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Chapter 13: Problem 13

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/s虏, (b) Approximately 1658 kg/m鲁.

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鈥檚 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鈥檚 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鈥檚 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

The mass of Venus is 81.5% that of the earth, and its radius is 94.9% that of the earth. (a) Compute the acceleration due to gravity on the surface of Venus from these data. (b) If a rock weighs 75.0 N on earth, what would it weigh at the surface of Venus?

An astronaut inside a spacecraft, which protects her from harmful radiation, is orbiting a black hole at a distance of 120 km from its center. The black hole is 5.00 times the mass of the sun and has a Schwarzschild radius of 15.0 km. The astronaut is positioned inside the spaceship such that one of her 0.030-kg ears is 6.0 cm farther from the black hole than the center of mass of the spacecraft and the other ear is 6.0 cm closer. (a) What is the tension between her ears? Would the astronaut find it difficult to keep from being torn apart by the gravitational forces? (Since her whole body orbits with the same angular velocity, one ear is moving too slowly for the radius of its orbit and the other is moving too fast. Hence her head must exert forces on her ears to keep them in their orbits.) (b) Is the center of gravity of her head at the same point as the center of mass? Explain.

A landing craft with mass 12,500 kg is in a circular orbit 5.75 \(\times\) 10\(^5\) m above the surface of a planet. The period of the orbit is 5800 s. The astronauts in the lander measure the diameter of the planet to be 9.60 \(\times\) 10\(^6\) m. The lander sets down at the north pole of the planet. What is the weight of an 85.6-kg astronaut as he steps out onto the planet's surface?

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?

The acceleration due to gravity at the north pole of Neptune is approximately 11.2 m/s\(^2\). Neptune has mass 1.02 \(\times\) 10\(^{26}\) kg and radius 2.46 \(\times\) 10\(^4\) km and rotates once around its axis in about 16 h. (a) What is the gravitational force on a 3.00-kg object at the north pole of Neptune? (b) What is the apparent weight of this same object at Neptune's equator? (Note that Neptune's "surface" is gaseous, not solid, so it is impossible to stand on it.)
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