard notation.

Short Answer

Expert verified
\(3 \times 10^{27}\)

Step by step solution

01

Identify the Given Densities

The density of the universe is given as \(4 \times 10^{-28} \mathrm{kg} / \mathrm{m}^{3}\). The density of Earth's atmosphere at sea level is given as \(1.2 \mathrm{kg} / \mathrm{m}^{3}\).
02

Set Up the Ratio

Set up the ratio of the density of Earth's atmosphere to the density of the universe: \(\text{Ratio} = \frac{\text{Density of Earth's atmosphere}}{\text{Density of the universe}}\).
03

Substitute the Values

Substitute the given densities into the ratio: \(\text{Ratio} = \frac{1.2 \mathrm{kg} / \mathrm{m}^{3}}{4 \times 10^{-28} \mathrm{kg} / \mathrm{m}^{3}}\).
04

Calculate the Ratio

Perform the division: \[\text{Ratio} = \frac{1.2}{4 \times 10^{-28}} = 0.3 \times 10^{28} = 3 \times 10^{27}\].
05

Express in Standard Notation

Express the ratio in standard notation: \(3 \times 10^{27}\).

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Key Concepts

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

universe density
The density of the universe is an incredibly small value. When we talk about the universe's density, we refer to the typical mass spread over a large volume. This density is given as \(4 \times 10^{-28} \mathrm{kg} / \mathrm{m}^{3}\). This means that for every cubic meter of space, there is only \(4 \times 10^{-28} \mathrm{kg}\) of mass. To put this into perspective:
  • This value is almost zero, showing how empty space can be in the universe.
  • Even the emptiest regions of space still contain some particles and energy.
  • This density is much less than anything we encounter on earth.
Understanding the universe's density highlights how vast and sparse the cosmos is.
Earth's atmosphere density
In contrast to the universe, the Earth's atmosphere at sea level is much denser. The density given is \(1.2 \mathrm{kg} / \mathrm{m}^{3}\). This means that for every cubic meter of air, there are \(1.2 \mathrm{kg}\) of mass. Here's why:
  • The Earth's gravity pulls molecules of air towards its center, increasing the density.
  • The atmosphere is composed of various gases including nitrogen, oxygen, and other trace gases.
Compared to the universe, the density of Earth's atmosphere might not seem high, but it is substantial enough to sustain life by providing essential gases and creating pressure.
ratio calculation
To understand how much denser Earth's atmosphere is compared to the universe, we calculate the ratio of their densities. Here is a simple step-by-step process:
  • First, set up the ratio: \[ \text{Ratio} = \frac{\text{Density of Earth's atmosphere}}{\text{Density of the universe}} \]
  • Next, substitute the given densities: \[ \text{Ratio} = \frac{1.2 \mathrm{kg}/\mathrm{m}^{3}}{4 \times 10^{-28} \mathrm{kg}/\mathrm{m}^{3}} \]
  • Then, perform the division: \[ \text{Ratio} = \frac{1.2}{4 \times 10^{-28}} = 0.3 \times 10^{28} = 3 \times 10^{27} \]
  • Finally, express it in standard notation: \[3 \times 10^{27}\]
This ratio tells us that Earth's atmosphere is \[3 \times 10^{27}\] times denser than the universe. This large figure emphasizes how incredibly rarefied space is compared to our everyday environment.

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Most popular questions from this chapter

Cosmological redshifts are calculated from observations of spectral lines from a. individual stars in distant galaxies. b. clouds of dust and gas in distant galaxies. c. spectra of entire galaxies. d. rotations of the disks of distant galaxies.

Some galaxies have redshifts \(z\) that if equated to \(v / c\) correspond to velocities greater than the speed of light. Special relativity is not violated in this case a. because of relativistic beaming. b. because of superluminal motion. c. because redshifts carry no information. d. because these velocities do not measure motion through space.

The spectrum of a distant galaxy shows the Ho line of hydrogen \(\left(\lambda_{\text {rest }}=656.28 \mathrm{nm}\right)\) at a wavelength of \(750 \mathrm{nm}\) Assume that \(H_{\mathrm{o}}=70 \mathrm{km} / \mathrm{s} / \mathrm{Mpc}\) a. What is the redshift (z) of this galaxy? b. What is its recessional velocity \(\left(v_{r}\right)\) in kilometers per second? c. What is the distance of the galaxy in megaparsecs?

The average density of normal matter in the universe is \(4 \times 10^{-28} \mathrm{kg} / \mathrm{m}^{3} .\) The mass of a hydrogen atom is \(1.66 \times 10^{-27} \mathrm{kg}\) On average, how many hydrogen atoms are there in each cubic meter in the universe?

Go to the "Astronomy" page of the Phys.Org website (http://phys.org/space- news/astronomy), click on "Search," and enter "Type Ia supernova" in the Search box. Find a recent story about one of these supernovae. What is its distance and brightness? What type of star produced the explosion?
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