Question

Suppose we wish to put a satellite with mass \(m\) into an elliptical orbit around Earth. Assume that the only force acting on the object is Earth's gravity, given by $$ \mathbf{F}(r, \theta)=-m g\left(\frac{R^{2}}{r^{2}}\right)(\cos \theta \mathbf{i}+\sin \theta \mathbf{j}) $$ where \(R\) is Earth's radius, \(g\) is the acceleration due to gravity at Earth's surface, and \(r\) and \(\theta\) are polar coordinates in the plane of the orbit, with the origin at Earth's center. (a) Find the eccentricity required to make the aphelion and perihelion distances equal to \(R \gamma_{1}\) and \(R \gamma_{2},\) respectively, where \(1<\gamma_{1}<\gamma_{2}\). (b) Find the initial conditions $$ r(0)=r_{0}, \quad r^{\prime}(0)=r_{0}^{\prime}, \text { and } \theta(0)=\theta_{0}, \quad \theta^{\prime}(0)=\theta_{0}^{\prime} $$ required to make the initial point the perigee, and the motion along the orbit in the direction of increasing \(\theta .\) HINT: Use the results of Exerise \(2 .\)

Short answer

Answer: The eccentricity, e, is given by the equation: e = (γ₂ - γ₁) / (γ₂ + γ₁)

Step-by-step solution

Recall the gravitational force formula

The formula given by the exercise is the gravitational force acting on the satellite: $$ \mathbf{F}(r, \theta)=-m g\left(\frac{R^{2}}{r^{2}}\right)(\cos \theta \mathbf{i}+\sin \theta \mathbf{j}) $$ Newton's second law of motion in polar coordinates states that: $$ m\left(\ddot{r}-r\dot{\theta}^{2}\right)=-mg\frac{R^2}{r^2}\cos\theta \qquad(1) $$ and $$ m\left(r\ddot{\theta}+2\dot{r}\dot{\theta}\right)=-mg\frac{R^2}{r^2}\sin\theta \qquad(2) $$ where the dots are derivatives with respect to time.

Find the eccentricity

First, we will look for the eccentricity required for given values of aphelion and perihelion distances. To find the eccentricity, we will need to use the polar equation of an ellipse centered at one of its foci (in this case, Earth's center): $$ r=\frac{a(1-e^2)}{1+e\cos\theta} $$ where \(a\) is the semimajor axis, \(e\) is the eccentricity, and \(r\) and \(\theta\) are the polar coordinates of the satellite's position. For an ellipse, \(0<e<1\). For perihelion, \(r=R\gamma_1\), which is the minimum distance from Earth, so \(\theta=0\): $$ R\gamma_1=\frac{a(1-e^2)}{1+e} $$ And for aphelion, \(r=R\gamma_2\), which is the maximum distance from Earth, so \(\theta=\pi\): $$ R\gamma_2=\frac{a(1-e^2)}{1-e} $$ Now we solve these equations simultaneously for \(e\) and obtain: $$ e=\frac{\gamma_2-\gamma_1}{\gamma_2+\gamma_1} $$

Find the initial polar coordinates and velocity components

Now we will find the initial conditions for the satellite's position and velocity. Since its initial position is at the perihelion, its polar coordinates are: $$ r(0)=R\gamma_1 \qquad \theta(0)=0 $$ Next, we need to find the initial polar component of velocities, \(r'(0)=r_0'\) and \(\theta'(0)=\theta_0'\). We can find these by using conservation of angular momentum (\(L\)) and conservation of mechanical energy. Conservation of angular momentum states that: $$ m\thinspace(r\thinspace^2\dot{\theta})=L \Rightarrow m\thinspace(R\gamma_1\thinspace^2\theta_{0}^{\prime})=L $$ Therefore, $$ \theta_{0}^{\prime} = \frac{L}{m\thinspace R^2 \gamma_1^2}\qquad(3) $$ To find the conservation of mechanical energy, we can write the total mechanical energy as the sum of kinetic and potential energies: $$ E=K+U=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-\frac{G\thinspace m\thinspace M}{r} \Rightarrow E=\frac{1}{2}m(\dot{r}^2+r^2\dot{\theta}^2)-\frac{m\thinspace g\thinspace R^2}{r} $$ At \(r(0)=R\gamma_1\) and \(\theta(0)=0\), we obtain: $$ E=\frac{1}{2}m(r_{0}'^2+\gamma_1^2R^2\theta_{0}'^2)-m\thinspace g\thinspace R $$ Now we can use equations (1) and (2) to find an expression for \(\ddot{r}\) and \(\ddot{\theta}\) and substitute them into the energy equation to solve for \(r_0'\). Finally, we will have the following initial conditions for the satellite: $$ r(0)=R\gamma_1, \qquad \theta(0)=0, \qquad r^{\prime}(0)=r_{0}^{\prime}, \qquad \theta^{\prime}(0)=\theta_{0}^{\prime} $$

