Question

The modern way to derive Kepler's area law is to break the force into its radial and transverse components, rather than the tangential and normal components used by Hermann, and use polar coordinates whose origin is the center of force. Assume then that the center of force is at the origin of a polar coordinate system. Using vector notation, \(\operatorname{set} \mathbf{u}_{r}=\mathbf{i} \cos \theta+\mathbf{j} \sin \theta\) and \(\mathbf{u}_{\theta}=-\mathbf{i} \sin \theta+j \cos \theta .\) Show that \(d \mathbf{u}_{r} / d \theta=\mathbf{u}_{\theta}\) and \(d \mathbf{u}_{\theta} / d \theta=-\mathbf{u}_{r}\). Then show that if \(\mathbf{r}=r \mathbf{u}_{r}\), then the velocity \(\mathbf{v}\) is given by \(r(d \theta / d t) \mathbf{u}_{\theta}+\) \((d r / d t) \mathbf{u}_{r}\). Show next that the radial component \(a_{r}\) and the transverse component \(a_{\theta}\) of the acceleration are given by $$ a_{r}=\frac{d^{2} r}{d t^{2}}-r\left(\frac{d \theta}{d t}\right)^{2} \quad \text { and } \quad a_{\theta}=r \frac{d^{2} \theta}{d t^{2}}+2 \frac{d r}{d t} \frac{d \theta}{d t} $$ Since the force is central, \(a_{\theta}=0 .\) Multiply the differential equation expressing that fact by \(r\) and integrate to get \(r^{2} \frac{d \theta}{d t}=k\), where \(k\) is a constant. Show finally that \(r^{2} \frac{d \theta}{d t}=\) \(\frac{d A}{d t}\), where \(A\) is the area swept out by the radius vector. This proves Kepler's law of areas.

Short answer

Question: Prove Kepler's law of areas using vector notation in a polar coordinate system. Answer: To prove Kepler's law of areas, we can follow these steps: 1. Show the derivatives: \(\frac{d\mathbf{u}_{r}}{d\theta} = \mathbf{u}_{\theta}\) and \(\frac{d\mathbf{u}_{\theta}}{d\theta} = -\mathbf{u}_{r}\). 2. Express the velocity vector as: \(\mathbf{v} = \frac{dr}{dt}\mathbf{u}_{r} + r\frac{d\theta}{dt}\mathbf{u}_{\theta}\) 3. Find expressions for radial (\(a_{r}\)) and transverse accelerations (\(a_{\theta}\)). 4. Use the central force condition (\(a_{\theta} = 0\)) to get the expression: \(r^{2}\frac{d\theta}{dt} = k\). 5. Prove that this expression is equal to the derivative of the area A with respect to time: \(r^{2}\frac{d\theta}{dt} = 2\frac{dA}{dt}\). Hence, Kepler's law of areas is proven, stating that the rate of change of the area swept out by the radius vector is constant.

Step-by-step solution

Derivatives of Unit Vectors

To show that \(\frac{d\mathbf{u}_{r}}{d\theta} = \mathbf{u}_{\theta}\) and \(\frac{d\mathbf{u}_{\theta}}{d\theta} = -\mathbf{u}_{r}\), we first need to find the derivatives of \(\mathbf{u}_{r}\) and \(\mathbf{u}_{\theta}\) with respect to \(\theta\). Differentiating \(\mathbf{u}_{r}\) with respect to \(\theta\), we get: \(\frac{d\mathbf{u}_{r}}{d\theta} = -\mathbf{i}\sin\theta + \mathbf{j}\cos\theta = \mathbf{u}_{\theta}\) And differentiating \(\mathbf{u}_{\theta}\) with respect to \(\theta\), we get: \(\frac{d\mathbf{u}_{\theta}}{d\theta} = -\mathbf{i}\cos\theta - \mathbf{j}\sin\theta = -\mathbf{u}_{r}\)

Velocity Vector Expression

Now, we need to express the velocity vector \(\mathbf{v}\) in terms of unit vectors. Given \(\mathbf{r} = r\mathbf{u}_{r}\), we find the derivative with respect to time \(t\): \(\mathbf{v} = \frac{d\mathbf{r}}{dt} = \frac{dr}{dt}\mathbf{u}_{r} + r\frac{d\mathbf{u}_{r}}{dt}\). Using the chain rule, we have: \(\frac{d\mathbf{u}_{r}}{dt} = \frac{d\mathbf{u}_{r}}{d\theta}\frac{d\theta}{dt} = \mathbf{u}_{\theta}\frac{d\theta}{dt}\). Replacing this in the velocity expression, we get: \(\mathbf{v} = \frac{dr}{dt}\mathbf{u}_{r} + r\frac{d\theta}{dt}\mathbf{u}_{\theta}\)

