Problem 48 - Calculus Early Transcendentals: Pearson New International Edition

Question

EXPL 48. In this exercise you will derive Kepler's First Law, that planets travel in elliptical orbits. We begin with the notation. Place the coordinate system so that the sun is at the origin and the planet's closest approach to the sun (the perihelion) is on the positive \(x\) -axis and occurs at time \(t=0\). Let \(\mathbf{r}(t)\) denote the position vector and let \(r(t)=\|\mathbf{r}(t)\|\) denote the distance from the sun at time \(t\). Also, let \(\theta(t)\) denote the angle that the vector \(\mathbf{r}(t)\) makes with the positive \(x\) -axis at time \(t\). Thus, \((r(t), \theta(t))\) is the polar coordinate representation of the planet's position. Let \(\mathbf{u}_{1}=\mathbf{r} / r=(\cos \theta) \mathbf{i}+(\sin \theta) \mathbf{j} \quad\) and \(\quad \mathbf{u}_{2}=(-\sin \theta) \mathbf{i}+(\cos \theta) \mathbf{j}\) Vectors \(\mathbf{u}_{1}\) and \(\mathbf{u}_{2}\) are orthogonal unit vectors pointing in the directions of increasing \(r\) and increasing \(\theta\), respectively. Figure 12 summarizes this notation. We will often omit the argument \(t\), but keep in mind that \(\mathbf{r}, \theta, \mathbf{u}_{1}\), and \(\mathbf{u}_{2}\) are all functions of \(t .\) A prime in. dicates differentiation with respect to time \(t\). (a) Show that \(\mathbf{u}_{1}^{\prime}=\theta^{\prime} \mathbf{u}_{2}\) and \(\mathbf{u}_{2}^{\prime}=-\theta^{\prime} \mathbf{u}_{1}\). (b) Show that the velocity and acceleration vectors satisfy $$ \begin{array}{l} \mathbf{v}=r^{\prime} \mathbf{u}_{1}+r \theta^{\prime} \mathbf{u}_{2} \\ \mathbf{a}=\left(r^{\prime \prime}-r\left(\theta^{\prime}\right)^{2}\right) \mathbf{u}_{1}+\left(2 r^{\prime} \theta^{\prime}+r \theta^{\prime \prime}\right) \mathbf{u}_{2} \end{array} $$ (c) Use the fact that the only force acting on the planet is the gravity of the sun to express a as a multiple of \(\mathbf{u}_{1}\), then explain how we can conclude that $$ \begin{aligned} r^{\prime \prime}-r\left(\theta^{\prime}\right)^{2} &=\frac{-G M}{r^{2}} \\ 2 r^{\prime} \theta^{\prime}+r \theta^{\prime \prime} &=0 \end{aligned} $$ (d) Consider \(\mathbf{r} \times \mathbf{r}^{\prime}\), which we showed in Example 8 was a constant vector, say D. Use the result from (b) to show that \(\mathbf{D}=r^{2} \theta^{\prime} \mathbf{k} .\) (e) Substitute \(t=0\) to get \(\mathbf{D}=r_{0} v_{0} \mathbf{k}\), where \(r_{0}=r(0)\) and \(v_{0}=\|\mathbf{v}(0)\|\). Then argue that \(r^{2} \theta^{\prime}=r_{0} v_{0}\) for all \(t\). (f) Make the substitution \(q=r^{\prime}\) and use the result from (e) to obtain the first-order (nonlinear) differential equation in \(q\) : $$ q \frac{d q}{d r}=\frac{r_{0}^{2} v_{0}^{2}}{r^{3}}-\frac{G M}{r^{2}} $$ (g) Integrate with respect to \(r\) on both sides of the above equation and use an initial condition to obtain $$ q^{2}=2 G M\left(\frac{1}{r}-\frac{1}{r_{0}}\right)+v_{0}^{2}\left(1-\frac{r_{0}^{2}}{r^{2}}\right) $$ (h) Substitute \(p=1 / r\) into the above equation to obtain $$ \frac{r_{0}^{2} v_{0}^{2}}{\left(\theta^{\prime}\right)^{2}}\left(\frac{d p}{d t}\right)^{2}=2 G M\left(p-p_{0}\right)+v_{0}^{2}\left(1-\frac{p^{2}}{p_{0}^{2}}\right) $$

Step-by-Step Solution

Answer

Planets travel in elliptical orbits due to gravitational forces, derived via polar coordinates and conservation laws.

Step 1: Evaluate Derivatives for Orthogonal Unit Vectors

Given vectors \( \mathbf{u}_1 \) and \( \mathbf{u}_2 \), take their time derivatives using the product rule and the chain rule. Since \( \mathbf{u}_1 = (\cos \theta) \mathbf{i} + (\sin \theta) \mathbf{j} \), differentiating gives: \( \mathbf{u}_1' = (-\sin \theta \cdot \theta') \mathbf{i} + (\cos \theta \cdot \theta') \mathbf{j} = \theta'\left( -\sin \theta \mathbf{i} + \cos \theta \mathbf{j} \right) = \theta' \mathbf{u}_2 \). Similarly, \( \mathbf{u}_2' = (\cos \theta \cdot \theta') \mathbf{i} + (\sin \theta \cdot \theta') \mathbf{j} = -\theta' \mathbf{u}_1 \)."

