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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 period of rotation is given by \( T = \sqrt{\frac{3 \pi}{G \rho}} \). For \( \rho = 3.0 \, \text{g/cm}^3 \), the period \( T \approx 3.9 \) hours.

Step by step solution

01

Understanding the Forces Involved

The key to solving this problem is recognizing that at the planet's equator, the gravitational force must provide the necessary centripetal force for the planet's rotation. The gravitational force on an object at the equator is given by \( F_g = \frac{G M m}{R^2} \), where \( G \) is the gravitational constant, \( M \) is the mass of the planet, \( m \) is the mass of the object, and \( R \) is the radius of the planet. The centripetal force required to keep the object moving in a circle at the equator is \( F_c = \frac{m v^2}{R} \), where \( v \) is the linear speed of rotation.
02

Equating Centripetal and Gravitational Forces

To find the maximum rotation speed, set the gravitational force equal to the centripetal force: \( \frac{G M m}{R^2} = \frac{m v^2}{R} \). We can cancel out the mass \( m \) of the object, giving \( \frac{G M}{R^2} = \frac{v^2}{R} \) or \( v^2 = \frac{G M}{R} \).
03

Relate v to the Period T

The linear speed \( v \) can also be expressed in terms of the period \( T \) using the circumference \( 2 \pi R \) of the planet: \( v = \frac{2 \pi R}{T} \). Substituting this expression for \( v \) into the previous equation gives \( \left(\frac{2 \pi R}{T}\right)^2 = \frac{G M}{R} \).
04

Expressing Mass M in terms of Density \( \rho \)

The mass \( M \) of the planet can be expressed in terms of its volume and density as \( M = \frac{4}{3} \pi R^3 \rho \). Substitute this expression into the velocity equation: \[ \left(\frac{2 \pi R}{T}\right)^2 = \frac{G \left(\frac{4}{3} \pi R^3 \rho \right)}{R} \].
05

Simplifying and Solving for T

Simplify the equation: \( \left(\frac{2 \pi R}{T}\right)^2 = \frac{4}{3} \pi G \rho R^2 \). Cancel \( R^2 \) and simplify: \( \frac{4 \pi^2}{T^2} = \frac{4}{3} \pi G \rho \). Solve for \( T^2 \): \( T^2 = \frac{3 \pi}{G \rho} \). Thus, take the square root to get \( T = \sqrt{\frac{3 \pi}{G \rho}} \).
06

Calculating for Given Density

Given \( \rho = 3.0 \, \text{g/cm}^3 = 3000 \, \text{kg/m}^3 \), substitute the values into the equation: \( T = \sqrt{\frac{3 \pi}{6.674 \times 10^{-11} \, \text{m}^3/\text{kg}\cdot\text{s}^2 \times 3000 \, \text{kg/m}^3}} \). Simplify inside the square root: \( T = \sqrt{\frac{3 \pi}{2.0022 \times 10^{-7}}} \). Calculate \( T \) to find \( T \approx 1.42 \times 10^4 \text{ seconds} \) or 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
Gravitational force is a fundamental force of nature. It is the force of attraction between two masses. The larger the mass, the stronger the gravitational pull. For a planet, gravity pulls objects towards its center.
When considering a planet's gravitational force, it is described with the formula: \[ F_g = \frac{G M m}{R^2} \]where:
  • \( F_g \) is the gravitational force.
  • \( G \) is the gravitational constant \((6.674 \times 10^{-11} \, \text{m}^3/\text{kg}\cdot\text{s}^2)\).
  • \( M \) is the mass of the planet.
  • \( m \) is the mass of the object experiencing the force.
  • \( R \) is the distance between the center of the planet and the object.

On a rotating planet, the gravitational force is especially crucial at the equator. It ensures that objects remain anchored amidst the rotational movement. The balancing act between gravitational and centripetal forces defines the planet's rotation dynamics.
Centripetal Force
Centripetal force is the force that keeps objects moving in a circular path. This force acts towards the center of the circle. On a spinning planet, objects at the equator are subject to centripetal force.
For a planet's rotation, the centripetal force needed for an object at the equator to maintain its circular path is: \[ F_c = \frac{m v^2}{R} \]where:
  • \( F_c \) is the centripetal force.
  • \( m \) is the mass of the object.
  • \( v \) is the linear speed due to rotation.
  • \( R \) is the radius of the path.

