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The fastest possible rate of rotation of a planet is that for which the gravitational force on material at the equator just barely provides the centripetal force needed for the rotation. (Why?) (a) Show that the corresponding shortest period of rotation is $$ T=\sqrt{\frac{3 \pi}{G \rho}}, $$ where \(\rho\) is the uniform density (mass per unit volume) of the spherical planet. (b) Calculate the rotation period assuming a density of \(3.0 \mathrm{~g} / \mathrm{cm}^{3}\), typical of many planets, satellites, and asteroids. No astronomical object has ever been found to be spinning with a period shorter than that determined by this analysis.

Short Answer

Expert verified
T = 3.9 hours for a density of 3.0 g/cm³.

Step by step solution

01

Understand the concept

We need to find the fastest rotation rate of a planet for which gravitational force on material at the equator provides the centripetal force needed for rotation. The balance between these forces allows us to derive the shortest rotation period, \(T\).
02

Gravitational and Centripetal Forces

At the equator, the gravitational force \(F_g\) is \(F_g = m g\), where \(g = \frac{G M}{R^2}\). The centripetal force required for rotation is \(F_c = m \frac{v^2}{R}\), where \(v = R \omega\) is the angular velocity. Thus, \(F_c = m R \omega^2\).
03

Equating Gravitational and Centripetal Forces

Set the gravitational force equal to the centripetal force: \(m \frac{G M}{R^2} = m R \omega^2\). Simplifying, we get \(\omega^2 = \frac{G M}{R^3}\).
04

Relate Mass to Density

The mass \(M\) of the planet in terms of its volume and density is \(M = \frac{4}{3} \pi R^3 \rho\). Substitute this into the previous equation: \(\omega^2 = \frac{G \left( \frac{4}{3} \pi R^3 \rho \right)}{R^3} = \frac{4 \pi G \rho}{3}\).
05

Solve for Period T

The period \(T\) is related to angular velocity by \(\omega = \frac{2 \pi}{T}\). Therefore, \(\left(\frac{2\pi}{T}\right)^2 = \frac{4 \pi G \rho}{3}\). Solve for \(T\) to find \(T = \sqrt{\frac{3 \pi}{G \rho}}\).
06

Calculate T for a given density

Convert \(\rho = 3.0 \text{ g/cm}^3\) to \(\text{ kg/m}^3\): \(\rho = 3000 \text{ kg/m}^3\). Substitute \(G = 6.674 \times 10^{-11} \text{ m}^3 \text{ kg}^{-1} \text{ s}^{-2}\) and \(\rho = 3000 \,\,\text{kg/m}^3\) into the formula: \(T = \sqrt{\frac{3 \pi}{G \rho}} = \sqrt{\frac{3 \pi}{6.674 \times 10^{-11} \times 3000}} \approx 1.41 \times 10^4 \text{ seconds}\), which converts to approximately 3.9 hours.

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Key Concepts

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

Gravitational Force
Gravity is a force that attracts two bodies toward each other. On a planet, gravitational force ensures that all objects are pulled toward its center. This force is crucial in planetary rotation as it helps maintain the orbit of any material on a planet's surface. The gravitational force at a planet's equator is given by the formula \( F_g = m g \), where \( g \) (acceleration due to gravity) is derived from Newton's law of universal gravitation: \( g = \frac{G M}{R^2} \). Here, \( G \) denotes the gravitational constant, \( M \) the mass of the planet, and \( R \) its radius. \( G \), a constant, is approximately \( 6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \). Gravitational force plays a pivotal role in ensuring that planetary rotation stays balanced and controlled.
Centripetal Force
Centripetal force is the force required to make an object follow a curved path. It is crucial in understanding planetary rotation because it balances the planetary materials during rotation, especially at the equator. The formula for centripetal force is given by \( F_c = m \frac{v^2}{R} \), where \( v \) represents the tangential velocity and \( R \) the radius. When considering angular velocity \( \omega \), the velocity \( v \) can be expressed as \( R \omega \), leading to the equation \( F_c = m R \omega^2 \). For a planet to rotate without tearing itself apart, its gravitational pull must match the centripetal requirements precisely at its equator.
Density of Planets
Density is a key factor in determining the rotation period and stability of a planet. It is defined as mass per unit volume \(( \rho )\). The uniform density of a planet plays a role in its mass distribution, affecting gravitational and centripetal forces. To relate a planet's mass to its density, the equation \( M = \frac{4}{3} \pi R^3 \rho \) is used, where \( M \) is the mass, \( \rho \) the density, and \( R \) the radius of the planet. Given that typical planetary densities hover around \(3.0 \text{ g/cm}^3\), this translates to \(3000 \text{ kg/m}^3\) in standard units, providing a baseline for calculating the shortest possible rotation period, \( T \).
Rotation Period
The rotation period of a planet is the time it takes for the planet to complete one full spin around its axis. It is determined by the balance between gravitational and centripetal forces at the equator, ensuring that these match perfectly. Based on theoretical calculations, the formula for the shortest rotation period \( T \) is \( T = \sqrt{\frac{3 \pi}{G \rho}} \). This formula derives from setting gravitational force equal to centripetal force and considering the planet's density \( \rho \) and gravitational constant \( G \).With a typical density of \( 3000 \text{ kg/m}^3 \), this results in a theoretical minimum achievable rotation period, ensuring that no planet or celestial body spins faster than this critical limit under normal circumstances.
Angular Velocity
Angular velocity \( \omega \) is a measure of the rate at which an object rotates, defined as the angle rotated per unit of time. For planets, it is linked directly to the rotation period. The relationship is represented as \( \omega = \frac{2 \pi}{T} \), where \( T \) is the rotation period. Higher angular velocity indicates a faster rotation. When gravitational force equates to centripetal force, angular velocity reaches a value that supports the shortest possible spin period. The calculation \( \omega^2 = \frac{4 \pi G \rho}{3} \) results from solving for \( \omega \) when the balance between gravitational and centripetal forces is exact, thereby providing a means to determine a planet's maximum sustainable speed.

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Most popular questions from this chapter

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