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.

Density is a core concept in understanding many physical phenomena. It measures how much mass is contained in a given volume. Mathematically, it is defined by the formula: \( \rho = \frac{m}{V} \), where \( \rho \) represents density, \( m \) denotes mass, and \( V \) stands for volume. A higher density indicates that there is more mass packed into a particular space. For example, metals like iron have a much higher density compared to substances like water or air.

The universe is immense and mostly empty, making it a fascinating area of study in astronomy. One of the striking characteristics of the universe is its incredibly low average density. On average, the density of the universe is around \( 4 \times 10^{-28} \text{kg/m}^3 \). This extremely low density is due to the vast spaces between celestial objects, such as stars and galaxies. Despite the occasional dense clusters of matter in the form of stars, planets, and galaxies, most of the universe is empty space, creating a stark contrast with the more familiar densities we encounter here on Earth.

Closer to home, the density of Earth's atmosphere varies greatly with altitude. At sea level, the density is about \( 1.2 \text{kg/m}^3 \). This density means that there is 1.2 kilograms of atmospheric gases, primarily nitrogen and oxygen, per cubic meter of space. This atmospheric layer is thick enough to support life by providing essential gases and protecting us from harmful solar radiation.

Ratio calculation helps us compare different quantities in a meaningful way. In this exercise, we compare the density of Earth's atmosphere to the density of the universe by calculating their ratio. We use the formula: \( \text{Ratio} = \frac{\rho_{\text{atmosphere}}}{\rho_{\text{universe}}} \). Plugging in the given values: \( \rho_{\text{atmosphere}} = 1.2 \text{kg/m}^3 \) and \( \rho_{\text{universe}} = 4 \times 10^{-28} \text{kg/m}^3 \), the ratio calculation becomes: \( \text{Ratio} = \frac{1.2}{4 \times 10^{-28}} = 0.3 \times 10^{28} = 3 \times 10^{27} \). This indicates that Earth's atmosphere is \( 3 \times 10^{27} \) times denser than the universe, highlighting just how sparse matter is distributed in the vast expanse of space.