Two Body Problem

Devendra Ghate

How this deck is organised

The two-body problem splits cleanly into two halves, and it pays to keep them apart.

Part I — what every orbit shares. The equation of motion, the three integrals ($\mathbf{h}$, $\mathbf{e}$, $\varepsilon$), the orbit equation $r = p/(1 + e\cos f)$, vis-viva, the velocity components, the speed, the flight path angle and the hodograph. Not one of these derivations asks what $e$ is. They are written in $(p, e, f)$ or $(h, e, f)$ and hold for the circle, the ellipse, the parabola and the hyperbola alike.

Parts II–IV — where the conics part company. Only one question forces a split: where is the body at time $t$? The integral

\[ t - t_p = \frac{p^{2}}{h}\int_{0}^{f}\frac{df'}{\left(1 + e\cos f'\right)^{2}} \]

is evaluated by a different substitution in each case — the eccentric anomaly $E$ for $e<1$, Barker’s cubic for $e=1$, the hyperbolic anomaly $H$ for $e>1$. Everything that is genuinely conic-specific hangs off that one fork, together with the handful of quantities (apoapsis, the period $T$, $v_\infty$, the turning angle) that simply do not exist for all three.

Rule of thumb for the whole deck: if a formula contains $p$ and $e$ it is universal; if it contains $a$ and needs $a > 0$, or an apoapsis, or a period, it is elliptic. The parameter $a$ itself is universal — it is only its sign that sorts the conics.

Part I — What Every Orbit Shares

Two body Problem

From earlier equations (from the N-body Problem), we have

\[ \ddot{\mathbf{R}}_2 - \ddot{\mathbf{R}}_1 = \frac{G m_1}{r_{21}^{3}} \mathbf{r}_{21} - \frac{G m_2}{r_{12}^{3}} \mathbf{r}_{12} = -G (m_1 + m_2)\,\frac{\mathbf{r}_{12}}{r_{12}^{3}} \]

which is clearly,

\[ \ddot{\mathbf{r}}_{12} + \frac{G (m_1 + m_2)}{r_{12}^{3}} \mathbf{r}_{12} = 0 \]

Dropping the subscript, we have

\[ \ddot{\mathbf{r}} + \frac{\mu}{r^{3}} \mathbf{r} = 0, \qquad \mu = G (m_1 + m_2) \]

Here $\mu$ is called the standard gravitational parameter. It is a measure of the strength of the gravitational field of the two-body system. So to analyse the two-body problem, we don’t need to know the individual masses, but only this constant.

In the early development of celestial mechanics, the standard gravitational parameter was often used instead of the individual masses. Also, since the mass of the Sun is so much larger than the mass of any planet, as long as they had a good estimate of the mass of the Sun, they could calculate the orbits of the planets without knowing their masses.

We need to solve only this nonlinear second order ODE to get the relative motion of the two bodies. The individual motions can be recovered from the relative motion after that. We will see this process in the Example 1.

Angular Momentum Conservation

We will proceed as the previous section, and calculate all the integrals of motion.

Taking a cross product of the equation with $\mathbf{r}$, we have

\[ \mathbf{r} \times \ddot{\mathbf{r}} = 0 \]

Since, $\frac{d}{dt}(\mathbf{r} \times \dot{\mathbf{r}}) = \mathbf{r} \times \ddot{\mathbf{r}}$, we have

\[ \frac{d}{dt}(\mathbf{r} \times \dot{\mathbf{r}}) = 0 \]

Defining $\mathbf{h} = \mathbf{r} \times \dot{\mathbf{r}}$, we have $\frac{d\mathbf{h}}{dt} = 0$. Therefore, $\mathbf{h}$ is a constant vector. This will only happen if the motion is planar. Therefore, the relative motion of the two bodies is planar. This only means that $\mathbf{r}(t)$ remains in a single plane.

This is a strong geometric statement: the trajectory in 3D collapses to a curve in a single 2D plane. From here on, we can do orbital geometry in two dimensions.

Now let us recall Kepler’s second law, which states that the line joining the two bodies sweeps out equal areas in equal times. This is a consequence of the conservation of angular momentum.

Since $\mathbf{h}$ is constant, the areal velocity is also constant.

Is $h=0$ possible? If yes, what kind of motion will this result in?

Answer

Yes. $\mathbf{h} = \mathbf{r}\times\dot{\mathbf{r}} = \mathbf{0}$ exactly when $\mathbf{r}$ and $\dot{\mathbf{r}}$ are parallel, so the motion is rectilinear: a straight line through the centre of attraction. Dropping a body from rest is the standard example.

The conic degenerates with it: $p = h^{2}/\mu = 0$ and $e = 1$ whatever the energy, so $(a, e)$ stops describing a shape. Energy still sorts the cases — $\varepsilon < 0$ rises, stops and falls back; $\varepsilon = 0$ is the rectilinear parabola; $\varepsilon > 0$ escapes. This is the $h = 0$ caveat that makes $e = 1$ ambiguous.

Example 1

Figure 1: Two-body motion at two mass ratios. Both panels share the same total mass, the same initial separation and the same eccentricity ($e = 0.70$); only $m_1/m_2$ differs. Green arrows are velocities (drawn to scale), purple arrows the gravitational forces (drawn to a compressed $F^{1/3}$ scale, since $F \propto 1/d^2$ spans too wide a range). Download the script.

What the animation shows

The centre of mass never moves. Taking the origin at the barycentre, the two bodies stay diametrically opposite about it: $m_1 \mathbf{R}_1 + m_2 \mathbf{R}_2 = \mathbf{0}$ at every instant.
With $\mathbf{r} = \mathbf{r}_{12} = \mathbf{R}_2 - \mathbf{R}_1$ and $M = m_1 + m_2$ as before, the relative coordinate obeys $\ddot{\mathbf{r}} = -GM\,\mathbf{r}/r^3$, and the individual orbits are recovered from \[\mathbf{R}_1 = -\frac{m_2}{M}\,\mathbf{r}, \qquad \mathbf{R}_2 = \frac{m_1}{M}\,\mathbf{r}.\] So each body traces a similar conic about the centre of mass: same shape, same eccentricity, sizes in the inverse ratio of the masses. Note the weights are crossed — body 1 is scaled by $m_2$ — so the heavier body gets the smaller orbit.

How to find $R_1$ and $R_2$

Define the change of variables $(\mathbf{R}_1, \mathbf{R}_2) \mapsto (\mathbf{R}_{cm}, \mathbf{r})$:

\[ \begin{aligned} \mathbf{R}_{cm} &= \frac{m_1}{M}\mathbf{R}_1 + \frac{m_2}{M}\mathbf{R}_2, \\ \mathbf{r} &= -\mathbf{R}_1 + \mathbf{R}_2 . \end{aligned} \]

In matrix form, acting componentwise on each Cartesian component,

\[ \begin{pmatrix} \mathbf{R}_{cm} \\ \mathbf{r} \end{pmatrix} = \underbrace{\begin{pmatrix} m_1/M & m_2/M \\ -1 & 1 \end{pmatrix}}_{\displaystyle A} \begin{pmatrix} \mathbf{R}_1 \\ \mathbf{R}_2 \end{pmatrix}. \]

It is invertible as \[ \det A = \frac{m_1}{M} + \frac{m_2}{M} = 1 \neq 0 . \]

Invert $A$ directly:

\[ A^{-1} = \frac{1}{\det A}\begin{pmatrix} 1 & -m_2/M \\ 1 & \phantom{-}m_1/M\end{pmatrix} = \begin{pmatrix} 1 & -m_2/M \\ 1 & \phantom{-}m_1/M\end{pmatrix}, \]

that is,

\[ \boxed{\; \begin{aligned} \mathbf{R}_1(t) &= \mathbf{R}_{cm}(t) - \frac{m_2}{M}\,\mathbf{r}(t), \\[4pt] \mathbf{R}_2(t) &= \mathbf{R}_{cm}(t) + \frac{m_1}{M}\,\mathbf{r}(t). \end{aligned}\;} \]

The two forces are equal and opposite (Newton’s third law), but the accelerations are not: the heavier body responds $m_1/m_2$ times less.
Velocities can also be calculated in a similar fashion.

Towards the restricted problem

As $m_1/m_2$ grows, the orbit of the heavier body shrinks and the centre of mass sinks towards its centre — this is the approximation of the previous section, now visible.
For the Sun-Jupiter pair, $m_1/m_2 \approx 1047$, so the Sun’s orbit about the barycentre is $1/1047$ of Jupiter’s. It is small, but not zero: the barycentre lies about $1.07$ solar radii from the centre of the Sun, i.e. just outside its surface.
This justifies the use of restricted two body problem as a model to predict planetary trajectories.

Eccentricity Vector

Let us post multiply the equation of motion with $\mathbf{h}$ to get

\[ \ddot{\mathbf{r}} \times \mathbf{h} = -\frac{\mu}{r^3} \mathbf{r} \times \mathbf{h} = -\frac{\mu}{r^3} \mathbf{r} \times \mathbf{r} \times \dot{\mathbf{r}} \]

Since $\mathbf{a} \times (\mathbf{b} \times \mathbf{c}) = (\mathbf{a} \cdot \mathbf{c})\mathbf{b} - (\mathbf{a} \cdot \mathbf{b})\mathbf{c}$, we have

\[ \ddot{\mathbf{r}} \times \mathbf{h} = -\frac{\mu}{r^3} \left[ (\mathbf{r} \cdot \dot{\mathbf{r}})\mathbf{r} - (\mathbf{r} \cdot \mathbf{r})\dot{\mathbf{r}} \right] = -\frac{\mu}{r^3} \left[ (\mathbf{r} \cdot \dot{\mathbf{r}})\mathbf{r} - r^2\dot{\mathbf{r}} \right] \]

Since $\mathbf{r} = r \hat{\mathbf{r}}$, we have $\dot{\mathbf{r}} = \dot{r} \hat{\mathbf{r}} + r \dot{\hat{\mathbf{r}}}$,

\[ \begin{aligned} \mathbf{r} \cdot \dot{\mathbf{r}} & = & r \hat{\mathbf{r}} \cdot (\dot{r} \hat{\mathbf{r}} + r \dot{\hat{\mathbf{r}}})\\ & = & r \dot{r} \hat{\mathbf{r}} \cdot \hat{\mathbf{r}} + r^2 \hat{\mathbf{r}} \cdot \dot{\hat{\mathbf{r}}}\\ & = & r \dot{r} + 0\\ \end{aligned} \]

We have used, $\frac{d}{dt}(\hat{\mathbf{r}} \cdot \hat{\mathbf{r}}) = 0$, which implies $\hat{\mathbf{r}} \cdot \dot{\hat{\mathbf{r}}} = 0$.

Therefore,

\[ \ddot{\mathbf{r}} \times \mathbf{h} = -\frac{\mu}{r^3} \left[ r \dot{r} \mathbf{r} - r^2\dot{\mathbf{r}} \right] = \mu \left( \frac{\dot{\mathbf{r}}}{r} - \frac{\dot{r} \mathbf{r}}{r^2} \right) \]

Since, \[ \frac{d}{dt}\left( \frac{\mathbf{r}}{r} \right) = \frac{\dot{\mathbf{r}}}{r} - \frac{\dot{r}}{r^2}\mathbf{r} \]

we have,

\[ \ddot{\mathbf{r}} \times \mathbf{h} = \mu \frac{d}{dt}\left( \frac{\mathbf{r}}{r} \right) \]

Since $\mathbf{h}$ is constant, integrating both sides gives

\[ \dot{\mathbf{r}} \times \mathbf{h} = \mu \left( \frac{\mathbf{r}}{r} + \mathbf{e} \right) \]

Here, $\mathbf{e}$ is a dimensionless constant vector. We note that it is perpendicular to both the $\mathbf{h}$ and $\dot{\mathbf{r}}$ vectors. Since $\mathbf{h}$ is a constant vector, $\mathbf{e}$ clearly lies in the orbital plane.

Rearranging previous equation we get,

\[ \mathbf{e} = \frac{\dot{\mathbf{r}} \times \mathbf{h}}{\mu} - \frac{\mathbf{r}}{r} \]

Differentiating with respect to time, and using $\dot{\mathbf{h}} = 0$,

\[ \dot{\mathbf{e}} = \frac{\ddot{\mathbf{r}} \times \mathbf{h}}{\mu} - \frac{d}{dt}\left( \frac{\mathbf{r}}{r} \right) \]

But we have already shown that $\ddot{\mathbf{r}} \times \mathbf{h} = \mu \frac{d}{dt}\left( \frac{\mathbf{r}}{r} \right)$, and hence the two terms cancel exactly,

\[ \dot{\mathbf{e}} = \mathbf{0} \]

Let us also show that both its magnitude and its direction are fixed.

Magnitude. Writing $e = |\mathbf{e}|$,

\[ \frac{d}{dt}\left( e^2 \right) = \frac{d}{dt}(\mathbf{e} \cdot \mathbf{e}) = 2\, \mathbf{e} \cdot \dot{\mathbf{e}} = 0 \quad \Longrightarrow \quad e = \text{constant} \]

Direction. With $\hat{\mathbf{e}} = \mathbf{e}/e$, and using $\dot{e} = 0$ from above,

\[ \dot{\hat{\mathbf{e}}} = \frac{\dot{\mathbf{e}}}{e} - \frac{\dot{e}}{e^2}\mathbf{e} = \mathbf{0} - \mathbf{0} = \mathbf{0} \quad \Longrightarrow \quad \hat{\mathbf{e}} = \text{constant} \]

So $\mathbf{e}$ picks out one fixed line in the orbital plane, for all time.

$\mathbf{h}$ fixes the plane of motion, while $\mathbf{e}$ additionally fixes an orientation within that plane. We shall soon see that this fixed direction points towards the point of closest approach (the periapsis), and that $e$ is precisely the eccentricity of the conic. Its constancy is therefore the statement that the orbit does not precess — the ellipse traced out is closed, and returns on itself.

Precession is the change in the direction of the rotational axis of a spinning object.

The vector $\mu\,\mathbf{e}$ is known as the Laplace–Runge–Lenz vector, and its conservation is special to the inverse-square law. Any perturbation to $1/r^2$ — oblateness of the central body, a third body, or the corrections of general relativity — makes $\dot{\mathbf{e}} \neq \mathbf{0}$, and the apsidal line (line passing through the periapsis and apoapsis) then slowly rotates. The residual perihelion precession of Mercury is the most famous instance of exactly this. Under certain conditions, it may change the shape of the orbit itself.

Equation of the Orbit

We now want the explicit geometric shape of the orbit — $r$ as a function of position angle. The cleanest route uses what we have already derived. We have $\mathbf{h}$ (which fixes the orbital plane) and $\mathbf{e}$ (which fixes the line of apsides). All we have to do is exploit a single dot-product identity to read off $r(f)$.

