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Chapter 13: Problem 14

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.

Short Answer

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

Step by step solution

01

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 \( M \).
02

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 \( M = 2.31 \times 10^{21} \, \mathrm{kg} \).
03

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 \( V \approx 1.879 \times 10^{18} \, \mathrm{m^3} \).
04

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 \( \rho = \frac{2.31 \times 10^{21}}{1.879 \times 10^{18}} \approx 1230 \, \mathrm{kg/m^3} \).

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Key Concepts

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

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鈥攊t 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.

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Most popular questions from this chapter

Your starship, the Aimless Wanderer, lands on the mysterious planet Mongo. As chief scientist-engineer, you make the following measurements: \(A 2.50-\) -kg stone thrown upward from the ground at 12.0 \(\mathrm{m} / \mathrm{s}\) returns to the ground in 6.00 s; the circumference of Mongo at the equator is \(2.00 \times 10^{5} \mathrm{km} ;\) and there is no appreciable atmosphere on Mongo. The starship commander, Captain Confusion, asks for the following information: (a) What is the mass of Mongo? (b) If the Aimless Wanderer goes into a circular orbit \(30,000 \mathrm{km}\) above the surface of Mongo, how many hours will it take the ship to complete one orbit?

Two satellites are in circular orbits around a planet that has radius \(9.00 \times 10^{6} \mathrm{m}\) . One satellite has mass 68.0 \(\mathrm{kg}\) , orbital radius \(5.00 \times 10^{7} \mathrm{m},\) and orbital speed 4800 \(\mathrm{m} / \mathrm{s} .\) The second satellite has mass 84.0 \(\mathrm{kg}\) and orbital radius \(3.00 \times 10^{7} \mathrm{m} .\) What is the orbital speed of this second satellite?

At the Galaxy's Core. Astronomers have observed a small, massive object at the center of our Milky Way galaxy (see Section 13.8 ). A ring of material orbits this massive object; the ring has a diameter of about 15 light-years and an orbital speed of about 200 \(\mathrm{km} / \mathrm{s} .\) (a) Determine the mass of the object at the center of the Milky Way galaxy. Give your answer both in kilograms and in solar masses (one solar mass is the mass of the sun). (b) Observations of stars, as well as theories of the structure of stars, suggest that it is impossible for a single star to have a mass of more than about 50 solar masses. Can this massive object be a single, ordinary star? (c) Many astronomers believe that the massive object at the center of the Milky Way galaxy is a black hole. If so, what must the Schwarzchild radius of this black hole be? Would a black hole of this size fit inside the earth's orbit around the sun?

Kirkwood Gaps. Hundreds of thousands of asteroids orbit the sun within the asteroid belt, which extends from about \(3 \times 10^{8} \mathrm{km}\) to about \(5 \times 10^{8} \mathrm{km}\) from the sun. (a) Find the orbital period (in years) of (i) an asteroid at the inside of the belt and (ii) an asteroid at the outside of the belt. Assume circular orbits. (b) In 1867 the American astronomer Daniel Kirkwood pointed out that several gaps exist in the asteroid belt where relatively few asteroids are found. It is now understood that these Kirkwood gaps are caused by the gravitational attraction of Jupiter, the largest planet, which orbits the sun once every 11.86 years. As an example, if an asteroid has an orbital period half that of Jupiter, or 5.93 years, on every other orbit this asteroid would be at its closest to Jupiter and feel a strong attraction toward the planet. This attraction, acting over and over on successive orbits, could sweep asteroids out of the Kirkwood gap. Use this hypothesis to determine the orbital radius for this Kirkwood gap. (c) One of several other Kirkwood gaps appears at a distance from the sun where the orbital period is 0.400 that of Jupiter. Explain why this happens, and find the orbital radius for this Kirkwood gap.

Binary Star-Different Masses. Two stars, with masses \(M_{1}\) and \(M_{2},\) are in circular orbits around their center of mass. The star with mass \(M_{1}\) has an orbit of radius \(R_{1} ;\) the star with mass \(M_{2}\) has an orbit of radius \(R_{2} .\) (a) Show that the ratio of the orbital radii of the two stars equals the reciprocal of the ratio of their masses - that is, \(R_{1} / R_{2}=M_{2} / M_{1}\) . (b) Explain why the two stars have the same orbital period, and show that the period \(T\) is given by \(T=2 \pi\left(R_{1}+R_{2}\right)^{3 / 2} / \sqrt{G\left(M_{1}+M_{2}\right)}\) . (c) The two stars in a certain binary star system move in circular orbits. The first star, Alpha, has an orbital speed of 36.0 \(\mathrm{km} / \mathrm{s}\) . The second star, Beta, has an orbital speed of 12.0 \(\mathrm{km} / \mathrm{s}\) . The orbital period is 137 \(\mathrm{d}\) . What are the masses of each of the two stars? (d) One of the best candidates for a black hole is found in the binary system called A \(0620-0090 .\) The two objects in the binary system are an orange star, V616 Monocerotis, and a compact object believed to be a black hole (see Fig. 13.27\() .\) The orbital period of \(\mathrm{A} 0620-0090\) is 7.75 hours, the mass of V616 Monocerotis is estimated to be 0.67 times the mass of the sun, and the mass of the black hole is estimated to be 3.8 times the mass of the sun. Assuming that the orbits are circular, find the radius of each object's orbit and the orbital speed of each object. Compare these answers to the orbital radius and orbital speed of the earth in its orbit around the sun.
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