# Orbits – Introduction

I wanted to write a post about Hohmann transfer orbits, but this requires a brief discussion of orbits in general. The aim of this post is to justify and explain some of the equations commonly used in orbital mechanics questions. The two most important results are the two conservation laws that orbital systems obey.

We will discuss the simplified case of something like an artificial satellite orbiting a planet, a planet orbiting a star, or a star orbiting a black hole. These are cases in which the central body is sufficiently heavy relative to its satellite that it is practically stationary. This simplifies the algebra involved. In general, we will take the greater mass $M$ to be situated at the origin and the smaller mass $m$ to have position vector $\bold{r}$.

The gravitational force $\bold{F}$ acting on $m$ by mass $M$ is given by

$\displaystyle \bold{F}=-\frac{GMm}{r^2}\hat {\bold{r}}$

where $G$ is the gravitational constant, $r$ is the magnitude of $\bold{r}$ and $\hat {\bold{r}}$ is a unit vector in the direction of $\bold{r}$. This is perhaps the most common form in which $\bold{F}$ is quoted, so as not to obscure its defining property: the inverse square relation. It is this particular power which gives orbits their characteristic shapes, those of the conic sections.

Bound orbits generally take the form of ellipses, of which circles are a special case. Unbound orbits are hyperbolic. As the energy of the satellite increases, its trajectory undergoes an elegant and continuous variation from one conic section to the other; the ellipse gradually stretches out and instantaneously splits into a parabola, thence becoming an increasingly flat hyperbola. An animation is shown at the bottom of this post illustrating this evolution.

We will first use the expression for $\bold{F}$ to calculate (not particularly rigorously) an expression for the gravitational potential energy of the small body $m$. It is common to adopt the convention that the system has zero potential energy when the two bodies are displaced by an infinite distance. Hence in pushing the satellite directly towards the primary ‘from infinity’ to a distance $r$, we supply it with energy $U$ given by

$\displaystyle U=-\int\limits_{\infty}^{r}\bold{F}\cdot d\bold{r}$

$\displaystyle U=\int\limits_{\infty}^{r}\frac{GMm}{r^2}dr$

$\displaystyle U=\bigg[\frac{GMm}{r}\bigg]^{\infty}_{r}$

$\displaystyle U=-\frac{GMm}{r}$

To complete the derivation, we must show that the gravitational force is conservative – this means that, no matter what path it takes, the body must be supplied with the same amount of energy to reach a distance $r$ from its primary. A concise mathematical proof can be provided by the use of vector calculus, but that is not the subject of this post. For the moment, we will have to settle for the reasoning that moving the satellite ‘around’ the primary (keep $r$ constant) requires no work, since $\bold{F}$ is radial and hence is perpendicular to the ’roundwards’ (azimuthal) direction ($\bold{F}\cdot d\bold{x}=0$). Assuming the gravitational force is conservative, we are at liberty to use the conservation of energy in considering the orbit of a satellite.

There is another important property of orbits we can show using the expression for $\bold{F}$ above. Consider the vector $\bold{L}$ given by

$\bold{L}=m(\bold{r}\times\bold{\dot{r}})$

where the $\times$ operator represents the vector cross product. Let’s calculate how $\bold{L}$ varies with time. The product rule for differentiation can be used on cross products much like conventional products:

$\bold{\dot{L}}=m(\bold{\dot{r}}\times\bold{\dot{r}}+\bold{r}\times\bold{\ddot{r}})$

From the definition of the cross product, $\bold{\dot{r}}\times\bold{\dot{r}}=0$, as the cross product of two parallel vectors is zero, so

$\bold{\dot{L}}=m(\bold{r}\times\bold{\ddot{r}})$

Using $\bold{F}$, we can relate $\bold{r}$ and $\bold{\ddot{r}}$:

$\displaystyle m\bold{\ddot{r}}=-\frac{GMm}{r^2}\bold{\hat{r}}$

$\displaystyle \bold{\dot{L}}=\frac{GMm}{r^2}(\bold{r}\times\bold{\hat{r}})$

Recall at $\bold{\hat{r}}$ points in the same direction as $\bold{r}$, hence

$\bold{\dot{L}=0}$

Hence this vector is constant in time. Note that, because $\bold{L}$ is a vector, both its direction and its magnitude is constant. Each of these properties tells us something important about orbits.

