Question

Rhea, one of Saturn's moons, has a radius of 764 km and an acceleration due to gravity of 0.265 m/s${}^2$ at its surface. Calculate its mass and average density.

Short answer

The mass of Rhea is approximately $2.31\times {10}^{21}$ kg and its density is $1236{\text{ kg/m}}^3$.

Step-by-step solution

Calculate the Mass of Rhea

To find the mass of Rhea, we need to use the formula for gravitational force:$$g=\frac{GM}{r^2}$$Where:- $g=0.265{\text{ m/s}}^2$ (acceleration due to gravity)- $G=6.674\times {10}^{-11}{\text{ Nm}}^2/{\text{kg}}^2$ (gravitational constant)- $r=764\times {10}^3\text{ m}$ (radius of Rhea)Solve for $M$ (mass of Rhea):$$M=\frac{g{r}^2}{G}$$Substitute the values:$$M=\frac{0.265\times (764\times {10}^3{)}^2}{6.674\times {10}^{-11}}$$After computing, $M\approx 2.31\times {10}^{21}\text{ kg}$.

Calculate the Volume of Rhea

The volume of a sphere is given by the formula:$$V=\frac{4}{3}\pi r^3$$Using the radius $r=764\times {10}^3\text{ m}$, substitute into the formula:$$V=\frac{4}{3}\pi (764\times {10}^3{)}^3$$After computing, $V\approx 1.87\times {10}^{18}{\text{ m}}^3$.

Calculate the Density of Rhea

Density is mass divided by volume:$$\text{Density}=\frac{M}{V}$$Using the mass $M\approx 2.31\times {10}^{21}\text{ kg}$ and volume $V\approx 1.87\times {10}^{18}{\text{ m}}^3$:$$\text{Density}=\frac{2.31\times {10}^{21}\text{ kg}}{1.87\times {10}^{18}{\text{ m}}^3}$$After computing, the density $\approx 1236{\text{ kg/m}}^3$.

Key concepts

Mass Determination

Finding the mass of a celestial body, like a moon or a planet, involves using the gravitational force it exerts. For Rhea, one of Saturn's moons, we make use of the formula that links gravity and mass. The formula is:$$g=\frac{GM}{r^2}$$By rearranging this formula, we can solve for mass $M$:$$M=\frac{g{r}^2}{G}$$- $g$ represents the acceleration due to gravity at Rhea's surface, which we know is 0.265 m/s${}^2$.- $G$ is the gravitational constant, $6.674\times {10}^{-11}{\text{ Nm}}^2/{\text{kg}}^2$.- $r$ is the radius of Rhea, converted to meters as $764\times {10}^3\text{ m}$. By plugging in these values, we compute the mass:$$M=\frac{0.265\times (764\times {10}^3{)}^2}{6.674\times {10}^{-11}}$$Thus, Rhea's mass $M\approx 2.31\times {10}^{21}\text{ kg}$. This helps us understand how massive Rhea is in comparison to other celestial bodies.

Density Calculation

Once we know the mass of a celestial body like Rhea, calculating its density becomes straightforward. Density is defined as mass per unit volume. To find Rhea's density:$$\text{Density}=\frac{M}{V}$$Where:- $M$ is the mass found in the previous step $\approx 2.31\times {10}^{21}\text{ kg}$.- $V$ is the volume, calculated using the formula for a sphere.By dividing the mass by the volume, we find Rhea's density:$$\text{Density}=\frac{2.31\times {10}^{21}\text{ kg}}{1.87\times {10}^{18}{\text{ m}}^3}$$This gives a density of approximately 1236 kg/m${}^3$. Identifying the density helps us understand Rhea’s composition and structure, offering insights into its geology or how it may have formed.

Gravitational Force Formula

The gravitational force formula ties the force of gravity to the mass of an object and the distance from its center:$$g=\frac{GM}{r^2}$$- This equation states that $g$, the gravitational acceleration on the surface, depends on $G$, the universal gravitational constant, $M$, the mass of the celestial body, and $r$, the radius.- $G$ is a constant measuring the strength of gravity, valued at $6.674\times {10}^{-11}{\text{ Nm}}^2/{\text{kg}}^2$.A deeper understanding of this formula shows us that:

• As mass $M$ increases, surface gravity $g$ also increases, assuming radius $r$ remains constant.
• If the radius $r$ increases, $g$ decreases for the same mass due to the squared term.

This relationship illustrates why large planets with more mass have stronger gravity fields than smaller planets or moons like Rhea.

Volume of a Sphere

To calculate the volume of a spherical object like Rhea, we use the sphere volume formula:$$V=\frac{4}{3}\pi r^3$$This formula tells us that to find a sphere's volume, we need the cube of its radius $r$ multiplied by $\pi$ and then scaled by $\frac{4}{3}$:- Rhea's radius is given as $764\times {10}^3\text{ m}$.- Plugging in this value, we find the volume:$$V=\frac{4}{3}\pi (764\times {10}^3{)}^3$$By calculating, we derive a volume $V\approx 1.87\times {10}^{18}{\text{ m}}^3$. Understanding volume is crucial for calculating density and for understanding the spatial dimensions of the moon.