Take the dot product of $\mathbf{r}$ with the eccentricity-vector integral $\dot{\mathbf{r}} \times \mathbf{h} = \mu(\hat{\mathbf{r}} + \mathbf{e})$:

\[ \mathbf{r} \cdot (\dot{\mathbf{r}} \times \mathbf{h}) = (\mathbf{r} \times \dot{\mathbf{r}}) \cdot \mathbf{h} = \mathbf{h} \cdot \mathbf{h} = h^2. \]

On the right side, let $f$ be the angle between $\mathbf{e}$ and $\mathbf{r}$ (the true anomaly):

\[ \mathbf{r} \cdot \mu(\hat{\mathbf{r}} + \mathbf{e}) = \mu(r + re\cos f). \]

Equating and solving for $r$ (Battin Eq. 3.20):

\[ \boxed{r = \frac{p}{1 + e\cos f}}, \qquad p = \frac{h^2}{\mu}. \]

This is the polar equation of a conic section with focus at the origin. The single number $e$ — the magnitude of the eccentricity vector — determines the qualitative shape of the orbit:

$e$	Conic	Orbit type
$0$	circle	bound
$0 < e < 1$	ellipse	bound
$e = 1$	parabola	escape, zero energy
$e > 1$	hyperbola	escape, positive energy

For $e>0$ and $h^2/\mu = \text{const}$, $f=0$ yields minimum $r$. We call this as periapsis. Similarly, $f=\pi$ has maximum $r$. This is termed as apoapsis.
This settles the question of the direction of $\mathbf{e}$ as well.
- Since $f$ is the angle between $\mathbf{r}$ and $\mathbf{e}$, $f=0$ yields the direction of $\mathbf{e}$.

\[ \boxed{\text{$\mathbf{e}$ always points in the direction of periapsis.}} \]

Reading the orbit equation

Before any dynamics, the single equation $r = p/(1 + e\cos f)$ already tells us which values of $f$ are allowed. Since $r > 0$ and $p > 0$, we need

\[ 1 + e\cos f > 0 . \]

$e < 1$. The bracket never vanishes ($1 + e\cos f \ge 1 - e > 0$), so $f$ sweeps the whole circle $[0, 2\pi)$ and the curve closes. Both apsides exist: \[r_p = \frac{p}{1+e}, \qquad r_a = \frac{p}{1-e} .\]
$e = 1$. The bracket vanishes only in the limit $f \to \pm\pi$, where $r \to \infty$. The curve is open, there is no apoapsis, and the single parameter $p$ fixes it completely: $r_p = p/2$.
$e > 1$. The bracket vanishes at a finite angle, \[\boxed{\;\cos f_\infty = -\frac{1}{e}\;}\] and the body exists only on the arc $|f| < f_\infty$. The two excluded directions are the asymptotes; $r \to \infty$ as $f \to \pm f_\infty$.

So the split is already visible in the geometry, and it is entirely about the denominator. Everything else in Part I ignores it.

As $e \to 1^{+}$, what happens to $f_\infty$? And as $e \to \infty$?

Answer

$f_\infty = \arccos(-1/e)$ decreases monotonically from $\pi$ to $\pi/2$.

At $e \to 1^{+}$ the asymptotes fold back onto the apse line — the hyperbola degenerates into the parabola, which “reaches infinity” only at $f = \pi$. At $e \to \infty$ the two asymptotes become perpendicular to the apse line: the trajectory is a straight line barely bent by the primary, which is the weak-deflection limit of a fast flyby.

Energy Integral (Vis-Viva)

We shall see, dotting the EOM with $\dot{\mathbf{r}}$ produces a scalar conservation law — the energy integral, also known as the vis-viva (literally, “living force”) equation.

This is arguably the single most useful formula in mission design.

Dot-multiply the equation of motion by $\dot{\mathbf{r}}$:

\[ \dot{\mathbf{r}} \cdot \ddot{\mathbf{r}} = -\frac{\mu}{r^3}\,\mathbf{r}\cdot\dot{\mathbf{r}}, \]

\[ \frac{d}{dt}\!\left(\frac{v^2}{2}\right) = -\frac{\mu\dot{r}}{r^2} = \frac{d}{dt}\!\left(\frac{\mu}{r}\right). \]

Integrating

\[ \frac{v^2}{2} - \frac{\mu}{r} = \varepsilon \]

where $\varepsilon$ is the constant of integration.

This is the first form of the vis-viva equation.

Casimir relations

We have now squeezed the two-body equation of motion three different ways — cross product, BAC–CAB, dot product — and harvested three integrals:

\[ \mathbf{h} = \mathbf{r} \times \dot{\mathbf{r}}, \qquad \mathbf{e} = \frac{\dot{\mathbf{r}} \times \mathbf{h}}{\mu} - \frac{\mathbf{r}}{r}, \qquad \varepsilon = \frac{v^2}{2} - \frac{\mu}{r} \]

That is $3 + 3 + 1 = 7$ scalar constants for a system whose state has only $6$ degrees of freedom. They cannot all be independent. Exactly two relations tie them together, and these are called the Casimir relations:

\[ \boxed{\;\mathbf{h} \cdot \mathbf{e} = 0\;} \qquad\qquad \boxed{\;\mu^{2}\left(e^{2} - 1\right) = 2\,\varepsilon\, h^{2}\;} \]

Neither is a new integral. They are identities — constraints satisfied by the integrals we already have, at every instant, on every orbit.

First relationship is trivial. So we will prove the second.

Square the expression for $\mathbf{e}$,

\[ \mu^{2} e^{2} = \underbrace{\left| \dot{\mathbf{r}} \times \mathbf{h} \right|^{2}}_{\text{(i)}} - 2\mu\, \underbrace{\hat{\mathbf{r}} \cdot \left( \dot{\mathbf{r}} \times \mathbf{h} \right)}_{\text{(ii)}} + \mu^{2}\, \underbrace{\hat{\mathbf{r}} \cdot \hat{\mathbf{r}}}_{= \,1} \]

Term (i): $\mathbf{h} = \mathbf{r} \times \dot{\mathbf{r}}$ is perpendicular to $\dot{\mathbf{r}}$, so the cross product has magnitude $v h$ with no $\sin$ factor to worry about:

\[ \left| \dot{\mathbf{r}} \times \mathbf{h} \right|^{2} = v^{2} h^{2} \]

Term (ii): cycle the triple product so that $\mathbf{h}$ meets $\mathbf{r} \times \dot{\mathbf{r}}$:

\[ \hat{\mathbf{r}} \cdot \left( \dot{\mathbf{r}} \times \mathbf{h} \right) = \frac{1}{r}\, \mathbf{r} \cdot \left( \dot{\mathbf{r}} \times \mathbf{h} \right) = \frac{1}{r}\, \mathbf{h} \cdot \left( \mathbf{r} \times \dot{\mathbf{r}} \right) = \frac{h^{2}}{r} \]

Substituting both terms,

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

The bracket is precisely twice the specific energy, $v^{2} - 2\mu/r = 2\varepsilon$. Hence

\[ \mu^{2} e^{2} = 2 \varepsilon h^{2} + \mu^{2} \qquad \Longrightarrow \qquad \mu^{2}\left( e^{2} - 1 \right) = 2 \varepsilon h^{2} \]

or, solved for the eccentricity,

\[ e = \sqrt{1 + \frac{2 \varepsilon h^{2}}{\mu^{2}}} \]

Since $h^{2} \ge 0$ and $\mu > 0$, the sign of $\varepsilon$ alone decides the shape of the conic:

Energy	Eccentricity	Orbit
$\varepsilon < 0$	$e < 1$	ellipse (bound)
$\varepsilon = 0$	$e = 1$	parabola (escape)
$\varepsilon > 0$	$e > 1$	hyperbola (unbound)

This is why the vis-viva constant is the natural “mission design” variable: it tells you at a glance whether you are captured or escaping.

Counting the integrals

The two Casimir relations settle the bookkeeping we started in the N-body section:

We possess $7$ scalar constants: $\mathbf{h}\,(3)$, $\mathbf{e}\,(3)$, $\varepsilon\,(1)$.
Relation 1 removes one; relation 2 removes another.
That leaves $5$ independent constants of the motion.

Five constants fix the orbit as a curve in space.

Its shape ($a, e$) or ($\varepsilon, e$) or ($h, e$).

What they cannot tell us is where on that curve the body is. One more constant, either in the form of $f$ or time of perapsis passage ($t_p$) fixes the position.

\[ 5 \;(\text{geometry}) \;+\; 1 \;(\text{timing, } t_p) \;=\; 6 \;=\; \text{order of the reduced system} \]

The size parameter $a$

The second Casimir relation, rearranged with $h^{2} = \mu p$, already gives the energy in terms of the shape alone:

\[ \mu^{2}\left(e^{2}-1\right) = 2\varepsilon h^{2} = 2\varepsilon\,\mu p \qquad\Longrightarrow\qquad \varepsilon = \frac{\mu\left(e^{2}-1\right)}{2p} . \]

This is exact for every conic. It is also awkward: mission designers quote neither $p$ nor $\varepsilon$, but a length. So define one:

\[ \boxed{\;a \;\equiv\; \frac{p}{1 - e^{2}}\;} \qquad\Longleftrightarrow\qquad \boxed{\;p = a\left(1 - e^{2}\right)\;} \]

and the energy collapses to a single, famous line:

\[ \boxed{\;\varepsilon = -\frac{\mu}{2a}\;} \]

Read the sign, and the conic falls out. $p > 0$ always, so $\operatorname{sign} a = \operatorname{sign}\left(1 - e^{2}\right)$:

$e$	$1-e^{2}$	$a$	$\varepsilon = -\mu/2a$
$<1$	$>0$	$a > 0$, finite	$\varepsilon < 0$
$=1$	$0$	$a \to \infty$ (useless — use $p$ or $r_p$)	$\varepsilon = 0$
$>1$	$<0$	$a < 0$	$\varepsilon > 0$

Two consequences we shall use constantly, both valid for all conics:

\[ h^{2} = \mu p = \mu a\left(1 - e^{2}\right), \qquad r_p = \frac{p}{1+e} = a(1 - e) . \]

$a$ is a number before it is a picture

At this stage $a$ is defined by $p = a(1-e^{2})$ — nothing more. That it is the semi-major axis of an ellipse (half the apse-to-apse distance) is a theorem about the case $e<1$, proved in Part II. For a hyperbola $a<0$ and $|a|$ is the distance from the centre to either vertex; for a parabola $a$ does not exist at all. Never write $r_a = a(1+e)$ outside $e<1$.

Evaluating $\varepsilon$ directly

The relation $\varepsilon = -\mu/2a$ came out of the Casimir identity, which is pure algebra. Here is the same result read off the physics, as a check.

Since $\varepsilon$ is constant, we may evaluate it anywhere on the orbit. Periapsis is the cheapest point: there $\dot r = 0$, the velocity is purely transverse, and hence $h = r_p v_p$, i.e. $v_p = h/r_p$:

\[ \varepsilon = \frac{v_p^{2}}{2} - \frac{\mu}{r_p} = \frac{h^{2}}{2 r_p^{2}} - \frac{\mu}{r_p}. \]

Substituting $h^{2} = \mu a\left(1 - e^{2}\right)$ and $r_p = a(1-e)$ from the previous slide (both valid on every conic):

\[ \begin{aligned} \varepsilon &= \frac{\mu a\left(1 - e^{2}\right)}{2 a^{2}(1-e)^{2}} - \frac{\mu}{a(1-e)} = \frac{\mu (1+e)}{2 a (1-e)} - \frac{\mu}{a(1-e)} \\[4pt] &= \frac{\mu}{a(1-e)}\left[\frac{1+e}{2} - 1\right] = \frac{\mu}{a(1-e)} \cdot \frac{e-1}{2} = -\frac{\mu}{2a}. \end{aligned} \]

The constant is therefore fixed by the size of the orbit alone:

\[ \boxed{\;\frac{v^{2}}{2} - \frac{\mu}{r} = -\frac{\mu}{2a}\;} \qquad\Longleftrightarrow\qquad \boxed{\;v^{2} = \mu\left(\frac{2}{r} - \frac{1}{a}\right)\;} \]

This is the vis-viva equation in its working form.

Two orbits with the same $a$ but wildly different shapes carry exactly the same energy — and, for $e<1$, the same period $T = 2\pi\sqrt{a^{3}/\mu}$ (Part II). Energy is blind to shape.
Putting $a = r$ (a circle) we get the circular speed $v_c = \sqrt{\mu/r}$
$\varepsilon = 0$ ($a \to \infty$) gives us the escape speed $v_{esc} = \sqrt{2\mu/r} = \sqrt{2}\, v_c$.

Bounded and unbounded motion

The sign of $\varepsilon$ was tied to the sign of $e-1$ by the Casimir relation. Here is the direct argument, straight from vis-viva and $v^{2}\ge 0$, with no reference to the conic at all.

Rearranging the vis-viva equation:

\[ \boxed{v^2=2\left(\varepsilon+\frac{\mu}{r}\right)}. \]

Since,

\[ v^2\geq 0, \]

This condition allows us to determine directly whether the radial distance $r$ can become arbitrarily large.

Negative specific energy: $\varepsilon<0$

If the specific mechanical energy is negative, write

\[ \varepsilon=-|\varepsilon|. \]

The vis-viva equation becomes

\[ v^2=2\left(\frac{\mu}{r}-|\varepsilon|\right). \]

Since $v^2\geq 0$,

\[ 2\left(\frac{\mu}{r}-|\varepsilon|\right)\geq 0. \]

Therefore,

\[ \frac{\mu}{r}\geq |\varepsilon|. \]

Because $r>0$ and $|\varepsilon|>0$, this gives

\[ r\leq\frac{\mu}{|\varepsilon|}. \]

Thus, there is a finite upper limit on the distance from the attracting body. Consequently,

\[ \boxed{\varepsilon<0\quad\Longrightarrow\quad\text{bounded motion}.} \]

Show that $r \le 2a$ for any orbit with $\varepsilon \le 0$.

Answer

From $\varepsilon = \dfrac{v^{2}}{2} - \dfrac{\mu}{r}$ and $v^{2} \ge 0$,

\[ \frac{\mu}{r} = \frac{v^{2}}{2} - \varepsilon \;\ge\; -\varepsilon = \frac{\mu}{2a} \qquad\Longrightarrow\qquad r \le 2a . \]

Equality demands $v = 0$. A genuine orbit never stops, so for $h \neq 0$ the bound is strict and the sharp value is $r_a = a(1+e) < 2a$; the case that does attain it is the rectilinear orbit $h = 0$, which comes to rest exactly at $r = 2a$ before falling back.

(For $\varepsilon = 0$, $a \to \infty$ and the bound says nothing — correctly, since a parabola is unbounded.)

Positive specific energy: $\varepsilon>0$

For positive energy, vis-viva gives

\[ v^2=2\left(\varepsilon+\frac{\mu}{r}\right). \]

$\varepsilon > 0$ implies $a \le 0$.

For every finite $r>0$, both terms inside the parentheses are positive:

\[ \varepsilon>0, \qquad \frac{\mu}{r}>0. \]

Therefore,

\[ v^2>0 \]

at every finite radius. In particular, there is no finite outer radius at which

\[ v=0. \]

Thus, the energy equation imposes no upper bound on $r$.

As $r\to\infty$,

\[ \frac{\mu}{r}\to0, \]

and hence

\[ v^2\to2\varepsilon. \]

Taking the positive square root for speed,

\[ \boxed{v\to\sqrt{2\varepsilon}>0}. \]

Therefore, the body can reach infinity while retaining a nonzero speed. This asymptotic speed is called the hyperbolic excess speed:

\[ \boxed{v_\infty=\sqrt{2\varepsilon}}. \]

Consequently,

\[ \boxed{\varepsilon>0\quad\Longrightarrow\quad\text{unbounded motion}.} \]

Zero specific energy: $\varepsilon=0$

The zero-energy case separates bounded and positive-energy unbounded motion. Setting $\varepsilon=0$ in vis-viva gives

\[ v^2=\frac{2\mu}{r}. \]

For every finite $r>0$, we have $v^2>0$. Therefore, there is no finite outer turning point.

As $r\to\infty$, $v^2\to0$ and hence $v\to0$. Thus the body can reach infinity, but its speed approaches zero:

\[ \boxed{\varepsilon=0\quad\Longrightarrow\quad\text{marginally unbounded motion}.} \]

This is the parabolic escape condition.

Summary

Conic	Specific energy	Behavior at large $r$	$a$
ellipse	$\varepsilon<0$	$r\leq \mu/\|\varepsilon\|$	$a > 0$
parabola	$\varepsilon=0$	$r\to\infty$ and $v\to0$	$a \to \infty$
hyperbola	$\varepsilon>0$	$r\to\infty$ and $v\to\sqrt{2\varepsilon}$	$a < 0$

Escape Velocity

For a body at distance $r$ from a primary of gravitational parameter $\mu$, the escape speed $v_{esc}(r)$ is the least speed at which the body, moving under the two-body gravitational field alone, has a trajectory that is unbounded — i.e. $r(t)\to\infty$ rather than returning to a finite apoapsis.

