## Introduction

First, let’s take it for granted that the differential equation for the two-body problem (e.g. the earth orbiting the sun in an otherwise empty universe) is

$\ddot{\vec{r}} + \frac{\mu}{r^3}\ \vec{r} = 0$

where $\vec{r}\$ is the vector from the primary (e.g. the sun with mass M) to the secondary (e.g. the earth, with mass m) and $\mu = G(M+m)\$; and a simple $r\$ is the magnitude of the vector $\vec{r}\$.

I want to derive

• conservation of energy
• conservation of angular momentum
• a neat equation for r and v
• the scalar equation for the orbit.

Furthermore, I want to use vector operations wherever possible. (I like vectors!)

If you’ve never seen any of this before, it will probably be a bit strange. I expect to talk in much more detail about the scalar equation for 2-body orbits. But I really like this set of derivations (which can be found, with a tad less explanation, in Bate, Mueller, and White’s “Fundamentals of Astrodynamics”; or partly in Prusing & Conway’s “Orbital Mechanics” — both of which are on my bibliography page.)

I’m going to need a few vector identities that you may have forgotten, and a few interesting observations about the derivatives of dot and cross products of vectors. Rather than spring them on you in the course of the derivation, let me collect them up front and show them to you before I begin.

I A.
The vector triple product $\vec{a} \times (\vec{b} \times \vec{c})\$ can be written as the difference of two dot products:

$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}\ (\vec{a} \cdot \vec{c}) - \vec{c}\ (\vec{a} \cdot \vec{b})\$

NOTE that we could write a similar, but different, expression for

$(\vec{a} \times \vec{b}) \times \vec{c}\$

The cross product is not associative, and the parentheses (the order) matters.

By the way, in the form I showed, it is called “the BAC CAB rule”, for the letters on the RHS. The problem is that I can never remember where the parentheses are in the triple cross product, nor where the dots are in BAC and CAB. So I look it up. The point is that I know it exists and where I can find it. (In my case, the Berkeley “Mechanics” or Marion’s “Classical Dynamics of Particles and Systems”, both on my bibliography page.)

I B.
The scalar triple product $\vec{a}\ \cdot\ \vec{b} \times \vec{c}\$ is almost unique: that is, we may take the cross product of any two, and dot with the third (but the order matters in the cross product). Here’s one set, and we would get another set by replacing $\vec{b}\ \times\ \vec{c}\$ by $\vec{c}\ \times\ \vec{b}\$ etc.:

$\vec{a} \cdot \vec{b} \times \vec{c} = \vec{a} \times \vec{b} \cdot \vec{c} = \vec{b} \cdot \vec{c}\times \vec{a}\$

By the way, we do not need parentheses for the expression $\vec{a}\ \cdot\ \vec{b} \times \vec{c}\$ because it only makes sense if we take the cross product first and the dot product second; taking the dot product first gives us a scalar — and makes the cross product impossible.

I C.

Let me remind you that the cross product of a vector with itself is zero:

$\vec{a} \times \vec{a} = 0$

II A.
Then we have an interesting observation about the derivative and magnitude of a vector. If we have a vector a, then its magnitude squared is

$\vec{a} \cdot \vec{a} = a^2$

and if we differentiate that equation with respect to time, we get

$2\ \vec{a} \cdot \dot{\vec{a}} = 2\ a\ \dot{a}$

$\vec{a} \cdot \dot{\vec{a}} = a\ \dot{a}$

The LHS is the dot product of a vector and its time derivative; the RHS is the product of two scalars, namely the magnitude of the vector and the derivative of the magnitude.

Please note that I am using the symbol $\dot{a}$ (which has no arrow on it) as

$\dot{a} = \frac{d}{dt}\ a$

and most emphatically NOT as the magnitude of the vector $\dot{\vec{a}}\$. In English, $\dot{a}$ is the derivative of the magnitude rather than the magnitude of the derivative.