Step 2: Expression for Velocity and Acceleration Vectors

For a position vector \( \mathbf{r}(t) = r(t) \mathbf{u}_1 \), the velocity \( \mathbf{v} = \frac{d \mathbf{r}}{dt} = r' \mathbf{u}_1 + r \mathbf{u}_1' \). Using the previous result for \( \mathbf{u}_1' \), this becomes \( \mathbf{v} = r' \mathbf{u}_1 + r \theta' \mathbf{u}_2 \). The acceleration \( \mathbf{a} \) is the derivative of velocity: \( \mathbf{a} = (r')' \mathbf{u}_1 + r' \mathbf{u}_1' + \frac{d}{dt}(r\theta' \mathbf{u}_2) \). This leads to \( \mathbf{a} = (r'' - r(\theta')^2) \mathbf{u}_1 + (2r'\theta' + r\theta'') \mathbf{u}_2 \)."

Step 3: Express Acceleration Due to Gravity

Since the only force is gravity, \( \mathbf{a} = -\frac{GM}{r^2} \mathbf{u}_1 \). From the acceleration expression, this means \( r'' - r(\theta')^2 = \frac{-GM}{r^2} \) and the \( \mathbf{u}_2 \) component must be zero, giving \( 2r'\theta' + r\theta'' = 0 \)."

Step 4: Angular Momentum Conservation

From physics, the angular momentum, a constant vector, is \( \mathbf{D} = \mathbf{r} \times \mathbf{v} = r^2 \theta' \mathbf{k} \). This confirms that the component of the cross product is a constant, consistent with our velocity expression."

Step 5: Evaluate Constant at \( t = 0 \)

At \( t = 0 \), the perihelion occurs, so \( \mathbf{D} = r_0 v_0 \mathbf{k} \). Since this value is constant for all \( t \), we can equate this to \( r^2 \theta' \mathbf{k} \) for any \( t \), showing \( r^2 \theta' = r_0 v_0 \)."

Step 6: Differential Equation for Rate of Change of \( r \)

Substitute \( q = r' \) into \( 2r'\theta' + r\theta'' = 0 \) to simplify and set up \( q \frac{dq}{dr} = \frac{r_0^2v_0^2}{r^3} - \frac{GM}{r^2} \) using the fact that \( r^2 \theta' = r_0 v_0 \)."

Step 7: Integration of Differential Equation

Integrate the resulting differential equation with respect to \( r \). The integration results in \( q^2 = 2GM\left(\frac{1}{r} - \frac{1}{r_0} \right) + v_0^2\left( 1 - \frac{r_0^2}{r^2} \right) \), using appropriate initial conditions."

Step 8: Substitute and Simplify for \( p \)

Substitute \( p = \frac{1}{r} \) into \( q^2 \) equation. This substitution gives \( \frac{r_0^2v_0^2}{(\theta')^2}\left(\frac{dp}{dt}\right)^2 = 2GM(p - p_0) + v_0^2\left(1 - \frac{p^2}{p_0^2}\right) \), where \( p_0 = \frac{1}{r_0} \)."

Key Concepts

Elliptical Orbits

In the study of celestial mechanics, one fundamental concept is that of the elliptical orbits described by Kepler's First Law. This law asserts that planets orbit the sun in elliptical paths, with the sun located at one of the two foci of the ellipse. The elliptical nature of orbits is fundamental because it describes the varying speeds at which planets travel; they move faster when closer to the sun and slower when farther away.

Why ellipses? The gravitational pull from the sun creates a balance between the inertia of the planet moving forward and the sun's gravitational attraction pulling inward. This results in an elliptical path instead of a circle. Each ellipse has two focal points, and as planets orbit, the distance from one focus to a planet and from the planet back to the other focus remains constant, highlighting the unique balance of forces at play.

• Kepler's First Law: Orbits are elliptical.
• The sun is at one focus of the ellipse.
• Orbits vary in speed—faster near the sun.

Polar Coordinates

To describe the motion of planets in elliptical orbits, polar coordinates provide a useful framework. In this coordinate system, the position of a planet is specified using a radius and an angle rather than the traditional Cartesian coordinates.

Polar coordinates are especially advantageous in celestial mechanics because they simplify the mathematical equations representing the planet's position and movement. The radius, denoted as \( r(t) \), signifies the distance from the sun to the planet, while the angle \( \theta(t) \) measures the planet's position in terms of rotation around the sun.

Using polar coordinates naturally aligns with the physics of planetary motion since it centers around radial and angular components, allowing for easier analysis of changes in distance and direction.

• Polar coordinates use radius \( r \) and angle \( \theta \).
• Flexible for describing rotational movements.
• Simplifies equations in elliptical orbits.

Differential Equations

Differential equations play a crucial role in analyzing and predicting the behavior of planetary motions. These mathematical equations describe how quantities change, encapsulating dynamics such as the velocity or acceleration of a planet over time.

In the context of Kepler's First Law, differential equations help in formulating the relationships between various expressions like radial distance \( r \), radial velocity \( r' \), and angular velocity \( \theta' \). By solving these equations, we gain insights into the elliptical paths of planets and the forces governing their trajectories.

Understanding how to work with these equations is essential as they provide a toolkit for discovering patterns and predicting future positions within the orbit, ultimately bridging abstract mathematical theory and real-world celestial mechanics.

• Used to model changes over time.
• Essential for calculating motion in orbits.
• Links mathematical theory to astronomy.

Gravitational Force

Gravitational force is the key factor in celestial movements, responsible for the unique paths planets follow around the sun. It's the attractive force that acts between any two masses, but in this context, it is primarily the pull between a planet and the sun.

According to Newton's law of universal gravitation, the force is proportional to the product of their masses and inversely proportional to the square of the distance between them. In simpler terms, this means that the force decreases with distance but increases with the mass of the objects involved.

In our study of elliptical orbits, the gravitational force is crucial for explaining the planet's constant curved path around the sun. The force is responsible for the centripetal acceleration that keeps the planet from flying off into space, maintaining its motion in a stable orbit.

• Depends on mass and distance.
• Keeps planets in steady orbit.
• Drives centripetal acceleration.