Key concepts

Gravitational Force Formula

Understanding the gravitational force is essential when dealing with satellite motion. The force formula provided in the exercise,
\[\begin{equation}\textbf{F}(r, \theta)=-m g\left(\frac{R^{2}}{r^{2}}\right)(\cos \theta \textbf{i}+\sin \theta \textbf{j}) \end{equation}\]
, is a manifestation of Newton's law of universal gravitation for an object in a non-circular (elliptical) orbit around Earth.

This formula signifies that gravity's pull lessens with distance, following an inverse square law. Earth's gravitational force keeps the satellite in its orbit, and despite being directed towards the center of the Earth (\(\theta\) is the angle relative to the perpendicular position), it acts along the radial direction in the plane of the orbit given by the polar coordinates (\(r\) and \(\theta\)). The vector components, \(\cos \theta \textbf{i}\) and \(\sin \theta \textbf{j}\), indicate that the force has both horizontal and vertical components relative to Earth's surface.

Orbit Eccentricity Calculation

Elucidating the concept of eccentricity is key for students to comprehend the shape of the satellite's orbit. Eccentricity (\(e\)) is a dimensionless number that defines the deviation of an ellipse from a perfect circle, with values between 0 (a circle) and 1 (a parabola).

For an elliptical orbit, where two specific points, aphelion (\(R \gamma_2\)) and perihelion (\(R \gamma_1\)) are given, the eccentricity calculation can be addressed analytically. Using the derived equations, the eccentricity can be found as \(e=\frac{\gamma_2-\gamma_1}{\gamma_2+\gamma_1}\).

This equation unveils that the larger the difference between the aphelion and perihelion distances, the more 'stretched out' the orbit is. In other words, the satellite's path will be markedly more elliptical if \(\gamma_?\) values greatly differ.

Polar Coordinates Motion

Polar coordinates (\(r\), \(\theta\)) provide a two-dimensional coordinate system where each point on a plane is determined by a distance from a reference point and an angle from a reference direction.

Motion of satellites in an elliptical orbit is naturally expressed in polar coordinates since it simplifies the math when dealing with central forces that are directed along the radius vector of the motion's path. The motion can be decomposed into radial (distance change, \(r\)) and angular (positional change over the orbit, \(\theta\)) components which makes it easier to apply physical laws like conservation of angular momentum and mechanical energy specifically tailored for radial systems.

Conservation of Angular Momentum

Angular momentum conservation is a pivotal principle that provides insight into the motion of the satellite along its orbit. For a satellite in an elliptical orbit around Earth, the angular momentum,
\[\begin{equation}L = m(ro^2 \dot{\theta})\end{equation}\]
, is conserved because the net external torque acting on the system is zero.

The satellite's speed and radial distance from Earth are inversely related due to this conservation law — as the satellite moves closer to Earth, its angular speed increases, and vice versa. This is because the product of the moment of inertia (\(m r^2\)) and angular velocity (\(\dot{\theta}\)) must remain constant if no external torques are applied. It's this conservation that causes the satellite's swift movement at the perihelion and slow procession at the aphelion.

Conservation of Mechanical Energy

Mechanical energy conservation involves the constant energy of a system in which all forces are conservative, like gravitational force in our satellite case. The satellite's mechanical energy (\(E\)), composed of its kinetic (\(K\)) and potential energy (\(U\)), is conserved in the absence of non-conservative forces ( like atmospheric drag).

In the context of an elliptical orbit, this principle allows the computation of the satellite's velocity at different points along its orbit. At the perihelion or aphelion where the speed is extremal, we can use the energy conservation to establish the velocity needed to keep the orbit stable. The mechanical energy would be: \(E = \frac{1}{2}m(r_0'^2 + \gamma_1^2R^2\theta_0'^2) - m g R\).

Finding the exact balance between kinetic and potential energies is crucial for the satellite not to fall back to Earth or escape its gravitational pull entirely.