Clearly that is parabolic trajectory.

The criterion is energetic, not kinematic. From the vis-viva equation,

\[ \varepsilon = \frac{v^{2}}{2} - \frac{\mu}{r}, \]

and $\varepsilon$ is constant along the orbit. Two one-line arguments settle it:

If $\varepsilon < 0$: since $v^2 \ge 0$, we need $\mu/r \ge -\varepsilon$, hence $r \le \mu/(-\varepsilon) = 2a$. The motion is trapped inside a sphere — bounded. (Indeed $r_a = a(1+e) \le 2a$.) If $\varepsilon \ge 0$: then $v^{2} = 2\varepsilon + 2\mu/r > 0$ for all finite $r$, so the body never comes to rest; past periapsis it recedes without a turning point. So escape $\iff \varepsilon \ge 0$, and the threshold $\varepsilon = 0$ gives

\[ \boxed{\;v_{esc}(r) = \sqrt{\frac{2\mu}{r}} = \sqrt{2}\,v_c(r)\;} \]

the marginal case being the parabolic orbit ($e = 1$, $a \to \infty$).

It is a speed, not a velocity. The name is a historical misnomer. $\varepsilon$ depends on $|\mathbf{v}|$ only, so the escape condition is indifferent to direction — straight up, sideways, or $45^\circ$ all escape at the same $11.2$ km/s, provided the trajectory does not intersect the planet’s surface or atmosphere. Direction changes where you go, not whether you leave.
It is local: $v_{esc}$ is a function of $r$. “The escape velocity of Earth is 11.2 km/s” is shorthand for at Earth’s surface. From LEO ($r \approx 6{,}700$ km) it is $10.9$ km/s; from GEO, $4.35$ km/s.
It is escape from one primary. Leaving Earth is not leaving the Sun. Escape from the Solar System starting at Earth’s orbit needs $v_{esc} = \sqrt{2\mu_\odot/1\,\text{AU}} = 42.1$ km/s heliocentric
Excess speed is what mission design actually buys. If $v > v_{esc}$ the orbit is hyperbolic and the residual speed at infinity is

\[ v_\infty^{2} = 2\varepsilon = v^{2} - v_{esc}^{2}, \qquad C_3 \equiv v_\infty^2 = -\mu/a, \]

the characteristic energy $C_3$ quoted on every launch-vehicle performance chart. Note $v_\infty^2$ is the difference of squares, not of speeds — a small $\Delta v$ above escape is disproportionately cheap (the Oberth effect in disguise).

Various Conic sections

All four numbers below describe the same orbit — move any one of them and watch the other three, and the curve, follow. Units are non-dimensional, with $\mu = G(m_1+m_2) = 1$; the attracting body sits at the amber focus and periapsis is to the right.

drive the size with ellipse

eccentricity e = 0.50

angular momentum h = 1.00

zoom

show the family swept by e at fixed h

p = a = h = ε = r_p = r_a = b = T =

Things to try

Hold $h$ and sweep $e$ from $0$ to $2.5$. Every curve in the family keeps the same semi-latus rectum $p = h^{2}/\mu$ — the conic simply unrolls from circle to ellipse to parabola to hyperbola, and $\varepsilon$ changes sign exactly at $e = 1$.
Hold $a$ (or, identically, $\varepsilon$) and sweep $e$. Now the size is frozen: every ellipse has the same major axis, the same energy and the same period $T$, but the focus slides from the centre out towards the periapsis. Energy is blind to shape.
Notice that the $a$ and $\varepsilon$ buttons do the same thing. They must: $\varepsilon = -\mu/2a$ is a one-to-one map, so $(a, e)$ and $(\varepsilon, e)$ are the same parametrisation, and only $(h, e)$ is genuinely different.
Try to cross $e = 1$ while holding $a$. You cannot — the slider stops. An ellipse cannot become a hyperbola without $a$ passing through infinity, which is the Casimir relation $\mu^{2}\left(e^{2}-1\right) = 2\varepsilon h^{2}$ forbidding $\operatorname{sign}\varepsilon \neq \operatorname{sign}(e-1)$. Hold $h$ instead and the passage is smooth.

Two spacecraft are on orbits with the same $a$ but $e = 0.1$ and $e = 0.9$. Which one is moving faster at periapsis, and which has the larger $h$?

Answer

\[ v_p = \sqrt{\frac{\mu}{a}}\,\sqrt{\frac{1+e}{1-e}}, \qquad h = \sqrt{\mu a\left(1 - e^{2}\right)} . \]

Faster at periapsis: $e = 0.9$, and not narrowly — $\sqrt{1.9/0.1} = 4.36$ against $\sqrt{1.1/0.9} = 1.11$, a factor of $3.9$.

Larger $h$: the other one. $\sqrt{1-e^{2}}$ is $0.995$ for $e = 0.1$ and only $0.436$ for $e = 0.9$.

So the eccentric orbit is much faster at periapsis yet carries less than half the angular momentum — because it is deep in ($r_p = 0.1a$) exactly when it is moving fast, and $h = r_p v_p$. Equal $a$ pins energy and period; it says nothing about $h$.

Velocity

Goal for this deck: given that $r(f)$ is known, obtain $\mathbf{v}$ as an explicit function of the true anomaly $f$ — magnitude and direction — without ever solving for $t$.

The two ingredients

We will use exactly two facts, both already proved:

The orbit equation. $r = p/(1 + e\cos f)$, with $f$ measured from periapsis at the focus.
Angular momentum. $h = r^{2}\dot f = \text{const}$, which is what converts any $d/df$ into a $d/dt$:

\[ \frac{d}{dt} = \dot f\,\frac{d}{df} = \frac{h}{r^{2}}\,\frac{d}{df} . \]

Working in the orbital plane with the rotating unit vectors $\hat{\mathbf{r}}$ (radially outward) and $\hat{\boldsymbol{\theta}}$ (transverse, in the direction of increasing $f$). They rotate with the body:

\[ \dot{\hat{\mathbf{r}}} = \dot f\,\hat{\boldsymbol{\theta}}, \qquad \dot{\hat{\boldsymbol{\theta}}} = -\dot f\,\hat{\mathbf{r}} . \]

Both statements are proved at the end of the deck — see the appendix, The rotating basis $(\hat{\mathbf{r}}, \hat{\boldsymbol{\theta}})$.

Differentiating $\mathbf{r} = r\,\hat{\mathbf{r}}$,

\[ \mathbf{v} = \dot{\mathbf{r}} = \dot r\,\hat{\mathbf{r}} + r\dot f\,\hat{\boldsymbol{\theta}} \;\equiv\; v_r\,\hat{\mathbf{r}} + v_\perp\,\hat{\boldsymbol{\theta}} . \]

So there are only two numbers to find: the radial speed $v_r = \dot r$ and the transverse speed $v_\perp = r\dot f$.

$v_r$ changes the size of $\mathbf{r}$; $v_\perp$ changes its direction. Only $v_\perp$ carries angular momentum — indeed $h = r v_\perp$.

The transverse component

Angular momentum gives it with no work at all:

\[ v_\perp = r\dot f = \frac{h}{r} = \frac{h}{p}\left(1 + e\cos f\right) . \]

Using $p = h^{2}/\mu$, the prefactor is $h/p = \mu/h$:

\[ \boxed{\;v_\perp = \frac{\mu}{h}\left(1 + e\cos f\right)\;} \]

$v_\perp > 0$ everywhere the conic exists, since $1 + e\cos f = p/r > 0$ — the body never reverses its swing.
$v_\perp$ is largest at periapsis and smallest at apoapsis, exactly as $h = rv_\perp = \text{const}$ demands.

The radial component

Differentiate the orbit equation. It is cleanest in reciprocal form:

\[ \frac{1}{r} = \frac{1 + e\cos f}{p} \qquad\Longrightarrow\qquad -\frac{\dot r}{r^{2}} = -\frac{e\sin f}{p}\,\dot f . \]

Hence

\[ \dot r = \frac{r^{2}\dot f}{p}\,e\sin f = \frac{h}{p}\,e\sin f , \]

where $r^{2}\dot f = h$ was used in the very last step. With $h/p = \mu/h$ again,

\[ \boxed{\;v_r = \frac{\mu}{h}\,e\sin f\;} \]

Both components carry the same prefactor $\mu/h = h/p = \sqrt{\mu/p}$. This is the natural speed unit of the orbit — the circular speed at radius $p$. Every velocity formula below is that number times a pure function of $e$ and $f$.

The velocity vector

Putting the two together:

\[ \boxed{\; \mathbf{v}(f) = \frac{\mu}{h}\left[\,e\sin f\;\hat{\mathbf{r}} \;+\;\left(1 + e\cos f\right)\hat{\boldsymbol{\theta}}\,\right] \;} \]

Clearly,

$v_r = 0$ at $f = 0$ and $f = \pi$ — the apsides are precisely the points where the radius stops growing or shrinking. The velocity there is purely transverse, i.e. perpendicular to $\mathbf{r}$.
$v_r > 0$ on the outbound half $0 < f < \pi$ and $v_r < 0$ on the inbound half.
$e = 0$ kills $v_r$ everywhere: circular motion, speed $\sqrt{\mu/p}$, forever perpendicular to the radius.

The speed

\[ v^{2} = v_r^{2} + v_\perp^{2} = \frac{\mu^{2}}{h^{2}}\left[e^{2}\sin^{2}f + \left(1 + e\cos f\right)^{2}\right] . \]

Expand the bracket:

\[ e^{2}\sin^{2}f + 1 + 2e\cos f + e^{2}\cos^{2}f = 1 + 2e\cos f + e^{2} . \]

So, using $\mu^2/h^2 = \mu/p$,

\[ \boxed{\;v(f) = \sqrt{\frac{\mu}{p}}\;\sqrt{1 + 2e\cos f + e^{2}}\;} \]

The speed depends on $f$ only through $\cos f$, so it is symmetric about the apse line — the body passes any given radius on the way out and on the way in with the same speed, and with flight path angles equal and opposite.

Flight Path Angle ($\gamma$)

The flight path angle $\gamma$ is the angle between the velocity and the local horizon — the direction perpendicular to $\mathbf{r}$, i.e. $\hat{\boldsymbol{\theta}}$. It is positive when the body is climbing.

\[ \tan\gamma = \frac{v_r}{v_\perp}, \qquad \cos\gamma = \frac{v_\perp}{v} = \frac{h}{r v}, \qquad \sin\gamma = \frac{v_r}{v} . \]

Substituting the two components derived above, the common factor $\mu/h$ cancels:

\[ \boxed{\;\tan\gamma(f) = \frac{e\sin f}{1 + e\cos f}\;} \]

Only $e$ and $f$ survive. Neither $a$ nor $\mu$ nor $h$ appears — $\gamma$ is a ratio of two velocity components that carry the same prefactor. Scale the orbit up ($a$) or change the central body ($\mu$): the direction of flight at a given true anomaly does not move.

Where $\gamma$ is zero, and where it is largest

At the apsides $\sin f = 0$, so $\gamma = 0$: the velocity is purely transverse and $v = h/r$ there — the two points where speed and radius decouple.

For the maximum, differentiate the tangent:

\[ \frac{d}{df}\!\left(\frac{e\sin f}{1 + e\cos f}\right) = \frac{e\cos f\,(1 + e\cos f) + e^{2}\sin^{2} f}{(1 + e\cos f)^{2}} = \frac{e\,(\cos f + e)}{(1 + e\cos f)^{2}} , \]

which vanishes at $\cos f = -e$. There $\sin f = \sqrt{1-e^{2}}$ and $1 + e\cos f = 1 - e^{2}$, so

\[ \tan\gamma_{\max} = \frac{e\sqrt{1-e^{2}}}{1-e^{2}} = \frac{e}{\sqrt{1-e^{2}}} \qquad\Longrightarrow\qquad \boxed{\;\gamma_{\max} = \arcsin e\;} \]

The location $\cos f = -e$ is exactly $r = p/(1-e^{2}) = a$: the ends of the minor axis. Same point as on the previous slide — where $v = \sqrt{\mu/a}$, the velocity is also tilted furthest from the horizon.
This is the first result that notices $e$. The stationary point needs $\cos f = -e$, which has a solution only for $e \le 1$. For $e > 1$ there is no interior maximum: $\gamma$ climbs monotonically along the whole arc and reaches its supremum $f_\infty - \pi/2$ only at the asymptote. Everything before this bullet is universal.
$\gamma > 0$ on the outbound half, $\gamma < 0$ on the inbound half, with equal and opposite values at the same radius. A circle has $\gamma \equiv 0$.
$h = r v \cos\gamma$ is the practical form: with a radar range, a range rate and a speed you have the angular momentum, and the orbit follows.

Picture

Figure 2: **Left:** the velocity splits into $v_r = \dot r = (\mu/h)e\sin f$ along $\hat{\mathbf{r}}$ and $v_\perp = h/r = (\mu/h)(1+e\cos f)$ along the local horizon; $\gamma$ is the tilt of $\mathbf{v}$ off that horizon. **Right:** the same velocity plotted in velocity space (perifocal components $v_P, v_Q$). As $f$ runs over the orbit the tip of $\mathbf{v}$ traces a *circle* of radius $\mu/h$ whose centre is displaced $\mu e/h$ perpendicular to the apse line. The topmost and bottommost points of that circle are periapsis and apoapsis.

Velocity Hodograph

The component form hides something remarkable. Go back to the eccentricity-vector integral,

\[ \dot{\mathbf{r}}\times\mathbf{h} = \mu\left(\hat{\mathbf{r}} + \mathbf{e}\right), \]

and cross it with $\mathbf{h}$ from the left. Using $\mathbf{h}\times(\mathbf{v}\times\mathbf{h}) = h^{2}\mathbf{v} - \mathbf{h}(\mathbf{h}\cdot\mathbf{v})$ and $\mathbf{h}\cdot\mathbf{v} = 0$,

\[ \boxed{\;\mathbf{v} = \frac{\mu}{h}\,\hat{\mathbf{h}}\times\left(\hat{\mathbf{r}} + \mathbf{e}\right)\;} \]

Now read it: $\hat{\mathbf{h}}\times$ is just a $90^{\circ}$ rotation inside the orbital plane. So $\mathbf{v}$ is

\[ \underbrace{\frac{\mu}{h}\,\hat{\mathbf{h}}\times\hat{\mathbf{r}}}_{\text{unit vector, rotating with }f} \;+\; \underbrace{\frac{\mu}{h}\,\hat{\mathbf{h}}\times\mathbf{e}}_{\text{fixed vector}} . \]

A vector of constant length $\mu/h$ sweeping a full turn, added to a constant vector. The tip of $\mathbf{v}$ therefore moves on a circle of radius $\mu/h$ centred at distance $\mu e/h$ from the origin of velocity space, perpendicular to $\mathbf{e}$. This curve is the hodograph.

The hodograph in components

In the perifocal frame $(\hat{\mathbf{P}}$ towards periapsis, $\hat{\mathbf{Q}}$ at $90^{\circ}$ in the direction of motion$)$, substituting $\hat{\mathbf{r}} = \cos f\,\hat{\mathbf{P}} + \sin f\,\hat{\mathbf{Q}}$:

\[ \boxed{\; v_P = -\frac{\mu}{h}\sin f, \qquad v_Q = \frac{\mu}{h}\left(e + \cos f\right) \;} \]

\[ \Longrightarrow\quad v_P^{2} + \left(v_Q - \frac{\mu e}{h}\right)^{2} = \left(\frac{\mu}{h}\right)^{2} . \]

Every Kepler orbit — ellipse, parabola or hyperbola — has a circular hodograph. Only the portion traversed differs: the ellipse goes round and round, the hyperbola covers only the arc $|f| < \arccos(-1/e)$ and the two missing endpoints are the incoming and outgoing $\mathbf{v}_\infty$.
$e < 1$ puts the velocity-space origin inside the circle (every direction of flight is achieved); $e > 1$ puts it outside (the flight direction is confined to a wedge — the turning angle of a flyby).