We will use this for a = r, to cancel some powers of r in the differential equation… we will also use it to recognize the derivative of the (specific, i.e. unit mass) kinetic energy:

$\vec{v} \cdot \dot{\vec{v}}= \frac{d}{dt}\ (\frac{v^2}{2})\$.

II B.
That the cross product of a vector with itself is zero has an interesting consequence when we take the derivative of $\vec{r} \times \vec{v}\$ (written as $\vec{r} \times \dot{\vec{r}}\$): one of the terms drops out, because it’s a vector crossed with itself:

$\frac{d}{dt}(\vec{r} \times \dot{\vec{r}}) = \dot{\vec{r}} \times \dot{\vec{r}} + \vec{r} \times \ddot{\vec{r}} = \ \vec{r} \times \ddot{\vec{r}}$.

That is,

$\frac{d}{dt}(\vec{r} \times \dot{\vec{r}}) = \vec{r} \times \ddot{\vec{r}}\$.

II C.
We will also want to recognize the derivative of a unit vector. From a vector $\vec{r}\$, we may form a unit vector $\hat{r}\$ as

$\hat{r} = \frac{\vec{r}}{r}\$.

Take the derivative:

$\frac{d}{dt}{\frac{\vec{r}}{r}} = \frac{\dot{\vec{r}}}{r} - \frac{\vec{r}\ \dot{r}}{r^2}\$.

As before, the symbol $\dot{r}\$ represents the derivative of the magnitude r.

## Summary

I A. The BAC-CAB rule for the vector triple product:
$\vec{a} \times (\vec{b} \times \vec{c}) = \vec{b}\ (\vec{a} \cdot \vec{c}) - \vec{c}\ (\vec{a} \cdot \vec{b})\$

I B. The scalar triple product:
$\vec{a} \cdot \vec{b} \times \vec{c} = \vec{a} \times \vec{b} \cdot \vec{c} = \vec{b} \cdot \vec{c}\times \vec{a}\$.

I C. $\vec{a} \times \vec{a} = 0\$.

II A. $\vec{a} \cdot \dot{\vec{a}} = a\ \dot{a}$

II B. $\frac{d}{dt}(\vec{r} \times \dot{\vec{r}}) = \vec{r} \times \ddot{\vec{r}}\$.

II C. The derivative of a unit vector:
$\frac{d}{dt}{\frac{\vec{r}}{r}} = \frac{\dot{\vec{r}}}{r} - \frac{\vec{r}\ \dot{r}}{r^2}\$.

## Conservation of Energy

Take the original differential equation…

$\ddot{\vec{r}} + \frac{\mu}{r^3}\ \vec{r} = 0$

Dot that with the velocity $\vec{v} = \dot{\vec{r}}\$

$\dot{\vec{r}} \cdot \ddot{\vec{r}} + \frac{\mu}{r^3}\ \dot{\vec{r}} \cdot \vec{r}$

$\vec{v} \cdot \dot{\vec{v}} + \frac{\mu}{r^3}\ \dot{\vec{r}} \cdot \vec{r}\$.

By II A we may replace the dot products by scalars…

$v\ \dot{v} + \frac{\mu}{r^3}\ r\ \dot{r} = 0$

$v\ \dot{v} + \frac{\mu}{r^2}\ \dot{r} = 0\$,

and we recognize both terms as derivatives:

$\frac{d}{dt}(\frac{v^2}{2}) - \mu\ \frac{d}{dt}(\frac{1}{r}) = 0\$

i.e.

$\frac{d}{dt}(\frac{v^2}{2} - \frac{\mu}{r}) = 0\$

i.e.

$\frac{v^2}{2} - \frac{\mu}{r} = \cal{E}\$, a constant.

The left term sure looks like kinetic energy — and the right term is potential!

