Question

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

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

Step-by-step solution

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.

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 \( \frac{1}{1700}M_e \).

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 \[ g_t = \frac{G \cdot M_e}{1700 \cdot \frac{R_e^2}{64}} = \frac{64G \cdot M_e}{1700 \cdot R_e^2} \].

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:\[ g_t = \frac{64}{1700} \cdot 9.8 \approx 0.369 \, \text{m/s}^2 \].

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:\[ \rho_t = \frac{\frac{M_e}{1700}}{\frac{4}{3} \pi \left(\frac{R_e}{8}\right)^3} \].

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:\[ \rho_t = \frac{512}{1700} \cdot 5500 \approx 1658 \, \text{kg/m}^3 \].

Key concepts

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.