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Chapter 13: Problem 80

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
The shortest rotation period is \( \approx 1.41 \) hours for density \(3 \text{ g/cm}^3\).

Step by step solution

01

Understanding the Problem

We need to find the condition for which the gravitational force at the equator of a planet provides the necessary centripetal force for its rotation. This occurs when the gravitational force equals the centripetal force required to keep material on the equator moving in a circular path.
02

Expression of Forces

At the equator, the gravitational force per unit mass is \(g = \frac{G M}{R^2}\), where \(G\) is the gravitational constant, \(M\) is the mass of the planet, and \(R\) is its radius. The centripetal force required for rotation is \(\frac{v^2}{R}\), where \(v\) is the linear velocity at the equator.
03

Equating Forces

For the fastest possible rotation, equate gravitational force to centripetal force: \( \frac{G M}{R^2} = \frac{v^2}{R} \). Simplify to find \(v^2 = \frac{G M}{R}\).
04

Relating Velocity to Period

The velocity \(v\) is related to the period \(T\) by the equation \(v = \frac{2 \pi R}{T}\). Substitute into the equation: \( \left( \frac{2 \pi R}{T} \right)^2 = \frac{G M}{R} \).
05

Solving for Period

Expand and simplify the equation \( \frac{4 \pi^2 R^2}{T^2} = \frac{G M}{R} \) to get \(T^2 = \frac{4 \pi^2 R^3}{G M}\).
06

Expressing Mass in Terms of Density

The density \(\rho\) is given by \(\rho = \frac{M}{V}\), where \(V\) is the volume \(\frac{4}{3} \pi R^3\). So, \(M = \rho \times \frac{4}{3} \pi R^3\). Substitute into the expression for \(T^2\).
07

Substitution and Simplification

Substitute \(M = \rho \times \frac{4}{3} \pi R^3\) into \(T^2 = \frac{4 \pi^2 R^3}{G M}\): \(T^2 = \frac{4 \pi^2 R^3}{G \left(\rho \times \frac{4}{3} \pi R^3\right)}\). Simplify to find \(T = \sqrt{\frac{3 \pi}{G \rho}}\).
08

Numerical Calculation

Given \(\rho = 3.0 \text{ g/cm}^3 = 3000 \text{ kg/m}^3\), substitute into \(T = \sqrt{\frac{3 \pi}{G \rho}}\) where \(G = 6.674 \times 10^{-11} \text{ m}^3/\text{kg}/\text{s}^2\). Calculate to find \(T \approx 5081 \text{ seconds} \approx 1.41 \text{ hours}\).

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Key Concepts

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

Gravitational Force
Gravitational force is a fundamental force of nature and is responsible for the attraction between two masses. It plays a crucial role in maintaining the motion of celestial objects, such as planets, stars, and moons, within their respective orbits.
  • According to Newton's law of universal gravitation, the force of gravity acting on a mass, such as a planet, is calculated using the formula: \( F = \frac{G M m}{R^2} \), where:
    • \( F \) is the gravitational force.
    • \( G \) is the gravitational constant \( (6.674 \times 10^{-11} \text{ m}^3/\text{kg}/\text{s}^2) \).
    • \( M \) and \( m \) are the masses of the two objects.
    • \( R \) is the distance between the centers of the two masses.
  • On a rotating planet, the gravitational force must balance with the centripetal force to keep material at the equator from flying off into space. This balance occurs when \( \frac{G M}{R^2} = \frac{v^2}{R} \), allowing us to study stable orbits and surface conditions.
  • Gravitational force ensures that planets remain in their orbits and plays an essential role in the formation and maintenance of planetary systems.
Planetary Rotation
Planetary rotation refers to the spin of a planet around its own axis, which influences many characteristics such as day and night cycles and the period of rotation.
  • For a planet, the fastest possible rotation is when the gravitational force at the equator equals the centripetal force needed for rotation. This ensures material at the equator remains in place.
  • The speed of rotation also affects factors such as atmospheric patterns and weather systems, contributing to climate and environmental conditions.
  • The relationship between the rotation period \( T \) and linear velocity \( v \) is captured by the equation: \( v = \frac{2 \pi R}{T} \). By manipulating such equations, we find limits on how fast a planet can rotate before destabilizing.
Understanding planetary rotation dynamics helps us to appreciate how planets interact with other celestial bodies and develop stable environments.
Density of Planets
The density of a planet, represented by \( \rho \), is a measure of its mass per unit volume and is a critical factor in determining the planet's gravitational force and stability.
  • Density formula: \( \rho = \frac{M}{V} \), where \( M \) is the mass, and \( V \) is the volume of the planet.
  • A planet's mean density can indicate its composition, suggesting whether it is more rock-like or gas-like, influencing its gravitational strength and, consequently, its atmosphere and potential to harbor life.
  • The density also affects rotation stability: i.e., planets with higher density can rotate faster without losing material at the equator. This density connection is evident in the formula for the shortest rotation period: \( T = \sqrt{\frac{3 \pi}{G \rho}} \).
Studying the density of planets allows scientists to estimate their internal structures and simulate their behavior within the cosmic ballet of the solar system.

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Most popular questions from this chapter

A certain triple-star system consists of two stars, each of mass \(m\), revolving in the same circular orbit of radius \(r\) around a central star of mass \(M\) (Fig. 13-53). The two orbiting stars are always at opposite ends of a diameter of the orbit. Derive an expression for the period of revolution of the stars.

A \(20 \mathrm{~kg}\) satellite has a circular orbit with a period of \(2.4 \mathrm{~h}\) and a radius of \(8.0 \times 10^{6} \mathrm{~m}\) around a planet of unknown mass. If the magnitude of the gravitational acceleration on the surface of the planet is \(8.0 \mathrm{~m} / \mathrm{s}^{2}\), what is the radius of the planet?

One model for a certain planet has a core of radius \(R\) and mass \(M\) surrounded by an outer shell of inner radius \(R\), outer radius \(2 R\), and mass \(4 M .\) If \(M=4.1 \times 10^{24} \mathrm{~kg}\) and \(R=6.0 \times 10^{6} \mathrm{~m}\) what is the gravitational acceleration of a particle at points (a) \(R\) and (b) \(3 R\) from the center of the planet?

The mean distance of Mars from the Sun is \(1.52\) times that of Earth from the Sun. From Kepler's law of periods, calculate the number of years required for Mars to make one revolution around the Sun; compare your answer with the value given in Appendix C.

Three point particles are fixed in place in an \(x y z\) coordinate system. Particle \(A\), at the origin, has mass \(m_{A} .\) Particle \(B\), at \(x y z\) coordinates \((2.00 d, 1.00 d, 2.00 d)\), has mass \(2.00 m_{A}\), and particle \(C\), at coordinates \((-1.00 d, 2.00 d,-3.00 d)\), has mass \(3.00 m_{A}\). A fourth particle \(D\), with mass \(4.00 m_{A}\), is to be placed near the other particles. In terms of distance \(d\), at what (a) \(x,(\mathrm{~b}) y\), and (c) \(z\) coordinate should \(D\) be placed so that the net gravitational force on \(A\) from \(B, C\), and \(D\) is zero?
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