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Titania, the largest moon of the planet Uranus, has \(\frac{1}{8}\) the radius of the earth and \(\frac{1}{1700}\) the mass of the earth. (a) What is the acceleration due to gravity at the surface of Titania? (b) What is the average density of Titania? (This is less than the density of rock, which is one piece of evidence that Titania is made primarily of ice.)

Short Answer

Expert verified
(a) Approximately 0.369 m/s², (b) Approximately 1658 kg/m³.

Step by step solution

01

Understand Gravity on Titania

The formula for gravitational acceleration at the surface of a sphere is given by \( g = \frac{G \cdot M}{R^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the celestial body, and \( R \) is its radius. We need to find the acceleration due to gravity on Titania using its mass and radius relative to Earth.
02

Calculate Radius and Mass of Titania

Let \( R_e \) and \( M_e \) be the radius and mass of the Earth respectively. Then, the radius of Titania \( R_t \) is \( \frac{1}{8}R_e \) and the mass of Titania \( M_t \) is \( \frac{1}{1700}M_e \).
03

Substitute Values into Gravity Equation

Substituting into the gravity equation, we have:\[ g_t = \frac{G \cdot \frac{M_e}{1700}}{(\frac{R_e}{8})^2} \]This simplifies to \[ g_t = \frac{G \cdot M_e}{1700 \cdot \frac{R_e^2}{64}} = \frac{64G \cdot M_e}{1700 \cdot R_e^2} \].
04

Simplify Gravity Equation

We know Earth's surface gravity \( g_e \) can be expressed as \( g_e = \frac{G \cdot M_e}{R_e^2} \), thus:\[ g_t = \frac{64}{1700} \cdot g_e \].Given \( g_e = 9.8 \, \text{m/s}^2 \), it follows that:\[ g_t = \frac{64}{1700} \cdot 9.8 \approx 0.369 \, \text{m/s}^2 \].
05

Determine Average Density

Density \( \rho \) is mass divided by volume, \( \rho = \frac{M}{V} \). Using the volume of a sphere \( V = \frac{4}{3}\pi R^3 \), the density of Titania is given by:\[ \rho_t = \frac{\frac{M_e}{1700}}{\frac{4}{3} \pi \left(\frac{R_e}{8}\right)^3} \].
06

Calculate Density in Terms of Earth's Density

Simplifying, the volume becomes \( \frac{4}{3} \pi \frac{R_e^3}{512} \), thus:\[ \rho_t = \frac{512 \cdot M_e}{1700 \cdot \frac{4}{3} \pi R_e^3} = \frac{512}{1700} \cdot \rho_e \].Assuming Earth’s density \( \rho_e = 5500 \, \text{kg/m}^3 \), we find:\[ \rho_t = \frac{512}{1700} \cdot 5500 \approx 1658 \, \text{kg/m}^3 \].