First consider its directional aspect. The vector $\bold{r}\times\bold{\dot{r}}$ points in a direction perpendicular to both the position vector of the satellite and its velocity vector. A little thought should convince you that this constrains the satellite to move in a plane. This means that we only ever require two coordinates to describe the position of a satellite in its orbit.

Now consider the scalar aspect. The quantity $\mid\bold{L}\mid=mvr\sin\theta$ is conserved, where $\theta$ is the angle between the satellite’s position and velocity vectors. This is in fact the satellite’s angular momentum about the primary. That it is conserved is a consequence of the gravitational force’s being a central force, one which acts along the vector joining the two bodies.

The conservation of angular momentum is a useful tool is determining the velocity of a satellite at various points in its orbit. It provides a nice mathematical expression for something quite intuitive about orbits: the closer the satellite is to the primary, the faster it whirls around it, and vice versa. This phenomenon manifests itself markedly in satellites with highly elliptical orbits, such as comets.

Here is another of its uses: consider the expression for the total energy of the satellite-primary system. It is given by the sum of its kinetic energy and its gravitational potential:

$\displaystyle E=\frac{1}{2}m\bold{\dot{r}}^2-\frac{GMm}{r}$

We can assume that mechanical energy is conserved given that the gravitational force is conservative. The term $\bold{\dot{r}}^2$ can be split into two. The first is its radial component, given by the rate at which the satellite is falling towards or receding from the primary. The second is its tangential or azimuthal component, contributed by its motion around the primary. Hence $E$ can be expressed as follows:

$\displaystyle E=\frac{1}{2}m\dot{r}^2+\frac{1}{2}mr^2\dot{\varphi}^2-\frac{GMm}{r}$

where $\varphi$ is an angle subtended by the satellite with an arbitrary axis through the origin. By using the angular momentum, we can eliminate $\varphi$ entirely from this equation so that $E$ is a function of $r$ and its time derivatives only. Consider again the expression for $L$:

$L=mvr\sin\theta$

$L=mr(v\sin\theta)$

The quantity $v\sin\theta$ is the tangential component of the satellite’s velocity. This can be expressed in terms of $\dot{\varphi}^2$ by

$L=mr(r\dot{\varphi})$

$L=mr^2\dot{\varphi}$

$\displaystyle \dot{\varphi}=\frac{L}{mr^2}$

Hence we can replace $\dot{\varphi}$ in the expression $E$ with the relation above, yielding the equation

$\displaystyle E=\frac{1}{2}m\dot{r}^2+\frac{L^2}{2mr^2}-\frac{GMm}{r}$

We will use this expression, and the conservation of angular momentum, when considering Hohmann transfers in the next post.

This animation shows how the shape of the simplified two-body orbit changes as the energy of the satellite varies.

The mass of the satellite $m$ and the gravitational parameter of the primary $GM=\mu$ have both been taken to equal 1.

For $E=-0.5$, the orbit is circular. For $-0.5, the orbit is elliptical. When the net energy of the satellite is precisely $0$, the orbit is parabolic. For $E>0$, the orbit is hyperbolic.

That the orbit becomes unbound when $E\ge0$ is not a coincidence; it is a consequence of definition of the system’s potential energy. The way to think about it is this: the energy of the satellite is the sum of its kinetic and potential energy:

$E=\frac{1}{2}mv^2-\frac{GMm}{r}$

Take the case where the satellite’s total energy is greater than 0. We can then write

$\frac{1}{2}mv^2>\frac{GMm}{r}$

Let the satellite’s velocity be $V$ in the limit that $r$ tends to infinity. This gives

$\frac{1}{2}mV^2>0$

Hence when the satellite is ‘at infinity’, its velocity is non-zero. This means that the satellite is still ‘going’ even as it reaches an infinite distance from the primary! It can go to infinity … and beyond. Hence the orbit is unbound for $E\ge0$.