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Chapter 13: Problem 14

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

Short Answer

Expert verified
After performing the calculations, one would find the mass (M) and average density (蟻) of Rhea. Both results should be expressed in appropriate units: kilograms (kg) for mass and kilograms per cubic metre (kg/m鲁) for density.

Step by step solution

01

Calculate Mass

First, rearrange the gravitational formula to solve for mass: \(M = \frac{g \cdot r^2}{G}\). Substitute the given values: \(M = \frac{0.265 \mathrm{m/s^2} \cdot (764 \mathrm{km} \cdot 10^3 \mathrm{m/km})^2}{6.674 \times 10^{-11} \mathrm{Nm^2/kg^2}}\). Simplify to find M.
02

Calculate Volume

Next, compute the volume of Rhea using the spherical volume formula: \(V = \frac{4}{3}\pi r^3\). Substitute the given radius: \(V = \frac{4}{3} \pi (764 \mathrm{km} \cdot 10^3 \mathrm{m/km})^3\). Solve for V.
03

Calculate Average Density

Finally, use the mass and volume calculated in the previous steps to determine the average density of Rhea using the formula: \(\rho = \frac{M}{V}\). By plugging in the previously calculated values for M and V, the average density of Rhea can be found.

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Key Concepts

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

Gravitational formula
When calculating the mass of an astronomical object like Rhea, Saturn's moon, the gravitational formula is pivotal. The gravitational formula used here is derived from Newton's law of universal gravitation. It is given as:\[ M = \frac{g \cdot r^2}{G} \]Where:
  • \( M \) is the mass of the Moon,
  • \( g \) represents the surface gravity,
  • \( r \) is the radius of Rhea,
  • \( G \) is the universal gravitational constant which is approximately \( 6.674 \times 10^{-11} \mathrm{Nm^2/kg^2} \).
The formula is a rearrangement of Newton鈥檚 original formula \( F = \frac{GMm}{r^2} \), solving for \( M \) when the gravitational force \( F \) is expressed as \( mg \). To find the mass of Rhea, simply plug in the known values of \( g \), \( r \), and \( G \) and solve. This approach uses the gravitational pull experienced at the surface to backtrack and find the mass.
Volume calculation
To find the volume of a spherical object like Rhea, we use the formula for the volume of a sphere:\[ V = \frac{4}{3}\pi r^3 \]This tells us how much three-dimensional space Rhea occupies. Here:
  • \( V \) represents the volume,
  • \( r \) is the radius of the sphere.
In this instance, Rhea鈥檚 radius is given as \( 764 \mathrm{~km} \). To use in the formula, convert kilometers to meters to ensure consistency with the units of the gravitational constant \( G \) used in calculating mass. Multiply 764 km by 1000 to get meters.

Substitute the radius in meters into the volume formula and solve to find the volume. This illustrates the sheer magnitude of space occupied by Rhea and is essential for subsequent calculations like density.
Average density
Average density gives us insight into the internal composition of an object like Rhea. It is a measure of how much mass is contained within a given volume. The formula for average density \( \rho \) is:\[ \rho = \frac{M}{V} \]Where:
  • \( \rho \) is the average density,
  • \( M \) is the mass calculated using the gravitational formula,
  • \( V \) is the volume found through the volume calculation.
By using the mass from the gravitational formula and the volume from the volume calculation, the average density provides a clear picture of how densely packed Rhea is. Understanding this concept can help astronomers infer whether Rhea is composed mostly of ice, rock, or a combination of materials. Density is crucial because it reveals details about the body's composition that can lead to broader insights regarding its origins and geological activity.

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Most popular questions from this chapter

At what distance above the surface of the earth is the acceleration due to the earth's gravity \(0.980 \mathrm{~m} / \mathrm{s}^{2}\) if the acceleration due to gravity at the surface has magnitude \(9.80 \mathrm{~m} / \mathrm{s}^{2} ?\)

Comets travel around the sun in elliptical orbits with large eccentricities. If a comet has speed \(2.0 \times 10^{4} \mathrm{~m} / \mathrm{s}\) when at a distance of \(2.5 \times 10^{11} \mathrm{~m}\) from the center of the sun, what is its speed when at a distance of \(5.0 \times 10^{10} \mathrm{~m} ?\)

A narrow uniform rod has length \(2 a\). The linear mass density of the rod is \(\rho,\) so the mass \(m\) of a length \(l\) of the rod is \(\rho l\). (a) A point mass is located a perpendicular distance \(r\) from the center of the rod. Calculate the magnitude and direction of the force that the rod exerts on the point mass. (Hint: Let the rod be along the \(y\) -axis with the center of the rod at the origin, and divide the rod into infinitesimal segments that have length \(d y\) and that are located at coordinate \(y\). The mass of the segment is \(d m=\rho d y\). Write expressions for the \(x\) - and \(y\) -components of the force on the point mass, and integrate from \(-a\) to \(+a\) to find the components of the total force. Use the integrals in Appendix B.) (b) What does your result become for \(a \gg r ?\) (Hint: Use the power series for \((1+x)^{n}\) given in Appendix B.) (c) For \(a \gg r,\) what is the gravitational field \(g=\boldsymbol{F}_{g} / m\) at a distance \(r\) from the rod? (d) Consider a cylinder of radius \(r\) and length \(L\) whose axis is along the rod. As in part (c), let the length of the rod be much greater than both the radius and length of the cylinder. Then the gravitational ficld is constant on the curved side of the cylinder and perpendicular to it, so the gravitational flux \(\Phi_{g}\) through this surface is cqual to \(g A\), where \(A=2 \pi r L\) is the area of the curved side of the cylinder (see Problem 13.59 ). Calculate this flux. Write your result in terms of the mass \(M\) of the portion of the rod that is inside the cylindrical surface. How does your result depend on the radius of the cylindrical surface?

A planet orbiting a distant star has radius \(3.24 \times 10^{6} \mathrm{~m}\). The escape speed for an object launched from this planet's surface is \(7.65 \times 10^{3} \mathrm{~m} / \mathrm{s}\). What is the acceleration due to gravity at the surface of the planet?

In 2005 astronomers announced the discovery of a large black hole in the galaxy Markarian 766 having clumps of matter orbiting around once every 27 hours and moving at \(30,000 \mathrm{~km} / \mathrm{s}\). (a) How far are these clumps from the center of the black hole? (b) What is the mass of this black hole, assuming circular orbits? Express your answer in kilograms and as a multiple of our sun's mass. (c) What is the radius of its event horizon?
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