Historical note

Hamilton published the circularity of the hodograph in 1846 and used it to give a purely geometric proof of Kepler’s laws; Möbius and Maxwell used it as a teaching device. It is the cleanest statement of “the inverse-square law is special”.

Speed and flight path angle across the conics

Figure 3: Speed (left) and flight path angle (right) against true anomaly, in units of $\sqrt{\mu/p}$. Circle: constant speed, $\gamma \equiv 0$. Ellipse: maximum at periapsis, minimum at apoapsis, $\gamma$ peaking at $\arcsin e$ where $\cos f = -e$. Parabola: speed falls to zero as $f\to\pm180^{\circ}$ (rest at infinity) and $\gamma = f/2$ exactly. Hyperbola: the curve stops at the asymptote $\cos f = -1/e$, where the speed has not decayed to zero but to $v_\infty = \sqrt{\mu/p}\,\sqrt{e^{2}-1} = \sqrt{\mu/|a|}$.

Limiting cases

Orbit	$e$	$v(f)$	$\gamma$	hodograph
circle	$0$	$\sqrt{\mu/a}$, constant	$0$	circle centred at origin
ellipse	$0<e<1$	$\sqrt{\mu/p}\sqrt{1+2e\cos f + e^{2}}$	$\lvert\gamma\rvert\le\arcsin e$	origin inside
parabola	$1$	$\sqrt{2\mu/r} = 2\sqrt{\mu/p}\,\left\|\cos\tfrac f2\right\|$	$f/2$	origin on the circle
hyperbola	$>1$	as above, $\to v_\infty$	$\to$ asymptote value	origin outside

For the hyperbola, at $\cos f = -1/e$,

\[ v_\infty^{2} = \frac{\mu}{p}\left(1 - 2 + e^{2}\right) = \frac{\mu}{p}\left(e^{2}-1\right) = \frac{\mu}{|a|} = C_3 , \]

recovering the characteristic energy from the escape-velocity slide.

From $r(f)$ to $r(t)$

We still just have the trajectory and not the exact location on the trajectory.

We currently know $r(f)$. We need $r(t)$. So clearly, we need $f(t)$.

Since,

\[ h = r^{2}\dot f \qquad\Longrightarrow\qquad dt = \frac{r^{2}}{h}\,df, \]

substituting the orbit equation $r = p/(1 + e\cos f)$ and integrating from periapsis ($f = 0$ at $t = t_p$),

\[ t - t_p = \frac{p^{2}}{h}\int_{0}^{f} \frac{df'}{\left(1 + e\cos f'\right)^{2}} . \]

This is the Kepler problem, and the integral is the obstruction. It is elementary, but ugly in $f$ — and, worse, the answer cannot be inverted to give $f(t)$ in closed form. The classical remedy is not to fight the integral but to change the variable.

And this is where the conics finally separate: the substitution that linearises $\left(1+e\cos f\right)^{-2}$ is a different one for $e<1$, $e=1$ and $e>1$. Everything up to this slide has been common ground.

Where the three conics part ways

That integral is the only place the two-body problem forces a choice. Take stock of what Part I delivered, none of it conditional on $e$:

\[ r = \frac{p}{1+e\cos f}, \quad h = r^{2}\dot f, \quad \varepsilon = -\frac{\mu}{2a}, \quad v^{2} = \mu\!\left(\frac{2}{r} - \frac{1}{a}\right), \]

\[ v_r = \frac{\mu}{h}e\sin f, \quad v_\perp = \frac{\mu}{h}\left(1 + e\cos f\right), \quad v = \sqrt{\frac{\mu}{p}}\sqrt{1 + 2e\cos f + e^{2}}, \quad \tan\gamma = \frac{e\sin f}{1+e\cos f} . \]

What is left is exactly three things, and each of them fails to make sense for at least one conic:

Question	$e<1$	$e=1$	$e>1$
an auxiliary angle that linearises $\int r^{2}df$	eccentric anomaly $E$	$D = \tan\frac f2$	hyperbolic anomaly $H$
the resulting Kepler equation	$M = E - e\sin E$	$M_p = D + \frac{D^{3}}{3}$	$M_h = e\sinh H - H$
is it invertible in closed form?	no (transcendental)	yes (a cubic)	no (transcendental)

and a short list of quantities that simply do not exist across the board: apoapsis $r_a$ and the period $T$ (ellipse only), $v_\infty$, $C_3$ and the turning angle $\delta$ (hyperbola only).

The next three parts take the conics one at a time. Nothing already derived is re-derived.

Part II — The Ellipse ($0 \le e < 1$)

What is new, and what is not

Bound motion. Everything in Part I still holds verbatim; what the constraint $e<1$ buys us is:

$a > 0$ is a genuine length on the figure — the semi-major axis;
an apoapsis exists, $r_a = a(1+e)$, and hence a closed curve of finite area;
a finite area plus a constant areal velocity gives a period $T$;
the closed curve carries an auxiliary circle, and with it the eccentric anomaly $E$ and Kepler’s equation.

The ellipse

Figure 4: Geometry of the elliptic orbit ($e = 0.6$). The attracting body sits at the occupied focus $F$; the other focus is empty. The centre-to-focus distance is $ae$, so the eccentricity is the fractional offset of the focus from the centre. The semi-latus rectum $p$ is the radius at $f = \pi/2$, i.e. $p = a(1-e^{2}) = b^{2}/a$, and the true anomaly $f$ is measured at the focus from the periapsis direction.

Why $a$ is the semi-major axis

In Part I, $a$ was defined by $p = a(1-e^{2})$ and nothing was claimed about its geometry. For $e<1$ we can now cash that in: $a$ is exactly half the apse-to-apse distance.

Evaluate $r$ at the two apsides, where the radius is stationary — and note that $r_a$ is finite only because $e<1$:

\[ f = 0: \quad r_p = \frac{p}{1 + e}, \qquad\qquad f = \pi: \quad r_a = \frac{p}{1 - e}. \]

The major axis runs from apoapsis to periapsis straight through the focus, so $2a = r_p + r_a$:

\[ 2a = p\left(\frac{1}{1+e} + \frac{1}{1-e}\right) = p\,\frac{(1-e) + (1+e)}{1 - e^{2}} = \frac{2p}{1 - e^{2}}, \]

that is,

\[ \boxed{\;p = a\left(1 - e^{2}\right)\;} \qquad\Longrightarrow\qquad \boxed{\;r = \frac{a\left(1 - e^{2}\right)}{1 + e\cos f}\;} \]

So the defined $a$ and the drawn semi-major axis are the same number — for bound orbits, and only for bound orbits. Three corollaries we use constantly:

\[ r_p = a(1 - e), \qquad r_a = a(1 + e), \qquad h^{2} = \mu p = \mu a\left(1 - e^{2}\right). \]

Period ($T$) of an orbit

Step 1 — the areal velocity. With $\mathbf{r} = r\hat{\mathbf{r}}$ and $\dot{\mathbf{r}} = \dot r\,\hat{\mathbf{r}} + r\,\dot{\hat{\mathbf{r}}}$ (used earlier in the eccentricity-vector derivation),

\[ \mathbf{h} = \mathbf{r} \times \dot{\mathbf{r}} = r\hat{\mathbf{r}} \times \left(\dot r\,\hat{\mathbf{r}} + r\,\dot{\hat{\mathbf{r}}}\right) = r^{2}\,\hat{\mathbf{r}} \times \dot{\hat{\mathbf{r}}} . \]

Because $\mathbf{e}$ is a fixed vector, the apsidal line does not move, so the true anomaly $f$ is the polar angle of $\hat{\mathbf{r}}$: writing $\dot{\hat{\mathbf{r}}} = \dot f\,\hat{\boldsymbol{\theta}}$ with $\hat{\boldsymbol{\theta}} \perp \hat{\mathbf{r}}$ in the orbital plane,

\[ h = r^{2}\dot f . \]

The radius vector sweeps a thin sector of area $dA = \tfrac{1}{2} r^{2}\,df$, hence

\[ \frac{dA}{dt} = \frac{1}{2} r^{2}\dot f = \frac{h}{2} = \text{const}, \]

which is Kepler’s second law once more. Constant rate is the whole point: the area is swept uniformly, so one revolution can be timed by simple division.

Step 2 — the area enclosed. For an ellipse $A = \pi a b$, so we need $b$; and $b$ follows from the orbit equation alone. Put $\cos f = -e$ in $r = a(1-e^{2})/(1 + e\cos f)$:

\[ r = \frac{a\left(1 - e^{2}\right)}{1 - e^{2}} = a, \qquad r\cos f = -ae . \]

That point lies a distance $ae$ to the left of the focus; but the centre is itself $ae$ to the left of the focus (Figure 4), so the point sits directly above the centre — it is the end of the minor axis. By Pythagoras,

\[ b^{2} = a^{2} - (ae)^{2} \qquad \Longrightarrow \qquad b = a\sqrt{1 - e^{2}} = \sqrt{a p} . \]

Step 3 — divide. One revolution sweeps the whole ellipse at the constant rate $h/2$:

\[ T = \frac{A}{dA/dt} = \frac{\pi a b}{h/2} = \frac{2\pi\, a \cdot a\sqrt{1 - e^{2}}}{\sqrt{\mu a \left(1 - e^{2}\right)}} = \frac{2\pi a^{2}}{\sqrt{\mu a}}, \]

\[ \boxed{\;T = 2\pi\sqrt{\frac{a^{3}}{\mu}}\;} \qquad\qquad n \equiv \frac{2\pi}{T} = \sqrt{\frac{\mu}{a^{3}}} \]

where $n$ is the mean motion, the average angular rate — the quantity we shall need when we come to Kepler’s equation.

Note once again that $e$ has cancelled: the period, like the energy, depends only on $a$. Indeed $T$ and $\varepsilon = -\mu/2a$ are two readings of the same number. And $T$ exists only for $a > 0$: a parabola or hyperbola never closes.

Kepler’s third law Modification

Squaring, $T^{2} = \dfrac{4\pi^{2}}{G(m_1 + m_2)}\,a^{3}$. Kepler asserted $T^{2} \propto a^{3}$ with the same constant for every planet; Newton’s version shows the constant carries $m_1 + m_2$. For the Sun and even its heaviest planet $m_2/m_1 \approx 10^{-3}$, so the discrepancy is a few parts in $10^{4}$ — comfortably below Kepler’s observational precision, which is why he never saw it.

Example 2

Take $G = 1$, so that $\mu = G(m_1 + m_2) = 1$:

\[ \begin{aligned} m_1 = m_2 &= \tfrac{1}{2}, \\[4pt] \mathbf{R}_1(0) &= \left(+\tfrac{1}{2},\, 0,\, 0\right), &\qquad \mathbf{V}_1(0) &= \tfrac{1}{2}\,\hat{\mathbf{y}}, \\ \mathbf{R}_2(0) &= \left(-\tfrac{1}{2},\, 0,\, 0\right), &\qquad \mathbf{V}_2(0) &= \tfrac{1}{2}\,\hat{\mathbf{z}} . \end{aligned} \]

The bodies sit on the $x$-axis, equidistant from the origin; one is launched along $y$, the other along $z$. The separation lies along one axis and the two velocities along the other two — all three directions mutually perpendicular.

Clearly, the total linear momentum

\[ \mathbf{P} = m_1\mathbf{V}_1 + m_2\mathbf{V}_2 = \tfrac{1}{4}\left(\hat{\mathbf{y}} + \hat{\mathbf{z}}\right) \neq \mathbf{0} \]

does not vanish, so the barycentre drifts. This is not the tidy equal-and-opposite setup of the previous animation.

The orbit, before any integration

The relative state at $t = 0$ is

\[ \mathbf{r}(0) = \mathbf{R}_2 - \mathbf{R}_1 = -\hat{\mathbf{x}}, \qquad \dot{\mathbf{r}}(0) = \mathbf{V}_2 - \mathbf{V}_1 = \tfrac{1}{2}\left(\hat{\mathbf{z}} - \hat{\mathbf{y}}\right), \]

so the integrals are fixed once and for all by these two vectors:

\[ \mathbf{h} = \mathbf{r} \times \dot{\mathbf{r}} = \tfrac{1}{2}\left(\hat{\mathbf{y}} + \hat{\mathbf{z}}\right), \qquad h = \tfrac{1}{\sqrt{2}}, \qquad \varepsilon = \frac{v^{2}}{2} - \frac{\mu}{r} = \frac{1}{4} - 1 = -\frac{3}{4}, \]

and — using three results now in hand (vis-viva, the period, and the second Casimir relation) — the whole geometry follows without solving anything:

\[ a = -\frac{\mu}{2\varepsilon} = \frac{2}{3}, \qquad e = \sqrt{1 + \frac{2\varepsilon h^{2}}{\mu^{2}}} = \frac{1}{2}, \qquad T = 2\pi\sqrt{\frac{a^{3}}{\mu}} = 3.4201 . \]

Since $\mathbf{r}\cdot\dot{\mathbf{r}} = 0$ at $t = 0$, the run starts at an apsis; and $r(0) = 1 = a(1+e)$ identifies it as apoapsis.

Helices, not closed curves

One more feature of this choice of initial data, which the animation is about to make obvious. The barycentre travels at

\[ \mathbf{V}_{cm} = \frac{\mathbf{P}}{M} = \tfrac{1}{4}\left(\hat{\mathbf{y}} + \hat{\mathbf{z}}\right) \qquad\text{while}\qquad \mathbf{h} = \tfrac{1}{2}\left(\hat{\mathbf{y}} + \hat{\mathbf{z}}\right), \]

so $\mathbf{V}_{cm} \parallel \mathbf{h}$: the drift is along the normal of the orbital plane. The plane therefore slides through space like a sheet pushed face-first, and neither body ever returns to a point it has already visited — each traces a helix, even though the relative orbit is a closed ellipse of period $T = 3.4201$.

Two bodies in 3D

Figure 5: **Left:** the inertial frame — the barycentre (amber) runs off along a straight line, dragging the orbital plane with it, and each body traces a helix. **Right:** the barycentric frame — the drift subtracted, the two bodies run on similar (here congruent, as $m_1 = m_2$) ellipses in one plane that never moves. Shaded patch: the plane $\perp \mathbf{h}$. Integrated with adaptive RKF7(8), $rtol = 10^{-12}$. Download the script.

The three speeds, plotted

Divide every component by $\sqrt{\mu/a}$ and the semi-major axis disappears:

\[ \frac{v_r}{\sqrt{\mu/a}} = \frac{e\sin f}{\sqrt{1-e^{2}}}, \qquad \frac{v_\perp}{\sqrt{\mu/a}} = \frac{1 + e\cos f}{\sqrt{1-e^{2}}}, \qquad \frac{v}{\sqrt{\mu/a}} = \frac{\sqrt{1 + 2e\cos f + e^{2}}}{\sqrt{1-e^{2}}} . \]

So $e$ fixes the shape of all three curves and $a$ only sets the scale.

Figure 6: Radial, transverse and total speed against true anomaly for a family of eccentricities, in units of $\sqrt{\mu/a}$ (the circular speed at radius $a$). The fourth panel repeats one eccentricity at three sizes of orbit: the curve is the same, stretched vertically by $\sqrt{\mu/a}$.

