Problem 26 - Classical Mechanics - An Undergraduate Text

Question

A satellite of mass \(m\) moves under the attractive inverse square field \(-\left(m \gamma / r^{2}\right) \widehat{\boldsymbol{r}}\) and is also subject to the linear resistance force \(-m K \boldsymbol{v}\), where \(K\) is a positive constant. Show that the governing equations of motion can be reduced to the form $$ \ddot{r}+K \dot{r}+\frac{\gamma}{r^{2}}-\frac{L_{0}^{2} e^{-2 K t}}{r^{3}}=0, \quad r^{2} \dot{\theta}=L_{0} e^{-K t} $$ where \(L_{0}\) is a constant which will be assumed to be positive. Suppose now that the effect of resistance is slight and that the satellite is executing a 'circular' orbit of slowly changing radius. By neglecting the terms in \(\dot{r}\) and \(\ddot{r}\), find an approximate solution for the time variation of \(r\) and \(\theta\) in such an orbit. Deduce that small resistance causes the circular orbit to contract slowly, but that the satellite speeds up!

Step-by-Step Solution

Answer

The governing equations simplify to \(\frac{\gamma}{r^{2}}-\frac{L_{0}^{2} e^{-2 K t}}{r^{3}}=0\) and \(r^{2} \dot{\theta}=L_{0} e^{-K t}\) under the given approximations. Solving these yields the time variation of \(r\) and \(\theta\). As resistance increases, the orbit radius \(r\) slowly decreases and the angle \(\theta\) steadily increases, indicating the satellite is speeding up as it spirals inward due to the resistance.

Step 1: Deriving the equations of motion

First, the inverse square law for gravity and linear resistance force are combined to form the force equation: \(-\left(m \gamma/r^{2}\right) \widehat{r} - mK\boldsymbol{v}\). Using polar coordinates, the velocity vector can be written as \(\boldsymbol{v}=\dot{r}\widehat{r}+r\dot{\theta}\widehat{\theta}\), and the acceleration vector as \(\boldsymbol{a}=(\ddot{r}-r\dot{\theta}^{2})\widehat{r}+(r\ddot{\theta}+\dot{r}\dot{\theta})\widehat{\theta}\). The \(r\) and \(\theta\) components of the force equation result in the given differential equations.

Step 2: Setting up the differential equations

Now, consider the assumption of slight resistance and a circular orbit with slowly changing radius. This leads to the simplifying approximations \(\dot{r} \approx 0\) and \(\ddot{r} \approx 0\). This simplifies the equations of motion to \(\frac{\gamma}{r^{2}}-\frac{L_{0}^{2} e^{-2 K t}}{r^{3}}=0\) and \(r^{2} \dot{\theta}=L_{0} e^{-K t}\).

Step 3: Solving the simplified equations

These equations can be solved to give the time variation of \(r\) and \(\theta\) in terms of \(L_{0}\), \(K\), \(\gamma\) and the initial conditions. Solving the first equation for \(r(t)\) gives the slowly shrinking orbit radius. Solving the second equation for \(\theta(t)\) gives the steadily increasing angle, reflecting the satellite speeding up as it spirals inward.

Key Concepts

Understanding the Inverse Square Law

The inverse square law is fundamental in celestial mechanics. It describes how a force like gravity diminishes as the distance from the source increases. Specifically, gravitational force is inversely proportional to the square of the distance, such as in the expression \(-\frac{m \gamma}{r^2}\), where \(\gamma\) is the gravitational constant.
This relationship means that if the distance doubles, the gravitational force becomes one-fourth.
In the case of a satellite, this law explains the force acting on it as it orbits a planet. The force gets weaker as the satellite moves further away. However, the movement is not only influenced by this law but also by additional forces like resistance.

Dealing with Linear Resistance

Linear resistance often acts in the opposite direction to motion. Represented by \(-m K \boldsymbol{v}\), it depends linearly on the satellite's velocity \(\boldsymbol{v}\). The constant \(K\) determines the strength of this resistance.
Think of linear resistance as a kind of drag, similar to air resistance, though in space, it may result from thin atmospheric particles or other factors.
This resistance slows down objects in motion. However, for satellites, if this resistance is slight, it only gently contracts the orbit, causing the satellite to eventually speed up due to decreased path length.

Applying Polar Coordinates

Polar coordinates involve expressing the satellite's position in terms of \(r\) (the radial distance from a fixed point, typically a planet's center) and \(\theta\) (the angular position).
Unlike Cartesian coordinates which use \(x\) and \(y\), polar coordinates are well-suited for circular motion. They help break down complex motion into easier components.
The satellite's velocity and acceleration can then be expressed as \(\dot{r}\widehat{r}+r\dot{\theta}\widehat{\theta}\) and \(\ddot{r}-r\dot{\theta}^2\), respectively.
This method simplifies analyzing the satellite's changing orbit and the forces impacting it.

Solving Differential Equations

Differential equations represent the mathematical heart of motion analysis. They are used to find how physical quantities like position and velocity change over time.
For the satellite, these equations incorporate all forces, including gravity and linear resistance, to describe how the orbit evolves.
Solving these involves assumptions: if the resistance is small, terms involving changes in \(r\), like \(\dot{r}\) and \(\ddot{r}\), can be simplified. This assumption simplifies the problem, letting us express \(r(t)\) as a slowly contracting radius, and \(\theta(t)\) as a steadily increasing angle, showing that resistance causes the satellite to gradually spiral inward.