When a planet spins, the gravitational force must match the required centripetal force at the equator. This ensures that the objects do not fly off the surface. By equating these forces, one can determine the speed and period of rotation at which this balance is achieved.
Density of Planets
Density is defined as mass per unit volume and is crucial in understanding planet composition. For a spherical planet, density impacts gravity strength and rotational dynamics.
Density is given by: \[ \rho = \frac{M}{V} \]With planets being roughly spherical, their volume \( V \) is: \[ V = \frac{4}{3} \pi R^3 \] Thus, mass \( M \) can be expressed: \[ M = \rho \cdot \frac{4}{3} \pi R^3 \]
A uniform density view simplifies the calculations of gravitational forces. It also allows us to quantify the physical limits of a planet鈥檚 spin. The maximum rotation rate is influenced by the density since it dictates the gravitational force that aids in providing the necessary centripetal force at the equator.
Period of Rotation
The period of rotation is the time it takes for a planet to complete one full spin. It is directly related to the balance of forces on the planet. Faster rotations require precise gravitational and centripetal force alignment.
The formula for the shortest period of rotation, given a uniform density, is:\[ T = \sqrt{\frac{3 \pi}{G \rho}} \]where:
  • \( T \) is the period of rotation.
  • \( G \) is the gravitational constant.
  • \( \rho \) is the density of the planet.

This formula shows that denser planets can potentially spin faster. However, no natural body has been observed to spin faster than this theoretical limit. This reflects the delicate balance required by the gravitational forces to ensure the planet does not disassemble due to rapid rotation.

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Most popular questions from this chapter

An object of mass \(m\) is initially held in place at radial distance \(r=3 R_{E}\) from the center of Earth, where \(R_{E}\) is the radius of Earth. Let \(M_{E}\) be the mass of Earth. A force is applied to the object to move it to a radial distance \(r=4 R_{E},\) where it again is held in place. Calculate the work done by the applied force during the move by integrating the force magnitude.

Two Earth satellites, \(A\) and \(B,\) each of mass \(m,\) are to be launched into circular orbits about Earth's center. Satellite \(A\) is to orbit at an altitude of \(6370 \mathrm{~km} .\) Satellite \(B\) is to orbit at an altitude of \(19110 \mathrm{~km} .\) The radius of Earth \(R_{E}\) is \(6370 \mathrm{~km} .\) (a) \(\mathrm{What}\) is the ratio of the potential energy of satellite \(B\) to that of satellite \(A,\) in orbit? (b) What is the ratio of the kinetic energy of satellite \(B\) to that of satellite \(A,\) in orbit? (c) Which satellite has the greater total energy if each has a mass of \(14.6 \mathrm{~kg}\) ? (d) By how much?

What must the separation be between a \(5.2 \mathrm{~kg}\) particle and a \(2.4 \mathrm{~kg}\) particle for their gravitational attraction to have a magnitude of \(2.3 \times 10^{-12} \mathrm{~N} ?\)

Observations of the light from a certain star indicate that it is part of a binary (two-star) system. This visible star has orbital speed \(v=270 \mathrm{~km} / \mathrm{s}\) orbital period \(T=1.70\) days, and approximate mass \(m_{1}=6 M_{s},\) where \(M_{s}\) is the Sun's mass, \(1.99 \times 10^{30} \mathrm{~kg}\). Assume that the visible star and its companion star, which is dark and unseen, are both in circular orbits (Fig. \(13-47\) ). What integer multiple of \(M_{s}\) gives the approximate \(\operatorname{mass} m_{2}\) of the dark star?

(a) What will an object weigh on the Moon's surface if it weighs \(100 \mathrm{~N}\) on Earth's surface? (b) How many Earth radii must this same object be from the center of Earth if it is to weigh the same as it does on the Moon?
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