What the plot says

Two knobs, and they do different jobs. $e$ chooses the curve, $a$ chooses the units. Two orbits of the same $e$ have identical velocity histories up to the factor $\sqrt{\mu/a}$ — which is why $e$, not $a$, is what makes an orbit “energetic” to fly.
$v_r$ is odd about the apse line. It vanishes at both apsides and is extremal at $f = 90^{\circ}, 270^{\circ}$ — the ends of the latus rectum — where $|v_r| = (\mu/h)\,e$. Outbound is the top half, inbound the bottom.
$v_\perp$ never changes sign. It is $h/r$, so it is largest at periapsis and smallest at apoapsis, in the ratio

\[ \frac{v_{\perp,p}}{v_{\perp,a}} = \frac{1+e}{1-e} = \frac{r_a}{r_p}, \]

which is just $h = r v_\perp = \text{const}$ read twice. - The apsidal speeds multiply to the circular value. At the apsides $v = v_\perp$, and

\[ v_p = \sqrt{\frac{\mu}{a}}\sqrt{\frac{1+e}{1-e}}, \qquad v_a = \sqrt{\frac{\mu}{a}}\sqrt{\frac{1-e}{1+e}}, \qquad v_p\,v_a = \frac{\mu}{a} . \]

The speed equals $\sqrt{\mu/a}$ exactly where $r = a$, i.e. at $\cos f = -e$ (the ends of the minor axis) — marked on the third panel. Vis-viva says so at a glance: $v^{2} = \mu(2/r - 1/a) = \mu/a$ when $r = a$. Inside that radius the body is faster than a circular orbit of the same radius would be, outside it slower. It is also, as the next slide shows, where the velocity is tilted furthest from the horizon.
Eccentric orbits are lopsided in time as well as speed. As $e \to 1$ at fixed $a$, $v_p \sim \sqrt{2\mu/(a(1-e))} \to \infty$ while $v_a \to 0$: the body hurries through periapsis and loiters at apoapsis. That is Kepler’s second law seen from the velocity side.

$\gamma$ across a family of ellipses

Figure 7: **Left:** $\gamma$ against true anomaly for a family of eccentricities; the markers sit at $\cos f = -e$, where $\gamma = \arcsin e$. **Centre:** the same eccentricity flown at three different semi-major axes — the curves lie exactly on top of one another, because $a$ and $\mu$ cancel in $v_r/v_\perp$. **Right:** the envelope $\gamma_{\max} = \arcsin e$.

$a$ does nothing to $\gamma$; $e$ does everything. Raising the orbit changes every speed but not a single flight path angle.
Near-circular orbits fly nearly horizontally. For small $e$, $\gamma \approx e\sin f$, so the Earth’s orbit ($e = 0.0167$) never tilts more than about $0.96^{\circ}$ — which is why “circular” is such a good first approximation for planets and for most operational satellites.
$\gamma_{\max} = \arcsin e$ grows slowly at first ($e = 0.2$ gives only $11.5^{\circ}$) and then sharply: $e = 0.8$ gives $53^{\circ}$, and the parabolic limit $e \to 1$ approaches $90^{\circ}$ — radial escape, all climb and no swing.

Apsidal speeds

Put $f = 0$ and $f = \pi$ in the boxed speed formula:

\[ v_p = \sqrt{\frac{\mu}{p}}\left(1 + e\right) = \sqrt{\frac{\mu}{a}\,\frac{1+e}{1-e}}, \qquad v_a = \sqrt{\frac{\mu}{p}}\left(1 - e\right) = \sqrt{\frac{\mu}{a}\,\frac{1-e}{1+e}} . \]

Two consequences worth remembering:

\[ \frac{v_p}{v_a} = \frac{1+e}{1-e} = \frac{r_a}{r_p}, \qquad v_p\,v_a = \frac{\mu}{p}\left(1-e^{2}\right) = \frac{\mu}{a} = n^{2}a^{2} . \]

The first is just $h = rv$ at the apsides (angular momentum). The second says the geometric mean of the apsidal speeds is the circular speed at radius $a$.

An orbit is raised by a small burn $\Delta v$ applied horizontally at periapsis. Using $v_p = \sqrt{\mu/p}\,(1+e)$ and $h = r_pv_p$, show that $r_a$ changes far more than $r_p$ does. Where does the extra energy go?

Answer

The impulse is tangential at periapsis, so the burn point stays on the new orbit and is still its periapsis — the velocity is still perpendicular to $\mathbf{r}$, and the speed there has only gone up. Hence

\[\delta r_p = 0 ,\]

and the whole change is forced onto the far side. With $\varepsilon = -\mu/2a$ and $\delta\varepsilon = v_p\,\delta v$,

\[ \delta a = \frac{2a^{2}}{\mu}\,v_p\,\delta v , \qquad r_a = 2a - r_p \;\Longrightarrow\; \delta r_a = 2\,\delta a = \frac{4a^{2}v_p}{\mu}\,\delta v . \]

For a LEO-sized orbit that coefficient is about $3400\ \mathrm{km}$ per $\mathrm{km\,s^{-1}}$: apoapsis does all the moving.

The energy goes into potential energy on the far side — the apoapsis climbs. And the leverage $4a^{2}v_p/\mu$ grows with $v_p$, which is the Oberth effect seen from the geometry side.

Worked example — a GTO

A geostationary transfer orbit: perigee altitude $300$ km, apogee at the geostationary radius. With $\mu_\oplus = 3.986\times10^{5}\ \mathrm{km^{3}/s^{2}}$,

\[ r_p = 6678\ \mathrm{km},\quad r_a = 42164\ \mathrm{km} \;\Longrightarrow\; a = \frac{r_p + r_a}{2} = 24421\ \mathrm{km},\quad e = \frac{r_a - r_p}{r_a + r_p} = 0.7266 . \]

\[ v_p = \sqrt{\frac{\mu}{a}\frac{1+e}{1-e}} = 10.15\ \mathrm{km/s}, \qquad v_a = 1.608\ \mathrm{km/s}, \qquad \frac{v_p}{v_a} = 6.31 = \frac{r_a}{r_p}\ \checkmark \]

The steepest climb occurs at $\cos f = -e$, i.e. $f = 136.6^{\circ}$, where $\gamma = \arcsin 0.7266 = 46.6^{\circ}$, $r = a = 24421$ km and $v = \sqrt{\mu/a} = 4.04$ km/s.

At which true anomaly is the spacecraft moving at exactly the local circular speed $\sqrt{\mu/r}$? (Hint: set $v^{2} = \mu/r$ in vis-viva — the answer is $r = a$, the same point as $\gamma_{\max}$. Coincidence?)

Answer

Vis-viva gives $v^{2} = \mu\!\left(\frac{2}{r} - \frac{1}{a}\right) = \frac{\mu}{r}$, i.e. $\frac{1}{r} = \frac{1}{a}$, so $r = a$. Then

\[ 1 + e\cos f = \frac{p}{a} = 1 - e^{2} \qquad\Longrightarrow\qquad \boxed{\cos f = -e} \]

— two points, symmetric about the apse line: the ends of the minor axis, each exactly $a$ from both foci.

Not a coincidence. From $\tan\gamma = \dfrac{e\sin f}{1 + e\cos f}$,

\[ \frac{d}{df}\tan\gamma = \frac{e\,\left(\cos f + e\right)}{\left(1 + e\cos f\right)^{2}} = 0 \qquad\Longleftrightarrow\qquad \cos f = -e , \]

the same condition. Both are the single fact $r = a$: inside that radius the body outruns a circular orbit of the same radius, outside it lags, and the crossing is where the velocity leans furthest from the local horizon.

The eccentric anomaly $E$

Circumscribe the ellipse with its auxiliary circle: the circle of radius $a$ centred at the ellipse’s centre $O$. Drop a perpendicular from the orbiting point $P$ to the major axis and extend it to meet the circle at $Q$. The eccentric anomaly $E$ is the angle $Q$ subtends at the centre, measured from periapsis — whereas the true anomaly $f$ is measured at the focus.

Clearly, $Q = (a\cos E,\; a\sin E)$. Now we try to calculate the coordinates of P. By definition, $x_P = x_Q = a \cos E$. To calculate $y_P$ we use the centre-based ellipse equation:

\[ \frac{x_P^{2}}{a^{2}} + \frac{y_P^{2}}{b^{2}} = 1 \;\xrightarrow{\;x_P \,=\, a\cos E\;}\; \cos^{2}E + \frac{y_P^{2}}{b^{2}} = 1 \quad\Longrightarrow\quad y_P^{2} = b^{2}\left(1 - \cos^{2}E\right) = b^{2}\sin^{2}E , \]

so $y_P = b\sin E$. $P = (a\cos E,\; b\sin E)$

The focus lies at $(ae, 0)$ relative to $O$, so measured from the focus

\[ x_P = a\left(\cos E - e\right), \qquad y_P = b \sin E = a\sqrt{1 - e^{2}}\,\sin E . \]

Now, \[ \begin{aligned} r^{2} = x_P^{2} + y_P^{2} &= a^{2}\left(\cos E - e\right)^{2} + a^{2}\left(1 - e^{2}\right)\sin^{2}E \\[2pt] &= a^{2}\left[\cos^{2}E - 2e\cos E + e^{2} + \sin^{2}E - e^{2}\sin^{2}E\right] \\[2pt] &= a^{2}\left[1 - 2e\cos E + e^{2}\cos^{2}E\right] = a^{2}\left(1 - e\cos E\right)^{2}, \end{aligned} \]

\[ \boxed{\; r = a\left(1 - e\cos E\right)\;} \]

Compare this with $r = a(1-e^{2})/(1 + e\cos f)$: the radius is now linear in $\cos E$ instead of reciprocal in $\cos f$. So $E$ is a much easier variable to work with instead of $f$.

Check: $E = 0, \pi$ give $r_p = a(1-e)$ and $r_a = a(1+e)$.

Answer

Straight substitution into $r = a\left(1 - e\cos E\right)$:

\[ E = 0: \quad r = a(1 - e) = r_p , \qquad E = \pi: \quad r = a(1 + e) = r_a . \quad\checkmark \]

Worth noticing why this is the right check to run: the apsides are the only two points where $E$ and $f$ agree, since both are measured from periapsis and the projection between the auxiliary circle and the ellipse does nothing on the apse line. Everywhere else the two anomalies differ.

Now we can go back to our integral and see what can be done.

The original differential equation of interest is:

\[ dt = \frac{r^2}{h}df \]

In plane polar coordinates $x = r\cos f,\; y = r\sin f$, so

\[ x\,dy - y\,dx = r^{2}\,df , \]

which is precisely the quantity we must integrate. Taking $x$ and $y$ in terms of $E$, $dx = -a\sin E\,dE$ and $dy = a\sqrt{1-e^{2}}\cos E\,dE$:

\[ \begin{aligned} r^{2}\,df &= a\left(\cos E - e\right)\cdot a\sqrt{1-e^{2}}\,\cos E\,dE \;-\; a\sqrt{1-e^{2}}\,\sin E \cdot \left(-a\sin E\,dE\right) \\[2pt] &= a^{2}\sqrt{1-e^{2}}\left[\cos^{2}E - e\cos E + \sin^{2}E\right] dE \\[2pt] &= a^{2}\sqrt{1-e^{2}}\left(1 - e\cos E\right) dE . \end{aligned} \]

The awkward $\left(1 + e\cos f\right)^{-2}$ has now turned into a linear factor.

Using $h = \sqrt{\mu p} = \sqrt{\mu a}\,\sqrt{1-e^{2}}$,

\[ dt = \frac{r^{2}\,df}{h} = \frac{a^{2}\sqrt{1-e^{2}}}{\sqrt{\mu a}\,\sqrt{1-e^{2}}}\left(1 - e\cos E\right) dE = \sqrt{\frac{a^{3}}{\mu}}\left(1 - e\cos E\right) dE . \]

Every trace of $e$ outside the bracket cancels. Integrating from periapsis ($E = 0$ at $t = t_p$),

\[ t - t_p = \sqrt{\frac{a^{3}}{\mu}}\left(E - e\sin E\right) . \]

The prefactor is just $1/n$, with $n = \sqrt{\mu/a^{3}} = 2\pi/T$ the mean motion from the period slide. Define the mean anomaly

\[ \boxed{\; M \equiv n\left(t - t_p\right) = \frac{2\pi}{T}\left(t - t_p\right) \;} \]

and the derivation closes as Kepler’s equation:

\[ \boxed{\; M = E - e\sin E \;} \]

Two immediate checks: setting $e = 0$ gives $M = E = f$ (a circle, swept uniformly), and $E = 2\pi$ gives $M = 2\pi$, i.e. $t - t_p = T$ — one full revolution, as it must.

What $M$ is. It is the only one of the three anomalies that is not an angle you can point at on the orbit. It is the angle of a fictitious body that circles at radius $a$ with the constant rate $n$, starting from periapsis together with the real one. Equivalently, by Kepler’s second law it is the swept area rescaled to an angle:

\[ M = 2\pi\,\frac{\text{area swept since periapsis}}{\pi a b} . \]

$M$, $E$ and $f$ agree at periapsis ($0$) and at apoapsis ($\pi$), and elsewhere $M \le E \le f$ on the outbound half — the body runs ahead of its fictitious twin after periapsis, because that is where it moves fastest.

Anomaly	Symbol	Vertex	Behaviour in time
true	$f$	focus	the physical angle; non-uniform
eccentric	$E$	centre	a computational intermediary
mean	$M$	—	linear in $t$ by construction

Before we proceed, lets calculate the relationship between $E$ and $f$.

Using the focus-based coordinates of the body,

\[ x = r\cos f,\qquad y = r\sin f. \]

From the eccentric-anomaly parametrization of the ellipse,

\[ x = a(\cos E-e),\qquad y = a\sqrt{1-e^2}\sin E, \]

and we assume

\[ r = a(1-e\cos E). \]

Therefore

\[ \cos f = \frac{x}{r} = \frac{\cos E-e}{1-e\cos E}, \]

and

\[ \sin f = \frac{y}{r} = \frac{\sqrt{1-e^2}\sin E}{1-e\cos E}. \]

Now use the half-angle identity

\[ \tan\frac{f}{2}=\frac{\sin f}{1+\cos f}. \]

Compute the denominator:

\[ 1+\cos f = 1+\frac{\cos E-e}{1-e\cos E} = \frac{(1-e)(1+\cos E)}{1-e\cos E}. \]

\[ \tan\frac{f}{2} = \frac{\sqrt{1-e^2}\sin E}{(1-e)(1+\cos E)} = \sqrt{\frac{1+e}{1-e}}\, \frac{\sin E}{1+\cos E}. \]

Finally,

\[ \frac{\sin E}{1+\cos E}=\tan\frac{E}{2}, \]

hence

\[ \boxed{ \tan\frac{f}{2} = \sqrt{\frac{1+e}{1-e}}\, \tan\frac{E}{2} } \]

This tangent form needs quadrant care near (E=); numerically the atan2 form is safer.

Using Kepler’s equation

The three anomalies give the standard propagation pipeline. Given the elements $(a, e, i, \Omega, \omega, t_p)$ and a time $t$:

\[ t \;\longrightarrow\; M = n(t - t_p) \;\longrightarrow\; E \;\longrightarrow\; f \;\longrightarrow\; \mathbf{r} \]

$t \to M$ is trivial — that is the entire point of defining $M$.
$M \to E$ is the hard step: $M = E - e\sin E$ is transcendental, with no solution in elementary functions. Newton’s method is the standard tool,

\[ E_{k+1} = E_{k} - \frac{E_{k} - e\sin E_{k} - M}{1 - e\cos E_{k}}, \]

starting from $E_{0} = M$ (or $E_{0} = \pi$ for $e$ near $1$, where the derivative $1 - e\cos E$ nearly vanishes at periapsis). The denominator is $1 - e\cos E = r/a > 0$, so the iteration is well posed for every $e < 1$ and converges quadratically. - $E \to f$ uses the half-angle form of the Step-2 relations,

\[ \tan\frac{f}{2} = \sqrt{\frac{1+e}{1-e}}\,\tan\frac{E}{2}, \]

which is quadrant-safe, unlike inverting $\cos f = (\cos E - e)/(1 - e\cos E)$. - $\to \mathbf{r}$: in fact $r = a(1 - e\cos E)$ needs no $f$ at all; $f$ is required only to orient $\mathbf{r}$ in the orbital plane.