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Key Concepts

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

Acceleration Due to Gravity
On celestial bodies like Titania, calculating the acceleration due to gravity involves understanding how mass and radius relate to gravitational pull. The formula used is \[ g = \frac{G \cdot M}{R^2} \] where:
  • \(G\) is the gravitational constant, approximately \(6.674 \times 10^{-11} \, \text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}\).
  • \(M\) stands for the mass of the object.
  • \(R\) is the radius of the object.
Titania, with \(\frac{1}{1700}\) of Earth's mass and \(\frac{1}{8}\) of Earth's radius, demonstrates how variations in these parameters affect gravity. To find Titania's surface gravity, substitute the relative values into the formula. Simplifying gives:\[ g_t = \frac{64}{1700} \cdot g_e \] where \(g_e\), Earth's gravity, is \(9.8 \, \text{m/s}^2\). This results in \(g_t \approx 0.369 \, \text{m/s}^2\), indicating much weaker gravity due to lower mass and smaller radius.
Density Calculation
Density is a helpful way to characterize different materials, offering insights into what they might consist of. The density \(\rho\) of an object is calculated using:\[ \rho = \frac{M}{V} \]where
  • \(M\) is mass.
  • \(V\) is volume.
For a spherical object like Titania, volume \(V\) is calculated with\[ V = \frac{4}{3} \pi R^3 \]. Substituting Titania's characteristics relative to Earth allows us to compare densities:\[ \rho_t = \frac{512}{1700} \cdot \rho_e \]Assuming Earth's density \(\rho_e\) is \(5500 \, \text{kg/m}^3\), Titania’s density \(\rho_t\) is approximately \(1658 \, \text{kg/m}^3\). This lower density suggests that Titania might have a composition less dense than rock, hinting at an icy makeup.
Gravitational Constant
The gravitational constant \(G\) is crucial for calculating gravitational effects between masses, serving as a universal factor in Newton's law of gravitation. Its value is about \(6.674 \times 10^{-11} \, \text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2}\).This constant ensures consistency in calculations involving gravity, linking two masses and the force between them with the equation:\[ F = \frac{G \cdot m_1 \cdot m_2}{r^2} \] where
  • \(m_1\) and \(m_2\) are two masses.
  • \(r\) is the distance between their centers.
Understanding \(G\) enables us to not only solve problems like calculating Titania’s gravity but also explore gravitational interactions universally. Despite its small size, \(G\) is vital for linking the vastness of space with forces between celestial bodies.

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Most popular questions from this chapter

On July 15, 2004, NASA launched the \(Aura\) spacecraft to study the earth's climate and atmosphere. This satellite was injected into an orbit 705 km above the earth's surface. Assume a circular orbit. (a) How many hours does it take this satellite to make one orbit? (b) How fast (in km/s) is the \(Aura\) spacecraft moving?

At a certain instant, the earth, the moon, and a stationary 1250-kg spacecraft lie at the vertices of an equilateral triangle whose sides are 3.84 \(\times\) 10\(^5\) km in length. (a) Find the magnitude and direction of the net gravitational force exerted on the spacecraft by the earth and moon. State the direction as an angle measured from a line connecting the earth and the spacecraft. In a sketch, show the earth, the moon, the spacecraft, and the force vector. (b) What is the minimum amount of work that you would have to do to move the spacecraft to a point far from the earth and moon? Ignore any gravitational effects due to the other planets or the sun.

An experiment is performed in deep space with two uniform spheres, one with mass 50.0 kg and the other with mass 100.0 kg. They have equal radii, \(r =\) 0.20 m. The spheres are released from rest with their centers 40.0 m apart. They accelerate toward each other because of their mutual gravitational attraction. You can ignore all gravitational forces other than that between the two spheres. (a) Explain why linear momentum is conserved. (b) When their centers are 20.0 m apart, find (i) the speed of each sphere and (ii) the magnitude of the relative velocity with which one sphere is approaching the other. (c) How far from the initial position of the center of the 50.0-kg sphere do the surfaces of the two spheres collide?

Planets are not uniform inside. Normally, they are densest at the center and have decreasing density outward toward the surface. Model a spherically symmetric planet, with the same radius as the earth, as having a density that decreases linearly with distance from the center. Let the density be 15.0 \(\times\) 10\(^3\) kg/m\(^3\) at the center and 2.0 \(\times\) 10\(^3\) kg/m\(^3\) at the surface. What is the acceleration due to gravity at the surface of this planet?

Consider a spacecraft in an elliptical orbit around the earth. At the low point, or perigee, of its orbit, it is 400 km above the earth's surface; at the high point, or apogee, it is 4000 km above the earth's surface. (a) What is the period of the spacecraft's orbit? (b) Using conservation of angular momentum, find the ratio of the spacecraft's speed at perigee to its speed at apogee. (c) Using conservation of energy, find the speed at perigee and the speed at apogee. (d) It is necessary to have the spacecraft escape from the earth completely. If the spacecraft's rockets are fired at perigee, by how much would the speed have to be increased to achieve this? What if the rockets were fired at apogee? Which point in the orbit is more efficient to use?

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