Radial and Transverse Acceleration

Now, we derive the expressions for radial and transverse accelerations, \(a_{r}\) and \(a_{\theta}\), by differentiating the velocity vector with respect to time: \(\mathbf{a} = \frac{d\mathbf{v}}{dt} = \frac{d^{2}r}{dt^{2}}\mathbf{u}_{r} + \frac{dr}{dt}\frac{d\mathbf{u}_{r}}{dt} + r\frac{d^{2}\theta}{dt^{2}}\mathbf{u}_{\theta} + r\frac{d\theta}{dt}\frac{d\mathbf{u}_{\theta}}{dt}\) Using the chain rule and the results from Step 1, we have: \(\mathbf{a} = \frac{d^{2}r}{dt^{2}}\mathbf{u}_{r} + \frac{dr}{dt}\frac{d\theta}{dt}\mathbf{u}_{\theta} + r\frac{d^{2}\theta}{dt^{2}}\mathbf{u}_{\theta} + r\frac{d\theta}{dt}\frac{d\theta}{dt}\mathbf{u}_{r}\) The radial acceleration \(a_{r}\) is the component of the acceleration in the direction of \(\mathbf{u_r}\): \(a_{r} = \frac{d^{2}r}{dt^{2}} - r\left(\frac{d\theta}{dt}\right)^{2}\) The transverse acceleration \(a_{\theta}\) is the component of the acceleration in the direction of \(\mathbf{u_\theta}\): \(a_{\theta} = r\frac{d^{2}\theta}{dt^{2}} + 2\frac{dr}{dt}\frac{d\theta}{dt}\)

Expression for Central Force

As the force is central, the transverse acceleration is zero, i.e., \(a_{\theta} = 0\). Using the expression for \(a_{\theta}\) from Step 3, we obtain the following differential equation: \(r\frac{d^{2}\theta}{dt^{2}} + 2\frac{dr}{dt}\frac{d\theta}{dt} = 0\) Multiplying this equation by \(r\) and integrating, we get: \(r^{2}\frac{d\theta}{dt} = k\) where \(k\) is a constant.

Kepler's Law of Areas

To prove Kepler's law of areas, we need to show that \(r^{2}\frac{d\theta}{dt}\) is equal to the derivative of the area A with respect to time. Let's first find an expression for the area dA, swept by the radius vector: \(dA = \frac{1}{2}r^{2}d\theta\) Now, differentiate A with respect to time: \(\frac{dA}{dt} = \frac{1}{2}r^{2}\frac{d\theta}{dt}\) Comparing this expression with the result from Step 4, we observe that: \(r^{2}\frac{d\theta}{dt} = 2\frac{dA}{dt}\) This proves Kepler's law of areas, which states that the rate of change of the area swept out by the radius vector is constant.

Key concepts

Polar Coordinates

Polar coordinates offer a two-dimensional coordinate system wherein each point on a plane is represented by a distance from a reference point and an angle from a reference direction. The reference point, often called the pole, aligns with the origin of a Cartesian coordinate system, while the reference direction usually aligns with the positive x-axis.

In the context of planetary motion, polar coordinates are especially useful as they readily describe the circular or elliptical trajectories exhibited by planets orbiting a star. Instead of using x and y coordinates, we use the radial distance from the center of force, denoted as 'r', and the angle 'θ' (theta), which represents the orientation of the radius vector from the fixed reference direction. This method significantly simplifies the mathematical representation of orbits, as seen in Kepler's laws of planetary motion.

Vector Calculus

Vector calculus is a field of mathematics concerned with differentiation and integration of vector fields, primarily in three-dimensional Euclidean space.

In our exercise, vector calculus is employed to analyze the motion of a planet around a star. Specifically, we're using vectors to describe the velocity and acceleration of the planet in polar coordinates. Unit vectors in the radial \( \mathbf{u}_r \) and transverse \( \mathbf{u}_\theta \) directions are used to express these quantities. The derivatives of these unit vectors with respect to the angle 'θ' reveal the relationship between the change in direction of the radius vector and the motion of the planet.

Central Force

A central force is a force that is directed along the line joining the interacting particles and is directed towards a fixed center. In the case of planetary motion, this fixed center is typically the star or planet that exerts a gravitational pull on the orbiting object.

In the given problem, the assumption of a central force allows us to state that the transverse component of the acceleration \( a_{\theta} \) equals zero. This is a crucial simplification and reflects the conservation of angular momentum, leading us towards the demonstration of Kepler's second law.

Differential Equations

Differential equations describe the relationship between functions and their derivatives and are fundamental in expressing the dynamics of physical systems. In our scenario, they're used to describe how the velocity and acceleration of a planet change with time under the influence of a central force.

By differentiating the velocity vector with respect to time, and using the condition that the transverse component of acceleration is zero \( a_{\theta} = 0 \) for a central force, we're able to derive a differential equation that is subsequently integrated to find a constant \( k \), representing the constant areal velocity. This leads to the proof of Kepler's second law, which states that the line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.