The sixth element. Rather than $t_p$, catalogues usually quote the mean anomaly at epoch, $M_{e} \equiv M(t_{0}) = n(t_{0} - t_p)$, for some stated epoch $t_{0}$; then $M(t) = M_{e} + n(t - t_{0})$. It is the same information in a form that does not blow up as $e \to 0$, where periapsis — and hence $t_p$ — becomes undefined. This is the timing constant promised in the integral count: five constants fix the curve, $M_e$ fixes where on it you are.

Aside — why there is no closed-form solution

Kepler posed this in Astronomia Nova (1609) and admitted he could not solve it: “I am sufficiently satisfied that it cannot be solved a priori, on account of the heterogeneous nature of the arc and the sine.” He was right — $E(M)$ is not an elementary function. Bessel’s series,

\[ E = M + 2\sum_{k=1}^{\infty} \frac{1}{k}\,J_{k}(ke)\,\sin kM, \]

converges for $e < 0.6627$ (the Laplace limit), and the Bessel functions $J_{k}$ were introduced by Bessel in 1817 for this very problem. Some four centuries of numerical methods have been aimed at this one scalar equation.

For unbound orbits the same derivation with $E \to iH$ gives the hyperbolic Kepler equation $M_{h} = e\sinh H - H$, with $r = a(e\cosh H - 1)$ and $n = \sqrt{\mu/|a|^{3}}$.

Relationship between $E$ $M$ and $f$?

Figure 9: The three anomalies against one another, for a range of eccentricities. **Left:** the map $E \mapsto f$ — monotone and pinned at $0^{\circ}$, $180^{\circ}$ and $360^{\circ}$ for every $e$; the circle $e = 0$ gives the diagonal $f = E$, and increasing $e$ bows the curve further above it on the outbound half. **Centre:** the same information as the lead $f - E$, which is odd about $E = 180^{\circ}$ — what the body gains on the way out it gives back on the way in. The maximum lead grows from $\approx 11^{\circ}$ at $e = 0.2$ to $\approx 93^{\circ}$ at $e = 0.95$, and always occurs at the point where $r = b$. **Right:** the map $M \mapsto f$, which is the composition of the other two. The dotted curve is the intermediate step $M \mapsto E$ at $e = 0.6$: Kepler’s equation bows the diagonal once, the focus offset bows it again, and the two bows add.

Two different questions. The left panel and the right one look alike and are not. $E \leftrightarrow f$ is a question about a picture: freeze the ellipse, and the two angles are the same point read off from two different vertices — the centre and the focus. No dynamics enter, and no clock. $M \leftrightarrow f$ is a question about time: $M$ is Kepler’s second law rescaled to an angle. So the extra bowing in the right panel is exactly what the areal law contributes on top of the focus offset.

Why the curves bow. Differentiate the half-angle relation $\tan\frac{f}{2} = \sqrt{\frac{1+e}{1-e}}\,\tan\frac{E}{2}$:

\[ \tfrac12\sec^{2}\tfrac f2\,df = \sqrt{\frac{1+e}{1-e}}\;\tfrac12\sec^{2}\tfrac E2\,dE \qquad\Longrightarrow\qquad \frac{df}{dE} = \sqrt{\frac{1+e}{1-e}}\;\frac{\cos^{2}\frac f2}{\cos^{2}\frac E2} = \sqrt{\frac{1+e}{1-e}}\;\frac{1 + \cos f}{1 + \cos E} . \]

The denominator computed on the way to the half-angle relation, $1 + \cos f = \dfrac{(1-e)(1 + \cos E)}{1 - e\cos E}$, collapses the last ratio:

\[ \frac{df}{dE} = \sqrt{\frac{1+e}{1-e}}\;\frac{1-e}{1 - e\cos E} = \frac{\sqrt{1 - e^{2}}}{1 - e\cos E} = \frac{b/a}{r/a} \qquad\Longrightarrow\qquad \boxed{\;\frac{df}{dE} = \frac{b}{r}\;} \]

A remarkably clean statement, and it reads off the whole left panel:

$r < b$ (near periapsis): $f$ outruns $E$. The focus is offset towards periapsis, so a small displacement there subtends a large angle at the focus but only a small one at the centre.
$r > b$ (near apoapsis): $f$ falls behind $E$.
$r = b$: the two run level, $df/dE = 1$, which is exactly the stationary point of $f - E$ marked on the centre panel. Setting $1 - e\cos E = \sqrt{1-e^{2}}$ gives its location in closed form,

\[ \cos E = \frac{1 - \sqrt{1 - e^{2}}}{e} . \]

$E = 0, \pi$ are fixed points for every $e$: at the apsides the centre, the focus and the body are collinear, so all three anomalies coincide. This is why the curves in the left panel form a pencil through $(180^{\circ}, 180^{\circ})$.

The same question for $M$. Kepler’s equation differentiates in one line,

\[ M = E - e\sin E \qquad\Longrightarrow\qquad \frac{dM}{dE} = 1 - e\cos E = \frac{r}{a} , \]

and the chain rule finishes it:

\[ \frac{dM}{df} = \frac{dM}{dE}\,\frac{dE}{df} = \frac{r}{a}\cdot\frac{r}{b} \qquad\Longrightarrow\qquad \boxed{\;\frac{dM}{df} = \frac{r^{2}}{ab}\;} \]

This is not a new fact — it is Kepler’s second law wearing different clothes. The area swept by the radius is $dA = \tfrac12 r^{2}\,df$, and $M = 2A/(ab)$ from the area definition of the mean anomaly, so $dM = r^{2}df/(ab)$ identically. Every one of the three derivatives is one power of $r$ away from its neighbour, and each has its own level-running radius:

relation	derivative	runs level where
$M$ against $E$	$dM/dE = r/a$	$r = a$
$f$ against $E$	$df/dE = b/r$	$r = b$
$f$ against $M$	$df/dM = ab/r^{2}$	$r = \sqrt{ab}$

and since $b \le \sqrt{ab} \le a$, the three crossings sit in that order as the body climbs from periapsis to apoapsis — the geometric mean $\sqrt{ab}$ sandwiched between the other two.

Small $e$ approximations

Derive the following small angle approximations, which are useful for low-eccentricity orbits:

\[ f = E + e\sin E + O\!\left(e^{2}\right), \qquad E = M + e\sin M + O\!\left(e^{2}\right) \;\Longrightarrow\; f = M + 2e\sin M + O\!\left(e^{2}\right), \]

Worked example — propagating a geosynchronous orbit

Not every geosynchronous orbit is geostationary. Take one of the eccentric, Tundra-type variety: the period is a sidereal day, but the orbit is far from circular. With $\mu_\oplus = 3.986\,004\times10^{5}\ \mathrm{km^{3}/s^{2}}$,

\[ a = 42\,164\ \mathrm{km},\qquad e = 0.25 \qquad\Longrightarrow\qquad r_p = 31\,623\ \mathrm{km},\quad r_a = 52\,705\ \mathrm{km},\quad b = 40\,825\ \mathrm{km}. \]

\[ n = \sqrt{\frac{\mu}{a^{3}}} = 7.29216\times10^{-5}\ \mathrm{rad/s}, \qquad T = \frac{2\pi}{n} = 86\,163.6\ \mathrm{s} = 23^{\mathrm{h}}56^{\mathrm{m}}04^{\mathrm{s}} , \]

the sidereal day — which is the whole point of the $42\,164$ km: $n$ is Earth’s rotation rate. At $t_{0}$ the spacecraft is at $f_{0} = 30^{\circ}$. Where is it 6 hours later?

Step 0 — run the pipeline backwards to get the clock. We are handed $f_0$, not $M_0$, so first $f \to E \to M$:

\[ \tan\frac{E_{0}}{2} = \sqrt{\frac{1-e}{1+e}}\,\tan\frac{f_{0}}{2} = \sqrt{\frac{0.75}{1.25}}\,\tan 15^{\circ} \qquad\Longrightarrow\qquad E_{0} = 23.4508^{\circ} = 0.409294\ \mathrm{rad}, \]

\[ M_{0} = E_{0} - e\sin E_{0} = 0.409294 - 0.25\sin(0.409294) = 0.309804\ \mathrm{rad} = 17.7504^{\circ} . \]

So $t_{0} - t_p = M_{0}/n = 4248.5\ \mathrm{s} = 1.180\ \mathrm{h}$ Its radius there is $r_{0} = a(1 - e\cos E_{0}) = 32\,494$ km.

Step 1 — $t \to M$.

\[ M_{1} = M_{0} + n\,\Delta t = 0.309804 + \left(7.29216\times10^{-5}\right)(21\,600) = 0.309804 + 1.575107 = 1.884910\ \mathrm{rad} = 107.997^{\circ}. \]

Step 2 — $M \to E$. Newton on $g(E) = E - e\sin E - M_{1}$, started at $E = M_{1}$:

$k$	$E_{k}$ [rad]	$g(E_{k})$	$\Delta E_{k}$
0	$1.884910$	$-2.378\times10^{-1}$	$+2.207\times10^{-1}$
1	$2.105629$	$+5.630\times10^{-3}$	$-4.994\times10^{-3}$
2	$2.100635$	$+2.685\times10^{-6}$	$-2.383\times10^{-6}$
3	$2.100633$	$+6.1\times10^{-13}$	—

\[ E_{1} = 2.100633\ \mathrm{rad} = 120.3574^{\circ} . \]

Four iterations to machine precision, and the residual squares each step ($10^{-1} \to 10^{-3} \to 10^{-6} \to 10^{-13}$) — quadratic convergence. This is the best thing about the Newton iterations. The derivative $1 - e\cos E_{1} = r_{1}/a = 1.126$ never came near zero. So this was a well conditioned problem. Otherwise, we may not get such good convergence behaviour.

Step 3 — $E \to f$.

\[ \tan\frac{f_{1}}{2} = \sqrt{\frac{1+e}{1-e}}\,\tan\frac{E_{1}}{2} = \sqrt{\frac{1.25}{0.75}}\,\tan(60.179^{\circ}) \qquad\Longrightarrow\qquad f_{1} = 132.118^{\circ}. \]

Step 4 — $\to r$. Either form, and they must agree:

\[ r_{1} = a\left(1 - e\cos E_{1}\right) = 47\,491\ \mathrm{km} = \frac{p}{1 + e\cos f_{1}} = \frac{39\,528}{1 + 0.25\cos 132.118^{\circ}}\ \checkmark \]

With the radius in hand the rest of the state follows: $v_{1} = \sqrt{\mu(2/r_{1} - 1/a)} = 2.708\ \mathrm{km/s}$ and $\tan\gamma_{1} = e\sin f_{1}/(1 + e\cos f_{1})$, giving $\gamma_{1} = 12.56^{\circ}$ — still climbing towards apogee.

Read the three anomalies off this one instant:

\[ M_{1} = 108.00^{\circ} \;<\; E_{1} = 120.36^{\circ} \;<\; f_{1} = 132.12^{\circ}, \]

the outbound ordering $M \le E \le f$, straight off the right-hand panel of Figure 9. The body is $24^{\circ}$ ahead of its fictitious uniformly circling twin, having spent the last seven hours near perigee where it moves fastest.

Flight Path Angle $\gamma$

The algebra of $\gamma$ was settled in Part I — it is universal, and none of Part II changed it. What is worth doing now that the ellipse is in front of us is to draw it, because $\gamma$ is one of those quantities that is obvious in a picture and easy to mis-sign in a formula.

Figure 10: **Left:** one sample point in each quadrant of an ellipse with $e = 0.6$. At each, the dotted line is the radius from the focus $F$, the green dashes are the **local horizon** (perpendicular to $\mathbf{r}$, i.e. along $\hat{\boldsymbol{\theta}}$), the blue and amber arrows are $v_\perp$ and $v_r$, and the maroon arrow is their resultant $\mathbf{v}$. The angle $\gamma$ is marked between the horizon and $\mathbf{v}$. **Right:** the same construction at $f = 135^{\circ}$ on its own, rotated so the local horizon lies flat — the velocity triangle with the orbit taken away.

Four things the picture makes obvious that the formula only implies:

The horizon is not horizontal. It is perpendicular to $\mathbf{r}$, so it swings right round with the body. $\gamma$ is measured against a reference that is itself rotating — which is why $\gamma$ is a local quantity and why it can be non-zero even though the orbit is closed.
$v_\perp$ never changes sign; $v_r$ does. The blue arrow points the same way round the orbit at all four points — it must, since $v_\perp = h/r > 0$ carries all the angular momentum. The amber arrow reverses between the second and third quadrants, and that reversal is the sign of $\gamma$: climbing on the way out ($0 < f < 180^{\circ}$), descending on the way back.
The two quadrant pairs are mirror images. $f = 45^{\circ}$ and $f = 315^{\circ}$ carry $\gamma = \pm16.6^{\circ}$, and $f = 135^{\circ}$ and $f = 225^{\circ}$ carry $\gamma = \pm36.4^{\circ}$: same radius, same speed, velocity reflected in the apse line. This is the symmetry noted back at the speed formula — $v$ depends on $f$ only through $\cos f$.
At the apsides the velocity lies flat along the horizon, $\gamma = 0$: the amber component has vanished, $\mathbf{v} \perp \mathbf{r}$, and $v = h/r$ exactly.

The $f = 135^{\circ}$ point is close to the steepest the orbit ever gets: with $e = 0.6$ the maximum is $\gamma_{\max} = \arcsin 0.6 = 36.87^{\circ}$ at $\cos f = -e$, i.e. $f = 126.87^{\circ}$, so $36.4^{\circ}$ at $135^{\circ}$ is already within half a degree of the ceiling. The curve is flat near its top, which is another way of saying that an eccentric orbit spends a long stretch of its climb at very nearly its steepest angle.

Parabolic Orbit ($e = 1$)

The one-parameter conic

Set $e = 1$ in the orbit equation and use $1 + \cos f = 2\cos^{2}\tfrac f2$:

\[ r = \frac{p}{1 + \cos f} = \frac{p}{2}\,\sec^{2}\frac{f}{2} \qquad\Longrightarrow\qquad \boxed{\;r_p = \frac{p}{2}\;} \quad\text{at } f = 0 . \]

There is no second parameter. An ellipse needs $(a, e)$ and a hyperbola needs $(p, e)$; a parabola is fixed by $p$ alone — equivalently by $r_p$, or by $h = \sqrt{\mu p}$. The eccentricity is not a free number here, it is the defining constraint.

Every Part I formula specialises by putting $e=1$:

\[ \varepsilon = \frac{\mu\left(e^{2}-1\right)}{2p} = 0, \qquad a = \frac{p}{1-e^{2}} \to \infty , \]

so never use $a$ for a parabola — $\varepsilon = -\mu/2a$ is the correct statement read the wrong way round. Vis-viva degenerates to

\[ \boxed{\;v^{2} = \frac{2\mu}{r}\;} \]

which is the escape-speed relation: a parabolic orbit is moving at exactly the local escape speed at every point of its path. It reaches infinity, and arrives there with nothing left.

Speed and direction on a parabola

From the universal speed formula with $e=1$:

\[ v = \sqrt{\frac{\mu}{p}}\sqrt{2 + 2\cos f} = 2\sqrt{\frac{\mu}{p}}\left|\cos\frac{f}{2}\right| , \]

which agrees with $v = \sqrt{2\mu/r}$, as it must. And the flight path angle collapses to something remarkably clean:

\[ \tan\gamma = \frac{\sin f}{1 + \cos f} = \frac{2\sin\frac f2\cos\frac f2}{2\cos^{2}\frac f2} = \tan\frac{f}{2} \qquad\Longrightarrow\qquad \boxed{\;\gamma = \frac{f}{2}\;} \]

The velocity direction turns at exactly half the rate of the position vector. Over the whole flight $f$ runs from $-\pi$ to $\pi$ and $\gamma$ from $-\pi/2$ to $\pi/2$: the velocity swings through a full half turn, from radially inbound at infinity to radially outbound at infinity.
Compare the ellipse, where $|\gamma| \le \arcsin e < \pi/2$: the parabola is the first case in which the bound is attained, in the limit.
The hodograph is still the circle of radius $\mu/h$, but the velocity-space origin now lies exactly on it — the point $v=0$ is reached, once, at infinity.

