Problem 14 - University Physics with Modern Physics

Question

Rhea, one of Saturn's moons, has a radius of 765 \(\mathrm{km}\) and an acceleration due to gravity of 0.278 \(\mathrm{m} / \mathrm{s}^{2}\) at its surface. Calculate its mass and average density.

Step-by-Step Solution

Answer

Rhea's mass is approximately \( 2.31 \times 10^{21} \mathrm{kg} \) and its density is \( 1230 \mathrm{kg/m^3} \).

Step 1: Calculate Rhea's mass using the formula for gravitational acceleration

The formula for gravitational acceleration at the surface of a spherical body is \( g = \frac{G \cdot M}{r^2} \), where \( g \) is the acceleration due to gravity, \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \, \mathrm{m^3 \, kg^{-1} \, s^{-2}}) \), \( M \) is the mass of the body, and \( r \) is its radius. Rearrange to find \( M \): \( M = \frac{g \cdot r^2}{G} \). Plug in the values: \( g = 0.278 \, \mathrm{m/s^2} \), \( r = 765,000 \, \mathrm{m} \). Substitute these values to find \( M \).

Step 2: Perform the mass calculation

Using the values from Step 1: \( g = 0.278 \), \( r = 765,000 \), and \( G = 6.674 \times 10^{-11} \), the mass \( M \) can be calculated as \( M = \frac{0.278 \times (765,000)^2}{6.674 \times 10^{-11}} \). Calculating this gives \( M = 2.31 \times 10^{21} \, \mathrm{kg} \).

Step 3: Calculate the volume of Rhea

The volume \( V \) of a sphere is calculated using the formula \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius. For Rhea, \( r = 765,000 \, \mathrm{m} \). Calculate \( V = \frac{4}{3} \pi (765,000)^3 \). This gives \( V \approx 1.879 \times 10^{18} \, \mathrm{m^3} \).

Step 4: Calculate the average density of Rhea

Density \( \rho \) is defined as mass per unit volume: \( \rho = \frac{M}{V} \). Using \( M = 2.31 \times 10^{21} \mathrm{kg} \) and \( V = 1.879 \times 10^{18} \mathrm{m^3} \), the density \( \rho \) can be calculated as \( \rho = \frac{2.31 \times 10^{21}}{1.879 \times 10^{18}} \approx 1230 \, \mathrm{kg/m^3} \).

Key Concepts

Acceleration due to Gravity

Acceleration due to gravity is a key concept in gravitational physics and describes how fast an object speeds up as it falls towards a massive body. It's measured in meters per second squared \(\mathrm{(m/s^2)}\). This acceleration depends on the mass and radius of the celestial body you are considering.

• The formula used is \(g = \frac{G \cdot M}{r^2}\), where \(g\) is the gravitational acceleration, \(G\) stands for the gravitational constant \((6.674 \times 10^{-11} \, \mathrm{m^3 \, kg^{-1} \, s^{-2}})\), \(M\) is the mass of the planet or moon, and \(r\) is its radius.
• This formula shows that gravity depends on both how heavy and how large a celestial object is.

For smaller celestial bodies like moons, the gravity is often much weaker compared to planets like Earth. On Rhea, one of Saturn's moons, the acceleration due to gravity is only about \(0.278 \, \mathrm{m/s^2}\). Understanding this concept helps us appreciate why different planets and moons have varying gravitational pulls.

Mass Calculation

Calculating the mass of a celestial object is crucial in understanding its gravity effects. By rearranging the formula for gravity's acceleration, we can determine the mass of the body:\[M = \frac{g \cdot r^2}{G}\]

• Here, \(g\) is the given acceleration due to gravity on the surface, \(r\) represents the radius, and \(G\) is the gravitational constant.
• This equation shows that knowing the gravity and size (radius) allows us to calculate how massive the object is.
• For Rhea, we find its mass to be \(2.31 \times 10^{21} \, \mathrm{kg}\) using an acceleration of \(0.278 \, \mathrm{m/s^2}\) and radius of \(765,000 \, \mathrm{m}\).

This calculated mass helps predict how Rhea interacts gravitationally with other bodies in space, and is essential for mission planning in space exploration.

Average Density

Average density is a measure of how much mass is contained in a given volume and is calculated using the formula: \[\rho = \frac{M}{V}\]where \(M\) is mass, and \(V\) is volume.

• An object's density provides insights into its composition—it can tell whether it is mostly rock, metal, ice, or a combination of these materials.
• For Rhea, with mass \(2.31 \times 10^{21} \, \mathrm{kg}\) and volume \(1.879 \times 10^{18} \, \mathrm{m^3}\), we calculate its average density to be approximately \(1230 \, \mathrm{kg/m^3}\).

A density lower than Earth's average implies Rhea might contain a significant amount of ice compared to rock. Insights into a celestial body's density assist scientists in hypothesizing its internal structure and surface characteristics.

Volume of Sphere

The volume of a sphere is an important geometric property used to assess the space inside a spherical object. The formula for volume is given by:\[V = \frac{4}{3} \pi r^3\]where \(r\) is the radius of the sphere.

• This formula shows us that any sphere's volume depends directly on the cube of its radius.
• For Rhea, using \(r = 765,000 \, \mathrm{m}\), the volume is approximately \(1.879 \times 10^{18} \, \mathrm{m^3}\).

Understanding volume is crucial in calculating other properties like density. It helps in analyzing the distribution of mass within a celestial body and plays a part in both theoretical models and practical applications, such as designing spacecraft trajectories.