Word Problems – 3

This one is a little different. Recently, I have had a few students struggling with literal equations. These are equations with many letters or symbols. For science and engineering wannabes, you need to develop the skill to work with these. Below is an example from orbital dynamics, but first some background.

Two Body Problem

Accurately determining orbits in the real world, requires computers. However, a good approximation that makes orbital calculations possible by hand (OK, using calculators), is to assume that the only two bodies that exist in the universe are the bodies orbiting each other. The shape of an orbit of a body in orbit around another in this universe can have 1 of 4 shapes: circular, elliptical, parabolic, or hyperbolic. I won’t talk about the last three, but let’s consider the circular orbit.

Here is a picture of an circular orbit:

In this kind of orbit, the earth is at the centre and the satellite follows an circular path around the earth.

In an orbit, such as a satellite orbiting the earth, we want to know a lot of things about the orbit, but two primary things are the satellite’s distance from the earth and its position. To measure its position, an arbitrary axis is agreed upon and the satellite’s position is its angle πœƒ measured from this axis. In orbital dynamics, the angle πœƒ is called the true anomaly. I don’t know why it is called that, but a definition of anomaly is “a deviation from the normal”. If we consider the normal being on the reference axis, this term makes a bit more sense.

Now I chose a circular orbit because the equations describing it are simpler than for other shapes. For example, the time it takes for a body to complete one orbit, either circular or elliptical is

\[T=\frac{2\pi}{\mu^2}\left(\frac{h}{\sqrt{1-e^2}}\right)^3\ \ \ \ \ \ \ \ \ (1)\]

where

T = the period, the time it takes to complete one orbit

ΞΌ = gravitational parameter. It is a combination of the gravitation constant (a constant in the universe) and the masses of the bodies

h = angular momentum per unit mass. Since the body has mass and is rotating around the other body, it has angular momentum

e = the eccentricity of the orbit. A measure of how elliptical the orbit is.

The eccentricity of a circular orbit is 0 and this simplifies equation (1).

So now the word problem.

The Word Problem

Equation (1) applies to elliptical and circular orbits. For a circular orbit, e = 0. The radius of a circular orbit is r. There is a relationship between the angular momentum and the radius of a circular orbit:

\[r=\frac{h^2}{\mu}\ \ \ \ \ \ \ \ \left(2\right)\]

If a satellite is at πœƒ = 0 at t = 0, the time it takes to travel πœƒ radians is

\[t=\frac{\theta}{2\pi}T\ \ \ \ \ \ \ \ \left(3\right)\]

So using equations (1), (2), and (3), develop an expression for t in terms of r and πœƒ and an expression for πœƒ in terms of r and t. Remember that πœ‹ and ΞΌ are constants so they can be in these expressions.

The Solution

First, we can simplify equation (1) by substituting e = 0:

\[T=\frac{2\pi}{\mu^2}\left(\frac{h}{\sqrt{1-e^2}}\right)^3\Longrightarrow\frac{2\pi}{\mu^2}h^3\ \ \ \ \ \ \ \ \ \ \left(4\right)\]

We can rearrange equation (2) to get h in terms of r:

\[r=\frac{h^2}{\mu}\Longrightarrow h=\sqrt{r\mu}\ \ \ \ \ \ \ \ \ \ \ \ (5)\]

Now substitute h from equation (5) into equation (4):

\[T=\frac{2\pi}{\mu^2}h^3=\frac{2\pi}{\mu^2}\left(({r\mu)}^\frac{1}{2}\right)^3=\frac{2\pi r^\frac{3}{2}\mu^\frac{3}{2}}{\mu^2}=\frac{2{\pi r}^\frac{3}{2}}{\sqrt\mu}\ \ \ \ \ \ \ (6)\]

To follow the development in (6), you need to remember the exponent rules and how to convert between a fractional exponent and square root notation.

Now substitute this into equation (3):

\[t=\frac{\theta}{2\pi}T=\frac{\theta}{2\pi}\frac{2{\pi r}^\frac{3}{2}}{\sqrt\mu}=\frac{{\theta r}^\frac{3}{2}}{\sqrt\mu}\ \ \ \ \ \ \ \ \ \ \ (7)\]

which is one of the answers. Rearranging this to solve for πœƒ gives the second answer:

\[t=\frac{{\theta r}^\frac{3}{2}}{\sqrt\mu}\Longrightarrow\theta=\frac{t\sqrt\mu}{r^\frac{3}{2}}\ \ \ \ \ \ \ \ \ \ \ (8)\]

Note that only basic algebra was used to find the answers. It doesn’t matter if you have equations with numbers or a lot of letters. The steps to find a solution are the same.