Why are there no parabolic orbits in nature?

Answer

$\varepsilon = 0$ is a single point in a continuum: the set of initial conditions giving exactly zero energy has measure zero, and any perturbation — a third body, a puff of outgassing, the finite mass of the “test” body — pushes $\varepsilon$ off zero. Long-period comets arrive on orbits with $e = 0.999\ldots$, which are elliptical with periods of $10^{5}$–$10^{7}$ years; astronomers fit them as parabolic because the observed arc near perihelion cannot distinguish $e = 1$ from $e = 1 \pm 10^{-3}$.

The parabola earns its place as the separatrix: it is the boundary whose crossing is the entire content of “escape”.

Time on a parabola — Barker’s equation

The Part I integral with $e = 1$:

\[ t - t_p = \frac{p^{2}}{h}\int_{0}^{f}\frac{df'}{\left(1+\cos f'\right)^{2}} = \frac{p^{2}}{4h}\int_{0}^{f}\sec^{4}\frac{f'}{2}\,df' . \]

Put

\[ D = \tan\frac{f}{2} \qquad\Longrightarrow\qquad dD = \tfrac{1}{2}\sec^{2}\frac{f}{2}\,df, \qquad \sec^{2}\frac{f}{2} = 1 + D^{2} . \]

$D$ is the parabolic anomaly. It plays the role $E$ plays for the ellipse: it is not an angle you can point at, but it linearises the integral. Substituting,

\[ \frac{p^{2}}{4h}\sec^{4}\frac f2\,df = \frac{p^{2}}{4h}\left(1+D^{2}\right)\cdot 2\,dD = \frac{p^{2}}{2h}\left(1 + D^{2}\right)dD , \]

and the integral is now a polynomial. With $h = \sqrt{\mu p}$ and $p = 2r_p$,

\[ t - t_p = \frac{p^{2}}{2h}\left(D + \frac{D^{3}}{3}\right) = \frac{1}{2}\sqrt{\frac{p^{3}}{\mu}}\left(D + \frac{D^{3}}{3}\right). \]

Defining the parabolic mean anomaly $M_p \equiv \sqrt{\dfrac{\mu}{2r_p^{3}}}\,\left(t - t_p\right)$,

\[ M_p = D + \frac{D^{3}}{3} \]

which is Barker’s equation (Barker, 1757).

$M_p = D + D^{3}/3$ is a cubic, so unlike $M = E - e\sin E$ it can be inverted in closed form. Write it as

\[ D^{3} + 3D - 3M_p = 0 , \]

Solution (as given by Cardano) is:

\[ D = \sqrt[3]{\frac{3M_p}{2} + \sqrt{\frac{9M_p^{2}}{4} + 1}} \;+\; \sqrt[3]{\frac{3M_p}{2} - \sqrt{\frac{9M_p^{2}}{4} + 1}}, \qquad f = 2\arctan D . \]

The parabola is the only conic whose Kepler problem has an elementary solution.
The root is unique because $dM_p/dD = 1 + D^{2} > 0$: $M_p(D)$ is strictly increasing

Hyperbolic Orbit ($e > 1$)

Geometry

With $e>1$ the definition $a = p/(1-e^{2})$ returns a negative number. Write $|a| = p/(e^{2}-1)$ and every ellipse formula reads correctly:

\[ r = \frac{|a|\left(e^{2}-1\right)}{1 + e\cos f}, \qquad r_p = \frac{p}{1+e} = |a|\left(e - 1\right), \qquad \varepsilon = -\frac{\mu}{2a} = +\frac{\mu}{2|a|} > 0 . \]

Clearly, the denominator goes to $0$ as $\cos f \rightarrow -1/e$. Therefore body exists only on $|f| < f_\infty$, with $\cos f_\infty = -1/e$. Geometrically (Figure 11):

$|a|$ is the distance from the centre of the hyperbola to the vertex (periapsis); the focus is a further $|a|e$ out, so $r_p = |a|(e-1)$.
$b = |a|\sqrt{e^{2}-1}$ is the semi-conjugate axis — not a distance from the centre to the curve, since the hyperbola has no perpendicular chord through its centre (next slide); the asymptotes make angle $\arctan(b/|a|)$ with the apse line.
The other branch is the trajectory of a hypothetical repulsive inverse-square force with the same $h$ and $\varepsilon$; it is not physical here.

Vis-viva with $a<0$ is best written with the sign made explicit:

\[ \boxed{\;v^{2} = \mu\left(\frac{2}{r} + \frac{1}{|a|}\right)\;} \]

— strictly larger than the local escape speed $\sqrt{2\mu/r}$ everywhere, by a margin that does not vanish as $r \to \infty$.

Speeds $v_\infty$, $C_3$

Speed at infinity. Let $r\to\infty$:

\[ \boxed{\;v_\infty = \sqrt{2\varepsilon} = \sqrt{\frac{\mu}{|a|}}\;}, \qquad C_3 \equiv v_\infty^{2} = \frac{\mu}{|a|} = -\frac{\mu}{a} . \]

This is the number on every launch-vehicle performance chart — the only thing that survives to interplanetary distances.

Turning angle

The incoming and outgoing asymptotes lie at $f = -f_\infty$ and $f = +f_\infty$ (Figure 11). The velocity direction turns through

\[ \delta = 2f_\infty - \pi, \qquad \sin\frac{\delta}{2} = \sin\!\left(f_\infty - \frac{\pi}{2}\right) = -\cos f_\infty \qquad\Longrightarrow\qquad \boxed{\;\sin\frac{\delta}{2} = \frac{1}{e}\;} \]

Everything in one line. Combining $r_p = |a|(e-1)$ with $|a| = \mu/v_\infty^{2}$:

\[ \boxed{\;e = 1 + \frac{r_p\,v_\infty^{2}}{\mu}\;} \]

so a flyby is steered entirely by how deep you go and how fast you arrive. Slow and deep turns hard ($e\to1$, $\delta\to\pi$); fast and shallow barely bends ($e\to\infty$, $\delta\to 0$).

Aiming radius

The perpendicular offset of the incoming asymptote from the primary (Figure 11) is the impact parameter

\[ \Delta = b = |a|\sqrt{e^{2}-1}, \qquad\text{and}\qquad h = \Delta\,v_\infty , \]

the last being $h = |\mathbf{r}\times\mathbf{v}|$ evaluated at infinity, where the lever arm is $\Delta$. Check: $\Delta^{2}v_\infty^{2} = a^{2}(e^{2}-1)\cdot\mu/|a| = \mu p = h^{2}$. $\checkmark$

A spacecraft passes Jupiter ($\mu = 1.267\times10^{8}\ \mathrm{km^{3}/s^{2}}$) with $v_\infty = 6\ \mathrm{km/s}$ at a periapsis radius of $r_p = 10^{6}$ km. How much is it turned?

Answer

\[ e = 1 + \frac{10^{6}\times 36}{1.267\times10^{8}} = 1.284, \qquad \delta = 2\arcsin\frac{1}{1.284} = 2\times 51.2^{\circ} = 102.4^{\circ} . \]

In Jupiter’s frame the speed is unchanged — $v_\infty$ in equals $v_\infty$ out — but the direction has swung by more than a right angle. In the Sun’s frame that rotation adds vectorially to Jupiter’s $13.1$ km/s orbital velocity, and the heliocentric speed changes by up to $2v_\infty = 12$ km/s. That momentum comes out of Jupiter’s orbit; the planet is slowed by a correspondingly absurd amount.

What $b$ is for a hyperbola

For the ellipse $b$ is a length you can measure: the semi-minor axis, centre to curve. A hyperbola never crosses the perpendicular to the apse line through its centre — the centre-based equation forces $|x| \ge |a|$ — so for $e>1$ the symbol $b$ has to be defined. Eliminating $f$ between $r = p/(1+e\cos f)$ and $x = r\cos f$ and shifting the origin from the focus to the centre gives

\[ \frac{x^{2}}{|a|^{2}} - \frac{y^{2}}{b^{2}} = 1, \qquad \boxed{\;b^{2} \;\equiv\; c^{2} - |a|^{2} \;=\; |a|^{2}\left(e^{2}-1\right)\;} \]

with $c = |a|e$ the centre-to-focus distance — the ellipse’s $b^{2} = a^{2} - c^{2}$ with the two terms exchanged, which is the whole of the difference. The segment of length $2b$ erected on the apse line at the centre is the conjugate axis; it meets the curve nowhere, and it is the only thing $b$ measures directly.

Three equivalent ways to pin the same number, all visible in Figure 11:

the asymptote slope. The asymptotes are $y = \pm(b/|a|)\,x$, so $b$ is the rise of an asymptote over one semi-transverse axis $|a|$, and the rectangle $|x|\le|a|$, $|y|\le b$ has them as its diagonals.
the same $p$ relation as the ellipse. $b^{2} = |a|\,p$, so $b = \sqrt{|a|p}$ — the geometric mean of $|a|$ and the semi-latus rectum, exactly as $b^{2} = ap$ for $e<1$.
the focus-to-asymptote distance. The perpendicular dropped from either focus onto either asymptote has length exactly $b$. That is why the aiming radius is $\Delta = b$, and why $h = \Delta v_\infty$ reads off the lever arm at infinity.

$b \lessgtr |a|$ according as $e \lessgtr \sqrt{2}$. Nothing makes the conjugate axis the shorter one — “semi-minor axis” is a misnomer for $e>1$, and the figure below is drawn at $e = 1.6$, where $b > |a|$.

The hyperbola in one picture

Figure 11: **Left — the conic.** The attracting body sits at the focus $F$, at distance $|a|e$ from the **centre** $O$ of the hyperbola; the vertex is $|a|$ from the centre, so the focus-to-vertex distance is $r_p = |a|e - |a| = |a|(e-1)$. The construction rectangle (dotted, half-sides $|a|$ and $b = |a|\sqrt{e^{2}-1}$) carries the asymptotes through its corners, so each asymptote makes an angle $\arctan(b/|a|) = \arccos(1/e)$ with the apse line — equivalently the body runs out along $f \to \pm f_\infty$ with $\cos f_\infty = -1/e$. The faint mirror branch has the same $h$ and $\varepsilon$ but belongs to a *repulsive* inverse-square force. **Right — the same orbit as a flyby.** The body arrives along the lower asymptote with $\mathbf{v}_\infty$ and leaves along the upper one at the same speed; carrying the incoming direction to $O$ shows that the velocity has swung through the turning angle $\delta = 2f_\infty - \pi$, whence $\sin(\delta/2) = 1/e$. The aiming radius $\Delta$ is the perpendicular offset of the incoming asymptote from the focus; it equals $b$ exactly, which is why the lever arm at infinity gives $h = \Delta v_\infty$. Drawn for $e = 1.6$.

The hyperbolic anomaly $H$

The elliptic trick was $x_P = a(\cos E - e)$, built on the auxiliary circle. Here the auxiliary curve is a rectangular hyperbola, and the circular functions become hyperbolic — formally the substitution $E \to iH$ (Figure 12). Measured from the focus,

\[ x_P = |a|\left(e - \cosh H\right), \qquad y_P = |a|\sqrt{e^{2}-1}\,\sinh H . \]

Figure 12: **Left:** why the hyperbolic functions appear at all. On the unit circle the parameter $E$ is twice the shaded sector area — and, uniquely for the circle, that is also an angle at the centre. On the rectangular hyperbola $u^{2}-v^{2}=1$ the point is $(\cosh H, \sinh H)$ and $H$ is *only* twice the shaded sector area: there is no angle to be had, which is exactly what $E \to iH$ costs us. **Right:** the same construction as the ellipse’s ($e = 1.2$ here). The auxiliary curve is the rectangular hyperbola $x^{2}-y^{2}=|a|^{2}$ centred at $O$; the orbit is that curve scaled by $b/|a| = \sqrt{e^{2}-1}$ in the transverse direction, so the body $P$ and the auxiliary point $Q$ share the abscissa $-|a|\cosh H$ — which is why $r$ comes out *linear* in $\cosh H$. Measured from the focus $F$ the coordinates are $x_P = |a|(e - \cosh H)$ (negative once $\cosh H > e$, as drawn) and $y_P = b\sinh H$.

Squaring and using $\cosh^{2}H - \sinh^{2}H = 1$:

\[ \begin{aligned} r^{2} &= a^{2}\left(e - \cosh H\right)^{2} + a^{2}\left(e^{2}-1\right)\sinh^{2}H \\[2pt] &= a^{2}\left[e^{2} - 2e\cosh H + \cosh^{2}H + \left(e^{2}-1\right)\left(\cosh^{2}H - 1\right)\right] \\[2pt] &= a^{2}\left[e^{2}\cosh^{2}H - 2e\cosh H + 1\right] = a^{2}\left(e\cosh H - 1\right)^{2}, \end{aligned} \]

\[ \boxed{\;r = |a|\left(e\cosh H - 1\right)\;} \]

exactly the elliptic $r = a(1 - e\cos E)$ with $\cos \to \cosh$ and the sign flipped — and, as there, linear in the new variable. The range is $H \in (-\infty, \infty)$, with $H = 0$ at periapsis and $H \to \pm\infty$ at the asymptotes.

The true anomaly follows from the same half-angle manipulation as for the ellipse:

\[ \boxed{\;\tanh\frac{H}{2} = \sqrt{\frac{e-1}{e+1}}\,\tan\frac{f}{2}\;} \]

The $\tanh$ on the left is bounded by $1$, and the right-hand side reaches $1$ precisely at $\tan\frac{f_\infty}{2} = \sqrt{(e+1)/(e-1)}$ — the asymptote, pushed off to $H = \infty$.

The three hyperbolic functions

$\cosh^{2}H - \sinh^{2}H = 1$ replaces $\cos^{2} + \sin^{2} = 1$, which is what puts $(\cosh H, \sinh H)$ on the rectangular hyperbola.
$\dfrac{d}{dH}\sinh H = \cosh H$ and $\dfrac{d}{dH}\cosh H = \sinh H$ — no sign flip. That is why $dM_h/dH = e\cosh H - 1 = r/|a| > 0$ and the hyperbolic Kepler equation is monotone, with no periodicity to wrap.
Near the origin $\sinh H, \tanh H = H + O(H^{3})$ and $\cosh H = 1 + H^{2}/2$: close to periapsis the three functions are nearly indistinguishable from their circular counterparts, which is the ill-conditioning the universal-variable formulation exists to cure.

The hyperbolic Kepler equation

Repeat the Part I integral with the same bookkeeping. From $r\,df$ in terms of $H$, and $h = \sqrt{\mu p} = \sqrt{\mu|a|\left(e^{2}-1\right)}$,

\[ dt = \frac{r^{2}}{h}\,df = \sqrt{\frac{|a|^{3}}{\mu}}\left(e\cosh H - 1\right)dH . \]

Integrating from periapsis ($H = 0$ at $t = t_p$) and defining the hyperbolic mean motion and mean anomaly

\[ n_h = \sqrt{\frac{\mu}{|a|^{3}}}, \qquad M_h = n_h\left(t - t_p\right), \]

gives the hyperbolic Kepler equation

\[ \boxed{\;M_h = e\sinh H - H\;} \]

Set against its elliptic twin:

	ellipse	hyperbola
radius	$r = a(1 - e\cos E)$	$r = \lvert a\rvert(e\cosh H - 1)$
Kepler	$M = E - e\sin E$	$M_h = e\sinh H - H$
derivative	$dM/dE = 1 - e\cos E = r/a$	$dM_h/dH = e\cosh H - 1 = r/\lvert a\rvert$
range	$E \in [0, 2\pi)$, periodic	$H \in (-\infty,\infty)$, monotone

Both derivatives equal $r/|a| > 0$, so both are strictly monotone and Newton–Raphson,

\[ H_{k+1} = H_k - \frac{e\sinh H_k - H_k - M_h}{e\cosh H_k - 1}, \]

is well posed. The practical difference is the starter: $\sinh$ grows exponentially, so $H_0 = M_h$ is a poor guess for large $M_h$ and $H_0 = \ln\!\left(2M_h/e + 1.8\right)$ is the standard choice.

