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Chapter 3: Problem 27

Consider a planet orbiting the fixed sun. Take the plane of the planet's orbit to be the \(x y\) plane, with the sun at the origin, and label the planet's position by polar coordinates \((r, \phi)\). (a) Show that the planet's angular momentum has magnitude \(\ell=m r^{2} \omega,\) where \(\omega=\dot{\phi}\) is the planet's angular velocity about the sun. (b) Show that the rate at which the planet "sweeps out area" (as in Kepler's second law) is \(d A / d t=\frac{1}{2} r^{2} \omega,\) and hence that \(d A / d t=\ell / 2 m .\) Deduce Kepler's second law. SECTION 3.5 Angular Momentum for Several Particles

Short Answer

Expert verified
The planet's angular momentum is \( \ell = m r^2 \omega \), and the area swept rate is \( \frac{dA}{dt} = \frac{\ell}{2m} \). Thus, the area swept per time is constant, proving Kepler's second law.

Step by step solution

01

Understanding Angular Momentum

The angular momentum \( \ell \) of a particle is given by \( \ell = r \times p \), where \( r \) is the position vector, and \( p \) is the linear momentum \( p = m v \). Since the motion is in the \( xy \)-plane and \( v \) is tangential, it can be expressed as \( v = r \omega \) with \( \omega = \dot{\phi} \). Thus, \( p = m r \omega \) and \( \ell = m r^2 \omega \).
02

Area Swept Derivation

Kepler's second law states that a line segment joining a planet and the sun sweeps out equal areas during equal intervals of time. To express \( \frac{dA}{dt} \) in this polar system, consider a small triangle swept out in time \( dt \); the area \( dA \) of this triangle is \( \frac{1}{2} r \cdot r d\phi \), giving \( \frac{dA}{dt} = \frac{1}{2} r^2 \dot{\phi} = \frac{1}{2} r^2 \omega \).
03

Relate Area Rate to Angular Momentum

We have derived \( \frac{dA}{dt} = \frac{1}{2} r^2 \omega \). From earlier, \( \ell = m r^2 \omega \), thus \( r^2 \omega = \frac{\ell}{m} \). Substituting in the area rate equation gives \( \frac{dA}{dt} = \frac{1}{2} \cdot \frac{\ell}{m} = \frac{\ell}{2m} \).
04

Deducing Kepler's Second Law

Kepler's second law states the area swept out per unit time \( \frac{dA}{dt} \) is constant. Since \( \frac{dA}{dt} = \frac{\ell}{2m} \) and \( \ell \) is constant for a given orbit (due to angular momentum conservation), it follows that \( \frac{dA}{dt} \) is constant, confirming Kepler's second law.

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Key Concepts

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

Angular Momentum
Angular momentum is a key concept in understanding planetary motion. It describes the rotational motion of a planet as it orbits around the sun. For a planet in orbit, angular momentum (\( \ell \) ) is calculated using the formula: \( \ell = m r^2 \omega \) . Here, \( m \) is the planet's mass, \( r \) is the distance from the sun, and \( \omega = \dot{\phi} \) represents the angular velocity, which is the rate at which the planet sweeps out angular distances over time.
Angular momentum remains constant during the planet's orbit due to the conservation of angular momentum unless acted upon by an external torque. This means as a planet gets closer to the sun, its speed increases, ensuring \( m r^2 \omega \) stays unaffected by these positional changes. This constancy is critical for understanding why the planet's orbit shape and pace remain consistent over time.
Polar Coordinates
To analyze planetary motion, polar coordinates provide a way to describe a planet's position in its orbit around the sun. In this system, each point is defined by the coordinates \((r, \phi)\) :
  • \( r \) : the radial distance from the origin (sun) to the planet.
  • \( \phi \) : the angle formed with the positive x-axis, showing the planet's direction.
Polar coordinates are more straightforward in this context than Cartesian coordinates since they naturally align with the circular or elliptical orbits of planets. Each change in \( \phi \) reflects a change in the planet's position in its trajectory, making it convenient for calculating angular velocity and understanding the geometric aspects of planetary pathways.
Area Swept Rate
Kepler's second law involves the rate at which a planet sweeps out an area as it orbits the sun. This area swept rate is important for explaining how, despite changes in orbital speed or distance from the sun, the motion remains consistent according to Kepler's observations.
The rate at which area is swept (\( \frac{dA}{dt} \) ) by a planet is given by \( \frac{1}{2} r^2 \omega \) , linking it directly to angular momentum with \( \frac{dA}{dt} = \frac{\ell}{2m} \) . Consequently, because \( \ell \) is conserved, so is \( \frac{dA}{dt} \) .This means a planet covers equal areas in equal time intervals. For students, understanding this connection makes it apparent why even when planets come closer or move farther from the sun during different phases of their orbit, their path adheres to consistent physical laws that maintain stability in motion.
Planetary Motion
Planetary motion, as described by Kepler's laws, offers a comprehensive understanding of how celestial bodies move in space. Kepler's second law is a crucial part of this framework, stating that a planet sweeps out equal areas in equal time intervals as it orbits the sun.
Incorporating the previously discussed concepts, planetary motion is governed by:
  • Conservation of angular momentum.
  • The use of polar coordinates for representing orbits.
  • The consistent area swept rate contributing to the elliptical nature of orbits.
Kepler's observations originally revolutionized how we understand the solar system. Reliance on angular momentum and sweep rates allows students to see why planetary orbits are stable and predictable, and why they vary in speed through their orbit while maintaining structural integrity within the mechanics of their motion.

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Most popular questions from this chapter

A uniform spherical asteroid of radius \(R_{\mathrm{o}}\) is spinning with angular velocity \(\omega_{\mathrm{o}}\). As the aeons go by, it picks up more matter until its radius is \(R\). Assuming that its density remains the same and that the additional matter was originally at rest relative to the asteroid (anyway on average), find the asteroid's new angular velocity. (You know from elementary physics that the moment of inertia is \(\frac{2}{5} M R^{2}\).) What is the final angular velocity if the radius doubles?

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[Computer] A grenade is thrown with initial velocity \(\mathbf{v}_{\mathrm{o}}\) from the origin at the top of a high cliff, subject to negligible air resistance. (a) Using a suitable plotting program, plot the orbit, with the following parameters: \(\mathbf{v}_{\mathrm{o}}=(4,4), g=1,\) and \(0 \leq t \leq 4\) (and with \(x\) measured horizontally and \(y\) vertically up). Add to your plot suitable marks (dots or crosses, for example) to show the positions of the grenade at \(t=1,2,3,4 .\) (b) At \(t=4,\) when the grenade's velocity is \(\mathbf{v},\) it explodes into two equal pieces, one of which moves off with velocity \(\mathbf{v}+\Delta \mathbf{v} .\) What is the velocity of the other piece? (c) Assuming that \(\Delta \mathbf{v}=(1,3),\) add to your original plot the paths of the two pieces for \(4 \leq t \leq 9 .\) Insert marks to show their positions at \(t=5,6,7,8,9\). Find some way to show clearly that the CM of the two pieces continues to follow the original parabolic path.

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