Problem 18 - Advanced Engineering Mathematics

Question

(a) Find the curvature of an elliptical orbit that is described by \(\mathbf{r}(t)=a \cos t \mathbf{i}+b \sin t \mathbf{j}+c \mathbf{k}, a>0, b>0, c>0\). (b) Show that when \(a=b\), the curvature of a circular orbit is the constant \(\kappa=1 / a\).

Step-by-Step Solution

Answer

(a) \( \kappa = \frac{ab}{(a^2 \sin^2 t + b^2 \cos^2 t)^{3/2}} \); (b) For \( a = b \), \( \kappa = \frac{1}{a}. \)

Step 1 - Find the Velocity Vector

Begin by calculating the velocity vector, which is the derivative of \( \mathbf{r}(t) \). Since \( \mathbf{r}(t) = a \cos t \mathbf{i} + b \sin t \mathbf{j} + c \mathbf{k} \), differentiate each component with respect to \( t \). This gives: \( \mathbf{v}(t) = \frac{d}{dt}(a \cos t) \mathbf{i} + \frac{d}{dt}(b \sin t) \mathbf{j} + \frac{d}{dt}(c) \mathbf{k} = -a \sin t \mathbf{i} + b \cos t \mathbf{j} + 0 \mathbf{k}. \) Hence, \( \mathbf{v}(t) = -a \sin t \mathbf{i} + b \cos t \mathbf{j}. \)

Step 2 - Find the Acceleration Vector

Next, find the acceleration vector, which is the derivative of the velocity vector. Differentiate the components of \( \mathbf{v}(t) \): \( \mathbf{a}(t) = \frac{d}{dt}(-a \sin t) \mathbf{i} + \frac{d}{dt}(b \cos t) \mathbf{j} = -a \cos t \mathbf{i} - b \sin t \mathbf{j}. \) So, \( \mathbf{a}(t) = -a \cos t \mathbf{i} - b \sin t \mathbf{j}. \)

Step 3 - Compute the Cross Product of Velocity and Acceleration

To find the curvature, compute the cross product of velocity and acceleration vectors: \( \mathbf{v}(t) \times \mathbf{a}(t) \). Since both vectors lie in the \( \mathbf{i} \) and \( \mathbf{j} \) plane, their cross product will only have a \( \mathbf{k} \) component: \(( -a \sin t)(-b \sin t) \mathbf{k} - (b \cos t)(-a \cos t) \mathbf{k} = (ab \sin^2 t + ab \cos^2 t) \mathbf{k} = ab \mathbf{k}. \)

Step 4 - Magnitude of Velocity and Cross Product

Calculate the magnitudes: \( \| \mathbf{v}(t) \| = \sqrt{(-a \sin t)^2 + (b \cos t)^2} = \sqrt{a^2 \sin^2 t + b^2 \cos^2 t }. \) The magnitude of the cross product is \( \| \mathbf{v} \times \mathbf{a} \| = |ab|. \)

Step 5 - Formula for Curvature

The formula for curvature \( \kappa \) is given by \( \kappa = \frac{\| \mathbf{v} \times \mathbf{a} \|}{\| \mathbf{v} \|^3}. \) Substitute the magnitudes: \( \kappa = \frac{ab}{(a^2 \sin^2 t + b^2 \cos^2 t)^{3/2}}. \)

Step 6 - Simplifications for Circular Orbit (a = b)

For part (b), set \( a = b \). Then, \( a^2 \sin^2 t + b^2 \cos^2 t = a^2 (\sin^2 t + \cos^2 t) = a^2. \) Thus, \( \kappa = \frac{a^2}{(a^2)^{3/2}} = \frac{a^2}{a^3} = \frac{1}{a}. \) This confirms that the curvature is constant and equals \( \frac{1}{a} \).

Key Concepts

Elliptical Motion

Elliptical motion refers to the path followed by an object moving in an ellipse, a shape that looks like a stretched circle. In the exercise, the given position vector \( \mathbf{r}(t) = a \cos t \mathbf{i} + b \sin t \mathbf{j} + c \mathbf{k} \) describes an elliptical orbit where the constants \( a \), \( b \), and \( c \) represent the semi-major axis, semi-minor axis, and the constant height along the \( \mathbf{k} \)-axis, respectively.

Elliptical orbits are common in astronomy, with planets and satellites often following these paths. The major and minor axes determine the shape and size of the ellipse. If \( a = b \), the orbit becomes circular, simplifying calculations and interpretations. Studying elliptical motion helps us understand the dynamics of bodies moving under gravitational forces, with applications beyond textbooks, into real-world scenarios like space exploration.

Velocity and Acceleration Vectors

In elliptical motion, understanding the velocities and accelerations is crucial for grasping how objects move along their paths. The velocity vector is obtained by differentiating the position vector \( \mathbf{r}(t) \) with respect to time \( t \). The exercise provides the velocity vector as \( \mathbf{v}(t) = -a \sin t \mathbf{i} + b \cos t \mathbf{j} \), which shows us how fast and in what direction the object is moving at any point.

The acceleration vector is similarly derived by differentiating the velocity vector. In this case, it shows as \( \mathbf{a}(t) = -a \cos t \mathbf{i} - b \sin t \mathbf{j} \). This vector gives insights into how the speed and direction of an object are changing over time. Together, these vectors are essential to describe motion, helping to predict future positions and paths.

Cross Product

The cross product of two vectors is a mathematical operation that helps to find a vector that is perpendicular to the plane formed by the initial vectors. In the context of the exercise, the cross product \( \mathbf{v}(t) \times \mathbf{a}(t) = ab \mathbf{k} \) is crucial in determining curvature, which indicates how sharply a path bends at any point.

By using the cross product, we isolate components that contribute to changes in direction and magnitude in the motion of an object on its elliptical path. This operation reduces computational complexity, focusing solely on the \( \mathbf{k} \)-component. It connects deeply to physics-related phenomena such as angular momentum and rotational motion, giving us tools to visualize and quantify twisting forces.

Curvature Formula

Curvature is a measure of how much a curve deviates from being a straight line. In this exercise, the curvature \( \kappa \) is given by the formula \( \kappa = \frac{\| \mathbf{v} \times \mathbf{a} \|}{\| \mathbf{v} \|^3} \). It uses the magnitudes of the velocity vector and the result of the cross product.

For an elliptical orbit, the curvature varies along the path due to changing speeds and directions. However, if the path is circular (when \( a = b \)), \( \kappa \) simplifies to a constant value \( \frac{1}{a} \), suggesting uniform bending or turning. This simplification underscores the close relationship between ellipse and circle characteristics, aiding in comprehension and practical calculation of circular orbits.