That is, (specific mechanical) energy is conserved, and is the sum of the specific kinetic energy and the specific potential energy. (Specific? That’s per unit mass — there’s no mass multiplying the kinetic energy…)

(In fact, we could write two constants, and use one of them to explicitly set the zero of potential energy. I have chosen PE = 0 at infinity, as is customary. It will mean that if specific mechanical energy is non-negative, the object is not bound to its primary; a bound orbit is characterized by negative specific mechanical energy. And since mass is positive, the total mechanical energy of the object has the same sign, of course.)

## Conservation of Angular Momentum

$\ddot{\vec{r}} + \frac{\mu}{r^3}\ \vec{r} = 0$

and take the cross product by $\vec{r}\$:

$\vec{r} \times \ddot{\vec{r}} + \frac{\mu}{r^3} \vec{r} \times \vec{r} = 0\$.

But the $\vec{r} \times \vec{r}\$ term vanishes, leaving

$\vec{r} \times \ddot{\vec{r}} = 0\$.

By IIB, we recognize that as a derivative, namely

$\vec{r} \times \ddot{\vec{r}} = \frac{d}{dt}(\vec{r} \times \dot{\vec{r}}) = \frac{d}{dt}(\vec{r} \times \vec{v}) = 0\$.

But $\vec{r} \times \vec{v}\$ is angular momentum! OK, there’s no mass term, so it’s specific angular momentum. And that equation says that the time derivative of angular momentum is 0.

Letting

$\vec{h} = \vec{r} \times \vec{v}$

we have just gotten conservation of (specific) angular momentum: it is constant.

Furthermore, as a cross product, it is perpendicular to the plane spanned by$\vec{r} \text{ and } \vec{v}\$. And because it’s constant, that plane is constant: the motion is confined to a fixed plane.

## a vector equation for r and v

$\ddot{\vec{r}} + \frac{\mu}{r^3}\ \vec{r} = 0$

and take the cross product by $\vec{h}\$:

$\ddot{\vec{r}} \times \vec{h} + \frac{\mu}{r^3}\ \vec{r} \times \vec{h} = 0\$.

This time we use the fact that h is constant, to write one term as a derivative:

$\frac{d}{dt}(\dot{\vec{r}} \times \vec{h}) + \frac{\mu}{r^3}\ \vec{r} \times \vec{h} = 0\$.

We would really like to see the other term as a derivative, too. Use the definition of $\vec{h} = \vec{r} \times \vec{v}\$

to get

$\vec{r} \times \vec{h} = \vec{r} \times (\vec{r} \times \vec{v})\$,

remembering that we need parentheses because the cross product is not associative.

Now use the BAC-CAB rule IA, and our equation becomes

$\frac{d}{dt}(\dot{\vec{r}} \times \vec{h}) + \frac{\mu}{r^3}\ (\vec{r}\ (\vec{r} \cdot \vec{v}) - \vec{v}\ (\vec{r} \cdot \vec{r}) ) = 0\$.

As before, use IIA to write write the dot products as products of scalars and cancel some r’s.

$\frac{d}{dt}(\dot{\vec{r}} \times \vec{h}) + \frac{\mu}{r^3}\ (\vec{r}\ r\ \dot{r} - \vec{v}\ r^2 ) = 0\$.

$\frac{d}{dt}(\dot{\vec{r}} \times \vec{h}) + \frac{\mu}{r^2}\ \vec{r}\ \dot{r} - \frac{\mu}{r}\ \vec{v} = 0\$.

$\frac{d}{dt}(\dot{\vec{r}} \times \vec{h}) + \frac{\mu}{r^2}\ \vec{r}\ \dot{r} - \frac{\mu}{r}\ \dot{\vec{r}} = 0\$.

By IIC, two of those terms are the negative of the derivative of the unit vector $\frac{\vec{r}}{r}\$, so we get

$\frac{d}{dt}(\dot{\vec{r}} \times \vec{h}) - \frac{d}{dt}(\frac{\mu\ \vec{r}}{r}) = 0\$.