There is no period, no mean motion in the physical sense, and no repetition: $M_h$ runs monotonically from $-\infty$ to $+\infty$. Everything about a hyperbolic pass happens once.

Part V — The Three Conics Side by Side

One table

	circle	ellipse	parabola	hyperbola
$e$	$0$	$0<e<1$	$1$	$>1$
$\varepsilon$	$<0$	$<0$	$0$	$>0$
$a = p/(1-e^{2})$	$a = r$	$a>0$	$\to\infty$	$a<0$
range of $f$	all	all	$\lvert f\rvert<\pi$	$\lvert f\rvert<\arccos(-1/e)$
$r_p$	$a$	$a(1-e)$	$p/2$	$\lvert a\rvert(e-1)$
$r_a$	$a$	$a(1+e)$	—	—
period	$2\pi\sqrt{a^{3}/\mu}$	$2\pi\sqrt{a^{3}/\mu}$	—	—
$v$ at $\infty$	—	—	$0$	$\sqrt{\mu/\lvert a\rvert}$
$\lvert\gamma\rvert_{\max}$	$0$	$\arcsin e$	$\pi/2$ (limit)	$f_\infty - \pi/2$ (limit)
auxiliary variable	$f$ itself	$E$	$D = \tan\frac f2$	$H$
Kepler equation	$M = f$	$M = E - e\sin E$	$M_p = D + \frac{D^{3}}{3}$	$M_h = e\sinh H - H$
closed-form inverse	trivial	no	yes (cubic)	no
hodograph	circle about origin	origin inside	origin on it	origin outside

Every row above the double line is a specialisation of one Part I formula. Every row below it is genuinely separate work.

What is universal after all

Three formulae from Part I are worth committing to memory precisely because they never need a case distinction:

\[ r = \frac{p}{1 + e\cos f}, \qquad v = \sqrt{\frac{\mu}{p}}\sqrt{1 + 2e\cos f + e^{2}}, \qquad \tan\gamma = \frac{e\sin f}{1 + e\cos f} . \]

Given $(p, e, f)$ they return the complete state — position, speed and direction — on any conic whatsoever, with no branch and no sign convention.

Universal variables. The same unification can be pushed through the time equation as well. Introducing $\chi$ with $d\chi = \sqrt{\mu}\,dt/r$ and the Stumpff functions

\[ C(z) = \frac{1-\cos\sqrt z}{z}, \qquad S(z) = \frac{\sqrt z - \sin\sqrt z}{z^{3/2}}, \qquad z = \frac{\chi^{2}}{a}, \]

turns all three Kepler equations into the single relation

\[ \sqrt{\mu}\left(t - t_p\right) = \frac{r_p\,\chi\,\left(1 - z S(z)\right) + \chi^{3}S(z)}{1} \;\;\text{(schematically)} , \]

with $z>0$ elliptic, $z=0$ parabolic and $z<0$ hyperbolic — $C$ and $S$ being entire functions of $z$, so nothing is singular at the parabolic boundary. This is how every production propagator (and every Lambert solver) is actually written. We take it up when we come to orbit determination.

Appendix

The rotating basis $(\hat{\mathbf{r}}, \hat{\boldsymbol{\theta}})$

Everything in the velocity derivation rested on two little formulae:

\[ \dot{\hat{\mathbf{r}}} = \dot f\,\hat{\boldsymbol{\theta}}, \qquad \dot{\hat{\boldsymbol{\theta}}} = -\dot f\,\hat{\mathbf{r}} . \]

They are worth proving twice — once with components, once without.

Setup. Let $\hat{\mathbf{P}}$ point from the focus towards periapsis and $\hat{\mathbf{Q}} = \hat{\mathbf{h}} \times \hat{\mathbf{P}}$ complete a right-handed pair in the orbital plane (the perifocal basis). Both are built from the constants $\mathbf{e}$ and $\mathbf{h}$, so both are fixed in space: $\dot{\hat{\mathbf{P}}} = \dot{\hat{\mathbf{Q}}} = \mathbf{0}$.

Since $f$ is measured from periapsis,

\[ \hat{\mathbf{r}} = \cos f\,\hat{\mathbf{P}} + \sin f\,\hat{\mathbf{Q}}, \qquad \hat{\boldsymbol{\theta}} = -\sin f\,\hat{\mathbf{P}} + \cos f\,\hat{\mathbf{Q}} . \]

Proof 1 — differentiate the components

Only $f$ depends on time, so the chain rule acts on the sines and cosines alone:

\[ \dot{\hat{\mathbf{r}}} = \frac{d}{dt}\left(\cos f\,\hat{\mathbf{P}} + \sin f\,\hat{\mathbf{Q}}\right) = \dot f\left(-\sin f\,\hat{\mathbf{P}} + \cos f\,\hat{\mathbf{Q}}\right) = \dot f\,\hat{\boldsymbol{\theta}} . \]

\[ \dot{\hat{\boldsymbol{\theta}}} = \dot f\left(-\cos f\,\hat{\mathbf{P}} - \sin f\,\hat{\mathbf{Q}}\right) = -\dot f\,\hat{\mathbf{r}} . \]

That is the whole proof — but it hides why it is true, so here is the same statement with no components at all.

Proof 2 — the unit circle argument

Two facts, and they are enough:

A unit vector’s derivative is perpendicular to it. From $\hat{\mathbf{u}} \cdot \hat{\mathbf{u}} = 1$, differentiating gives $2\,\hat{\mathbf{u}} \cdot \dot{\hat{\mathbf{u}}} = 0$. A unit vector can only turn; it can never stretch.
Its rate of turning is the rate of turning of the frame. Over a small interval the tip of $\hat{\mathbf{r}}$ moves along the unit circle by the chord

\[ \left|\Delta\hat{\mathbf{r}}\right| = 2\sin\frac{\Delta f}{2} \;\longrightarrow\; \Delta f \qquad\text{as}\qquad \Delta f \to 0, \]

and the chord’s direction tends to the tangent, which is $\hat{\boldsymbol{\theta}}$ — the direction of increasing $f$.

Dividing by $\Delta t$ and letting $\Delta t \to 0$ gives $\dot{\hat{\mathbf{r}}} = \dot f\,\hat{\boldsymbol{\theta}}$. Repeating the argument for $\hat{\boldsymbol{\theta}}$, whose tangent points at $-\hat{\mathbf{r}}$, gives the second formula. Note that the sign is the only thing that distinguishes them.

Figure 14: **Left:** the pair $(\hat{\mathbf{r}}, \hat{\boldsymbol{\theta}})$ carried by the body at two nearby true anomalies, alongside the fixed perifocal pair $(\hat{\mathbf{P}}, \hat{\mathbf{Q}})$ it is built from — the moving pair is the fixed pair rotated through $f$. **Right:** the two $\hat{\mathbf{r}}$’s redrawn from a common origin. Their tips lie on the unit circle, so the increment $\Delta\hat{\mathbf{r}}$ is a chord of length $2\sin(\Delta f/2) \to \Delta f$ whose direction tends to $\hat{\boldsymbol{\theta}}$.

Both formulae as one statement

The orbital plane’s basis rotates about $\hat{\mathbf{h}}$ with angular velocity

\[ \boldsymbol{\omega} = \dot f\,\hat{\mathbf{h}}, \]

and any vector $\hat{\mathbf{u}}$ fixed in that rotating frame obeys $\dot{\hat{\mathbf{u}}} = \boldsymbol{\omega} \times \hat{\mathbf{u}}$. Both formulae are then one line of cross products:

\[ \boldsymbol{\omega} \times \hat{\mathbf{r}} = \dot f\,\hat{\mathbf{h}} \times \hat{\mathbf{r}} = \dot f\,\hat{\boldsymbol{\theta}}, \qquad \boldsymbol{\omega} \times \hat{\boldsymbol{\theta}} = \dot f\,\hat{\mathbf{h}} \times \hat{\boldsymbol{\theta}} = -\dot f\,\hat{\mathbf{r}} , \]

using the right-handed triad $\hat{\mathbf{r}} \times \hat{\boldsymbol{\theta}} = \hat{\mathbf{h}}$.

This is the same rule that produces Coriolis and centrifugal terms in a rotating frame. The two-body problem is simply the case where the frame’s spin rate $\dot f = h/r^{2}$ is handed to us by a conservation law.

Payoff — acceleration in polar form

Differentiate $\mathbf{v} = \dot r\,\hat{\mathbf{r}} + r\dot f\,\hat{\boldsymbol{\theta}}$ once more, using the two formulae at every step:

\[ \ddot{\mathbf{r}} = \left(\ddot r - r\dot f^{2}\right)\hat{\mathbf{r}} + \left(r\ddot f + 2\dot r\dot f\right)\hat{\boldsymbol{\theta}} . \]

Feeding this into $\ddot{\mathbf{r}} = -\mu\hat{\mathbf{r}}/r^{2}$ splits the equation of motion into its two scalar components:

\[ \ddot r - r\dot f^{2} = -\frac{\mu}{r^{2}}, \qquad r\ddot f + 2\dot r\dot f = \frac{1}{r}\frac{d}{dt}\left(r^{2}\dot f\right) = 0 . \]

The transverse equation is the conservation of angular momentum, $r^{2}\dot f = h$ — rediscovered here as an integral of the motion rather than assumed. Substituting it into the radial equation leaves a single ODE for $r(t)$,

\[ \ddot r - \frac{h^{2}}{r^{3}} = -\frac{\mu}{r^{2}}, \]

which is the starting point for the Binet substitution $u = 1/r$ that yields the orbit equation directly.

\(e\)	\(1-e^{2}\)	\(a\)	\(\varepsilon = -\mu/2a\)
\(<1\)	\(>0\)	\(a > 0\), finite	\(\varepsilon < 0\)
\(=1\)	\(0\)	\(a \to \infty\) (useless — use \(p\) or \(r_p\))	\(\varepsilon = 0\)
\(>1\)	\(<0\)	\(a < 0\)	\(\varepsilon > 0\)

Conic	Specific energy	Behavior at large \(r\)	\(a\)
ellipse	\(\varepsilon<0\)	\(r\leq \mu/\|\varepsilon\|\)	\(a > 0\)
parabola	\(\varepsilon=0\)	\(r\to\infty\) and \(v\to0\)	\(a \to \infty\)
hyperbola	\(\varepsilon>0\)	\(r\to\infty\) and \(v\to\sqrt{2\varepsilon}\)	\(a < 0\)

Orbit	\(e\)	\(v(f)\)	\(\gamma\)	hodograph
circle	\(0\)	\(\sqrt{\mu/a}\), constant	\(0\)	circle centred at origin
ellipse	\(0<e<1\)	\(\sqrt{\mu/p}\sqrt{1+2e\cos f + e^{2}}\)	\(\lvert\gamma\rvert\le\arcsin e\)	origin inside
parabola	\(1\)	\(\sqrt{2\mu/r} = 2\sqrt{\mu/p}\,\left\|\cos\tfrac f2\right\|\)	\(f/2\)	origin on the circle
hyperbola	\(>1\)	as above, \(\to v_\infty\)	\(\to\) asymptote value	origin outside

Question	\(e<1\)	\(e=1\)	\(e>1\)
an auxiliary angle that linearises \(\int r^{2}df\)	eccentric anomaly \(E\)	\(D = \tan\frac f2\)	hyperbolic anomaly \(H\)
the resulting Kepler equation	\(M = E - e\sin E\)	\(M_p = D + \frac{D^{3}}{3}\)	\(M_h = e\sinh H - H\)
is it invertible in closed form?	no (transcendental)	yes (a cubic)	no (transcendental)

relation	derivative	runs level where
\(M\) against \(E\)	\(dM/dE = r/a\)	\(r = a\)
\(f\) against \(E\)	\(df/dE = b/r\)	\(r = b\)
\(f\) against \(M\)	\(df/dM = ab/r^{2}\)	\(r = \sqrt{ab}\)

Energy	Eccentricity	Orbit
\(\varepsilon < 0\)	\(e < 1\)	ellipse (bound)
\(\varepsilon = 0\)	\(e = 1\)	parabola (escape)
\(\varepsilon > 0\)	\(e > 1\)	hyperbola (unbound)

Anomaly	Symbol	Vertex	Behaviour in time
true	\(f\)	focus	the physical angle; non-uniform
eccentric	\(E\)	centre	a computational intermediary
mean	\(M\)	—	linear in \(t\) by construction

\(k\)	\(E_{k}\) [rad]	\(g(E_{k})\)	\(\Delta E_{k}\)
0	\(1.884910\)	\(-2.378\times10^{-1}\)	\(+2.207\times10^{-1}\)
1	\(2.105629\)	\(+5.630\times10^{-3}\)	\(-4.994\times10^{-3}\)
2	\(2.100635\)	\(+2.685\times10^{-6}\)	\(-2.383\times10^{-6}\)
3	\(2.100633\)	\(+6.1\times10^{-13}\)	—

	ellipse	hyperbola
radius	\(r = a(1 - e\cos E)\)	\(r = \lvert a\rvert(e\cosh H - 1)\)
Kepler	\(M = E - e\sin E\)	\(M_h = e\sinh H - H\)
derivative	\(dM/dE = 1 - e\cos E = r/a\)	\(dM_h/dH = e\cosh H - 1 = r/\lvert a\rvert\)
range	\(E \in [0, 2\pi)\), periodic	\(H \in (-\infty,\infty)\), monotone

Two Body Problem

How this deck is organised

Part I — What Every Orbit Shares

Two body Problem

Angular Momentum Conservation

Example 1

What the animation shows

How to find \(R_1\) and \(R_2\)

Towards the restricted problem

Eccentricity Vector

Equation of the Orbit

Reading the orbit equation

Energy Integral (Vis-Viva)

Casimir relations

Counting the integrals

The size parameter \(a\)

Evaluating \(\varepsilon\) directly

Bounded and unbounded motion

Negative specific energy: \(\varepsilon<0\)

Positive specific energy: \(\varepsilon>0\)

Zero specific energy: \(\varepsilon=0\)

Summary

Escape Velocity

Various Conic sections

Things to try

Velocity

The two ingredients

The transverse component

The radial component

The velocity vector

The speed

Flight Path Angle (\(\gamma\))

Where \(\gamma\) is zero, and where it is largest

Picture

Velocity Hodograph

The hodograph in components

Speed and flight path angle across the conics

Limiting cases

From \(r(f)\) to \(r(t)\)

Where the three conics part ways

Part II — The Ellipse (\(0 \le e < 1\))

What is new, and what is not

The ellipse

Why \(a\) is the semi-major axis

Period (\(T\)) of an orbit

Example 2

The orbit, before any integration

Helices, not closed curves

Two bodies in 3D

The three speeds, plotted

What the plot says

\(\gamma\) across a family of ellipses

Apsidal speeds

Worked example — a GTO

The eccentric anomaly \(E\)

Using Kepler’s equation

Relationship between \(E\) \(M\) and \(f\)?

Worked example — propagating a geosynchronous orbit

Flight Path Angle \(\gamma\)

Parabolic Orbit (\(e = 1\))

The one-parameter conic

Speed and direction on a parabola

Time on a parabola — Barker’s equation

Hyperbolic Orbit (\(e > 1\))

Geometry

Speeds \(v_\infty\), \(C_3\)

Turning angle

Aiming radius

What \(b\) is for a hyperbola

The hyperbola in one picture

The hyperbolic anomaly \(H\)

The three hyperbolic functions

The hyperbolic Kepler equation

Part V — The Three Conics Side by Side

One table

What is universal after all

Appendix

The rotating basis \((\hat{\mathbf{r}}, \hat{\boldsymbol{\theta}})\)

Proof 1 — differentiate the components

Proof 2 — the unit circle argument