That is,

$\frac{d}{dt}(\dot{\vec{r}} \times \vec{h} - \frac{\mu\ \vec{r}}{r}) = 0\$.

So we integrate, including a vector constant of integration. Let me say that again: $\vec{B}\$ is a constant vector.

$\dot{\vec{r}} \times \vec{h} - \frac{\mu\ \vec{r}}{r} = \vec{B}\$.

Weird? Well, we will eventually see that B points from the occupied focus to periapse, the point of closest approach. And since that’s going to be our x-axis in the orbit coordinate system, B is really handy — along with h, it will tell us how to change coordinate systems.

## The scalar equation

$\dot{\vec{r}} \times \vec{h} - \frac{\mu\ \vec{r}}{r} = \vec{B}\$.

Take a dot product with $\vec{r}$

$\vec{r} \cdot \dot{\vec{r}} \times \vec{h} - \frac{\mu}{r}\ \vec{r} \cdot \vec{r} = \vec{r} \cdot \vec{B}$

and simplify the $\vec{r} \cdot \vec{r}\$ term…

$\vec{r} \cdot \dot{\vec{r}} \times \vec{h} - \mu\ r = \vec{r} \cdot \vec{B}\$.

Use the invariance of the scalar triple product to rewrite the leftmost term…

$\vec{r} \times \dot{\vec{r}} \cdot \vec{h} - \mu\ r= \vec{r} \cdot \vec{B}\$.

But that’s

$\vec{r} \times \vec{v} \cdot \vec{h} - \mu\ r= \vec{r} \cdot \vec{B}\$,

and that in turn is

$\vec{h} \cdot \vec{h} - \mu\ r= \vec{r} \cdot \vec{B}\$.

Now, letting $\nu\$ be the angle between B and r, we get

$h^2 - \mu\ r = r B cos(\nu)\$.

OK, now solve for r:

$r = \frac{h^2/\mu}{1 - B/\mu\ cos(\nu)}\$.

Finally, with $e = B/\mu\$ and $p = h^2 / \mu \$,

we end up with

$r = \frac{p}{1 + e\ cos(\nu)}\$.

That will be our starting point when we pick this up again.

This is the polar form of the equation of a conic section. it can degenerate into a line (or two?), but for nice values it’s a nice conic. If e = 0, it’s a circle of constant radius… for e strictly between 0 and 1, it’s an ellipse… for e = 1, it’s a parabola… and for e > 1, it’s a hyperbola.

## Initial conditions?

Suppose we are given the position and velocity at some instant of time… call them $\vec{r_0} \text{ and } \vec{v_0}\$. Then, since angular momentum is constant, we compute it once and forever — for this orbit — from the initial conditions:

$h = \vec{r_0} \times \vec{v_0}\$.

To get $\vec{B}\$, we rewrite

$\dot{\vec{r}} \times \vec{h} - \frac{\mu\ \vec{r}}{r} = \vec{B}\$

as

$\vec{v} \times \vec{h} - \frac{\mu\ \vec{r}}{r} = \vec{B}\$

and plug in vo and ro:

$\vec{v_0} \times \vec{h} - \frac{\mu\ \vec{r_0}}{r_0} = \vec{B}\$.

and since B is constant, we have it once and forever — for this orbit.

Our equations

$\dot{\vec{r}} \times \vec{h} - \frac{\mu\ \vec{r}}{r} = \vec{B}\$

and

$r = \frac{p}{1 + e\ cos(\nu)}\$

are completely determined.

Finally, the maximum of r occurs when $cos\ \nu = 1\$, i.e. when $\nu = 0\$. So $\nu\$ , originally defined as the angle between B and r, is equal to zero at minimum distance, i.e. when B are r are parallel. Therefore, B points toward periapse.

Feel free to admire how much we got from the vector operations, even if you’re not quite sure just what we got.