Visual walkthrough — Specific angular momentum h = √(GMp)

We only need three raw ideas, and we will draw each one:

1. A moving dot near a big mass (position and velocity as arrows).
2. The "sweep rate" of the line from the mass to the dot.
3. A clever change of variable that turns the messy orbit into a clean sine wave.

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Step 1 — Draw the two arrows: $\vec{r}$ and $\vec{v}$

WHAT. Put a heavy mass $M$ at a fixed point $O$ (the focus). A tiny satellite sits at a point $P$. Draw the arrow from $O$ to $P$ — call it $\vec{r}$, the position vector. Draw a second arrow showing which way $P$ is moving right now — call it $\vec{v}$, the velocity vector.

WHY these two. Newton's law for the small mass is $\ddot{\vec{r}}=-\frac{GM}{r^2}\hat{r}$. The little mass $m$ has already cancelled out, so the only things the motion depends on are $\vec{r}$ and $\vec{v}$. If we want a conserved number, it must be built from these two arrows and nothing else.

PICTURE. Look at the figure: the amber arrow $\vec{v}$ is not pointing straight along the white arrow $\vec{r}$ — there is an angle $\varphi$ between them. That leftover sideways part of $\vec{v}$ is what "spinning around $O$" means.

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Step 2 — Build the spin number $\vec{h}=\vec{r}\times \vec{v}$

WHAT. We combine the two arrows with the cross product $\vec{r}\times \vec{v}$. Read this as: "take $\vec{r}$ and $\vec{v}$, and produce a new arrow that sticks straight out of the plane they span, with length equal to the area of the parallelogram they make."

WHY the cross product and not, say, the dot product? The dot product $\vec{r}\cdot \vec{v}$ measures how much $\vec{v}$ points along $\vec{r}$ — i.e. how fast the satellite is getting closer or farther. That is the opposite of what we want. We want the sideways motion, the part that carries the satellite around $O$. The cross product picks out exactly that: its size is $rv\sin \varphi$, and $\sin \varphi$ is largest when $\vec{v}$ is perpendicular (pure sideways) and zero when $\vec{v}$ points straight out (pure radial, no spin).

$$\vec{h}\equiv \underset{\text{where it is}}{\underbrace{\vec{r}}}\times \underset{\text{where it’s headed}}{\underbrace{\vec{v}}},h=|\vec{h}|=\underset{\text{length}}{\underbrace{r}}\underset{\text{speed}}{\underbrace{v}}\underset{\text{sideways fraction}}{\underbrace{\sin \varphi }}$$

PICTURE. The cyan parallelogram is spanned by $\vec{r}$ and $\vec{v}$. Its area is $h$. The purple arrow $\vec{h}$ pops straight out of the page — that is the axis the satellite spins around.

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Step 3 — Show that $\vec{h}$ never changes

WHAT. We check how fast $\vec{h}$ changes in time by differentiating $\vec{h}=\vec{r}\times \vec{v}$.

$$\frac{d\vec{h}}{dt}=\underset{\vec{v}\times \vec{v}=0}{\underbrace{\frac{d\vec{r}}{dt}\times \vec{v}}}+\underset{\vec{r}\times \vec{a}}{\underbrace{\vec{r}\times \frac{d\vec{v}}{dt}}}$$

WHY each piece dies.

• $\frac{d\vec{r}}{dt}=\vec{v}$, and $\vec{v}\times \vec{v}=0$ because any arrow crossed with itself gives zero — the parallelogram of an arrow with itself is a flat line, area $0$.
• $\vec{a}=-\frac{GM}{r^2}\hat{r}$ points straight along $\vec{r}$ (gravity pulls toward $O$). And $\vec{r}\times \hat{r}=0$ for the same reason — parallel arrows make a zero-area parallelogram.

So $\frac{d\vec{h}}{dt}=0$: $\vec{h}$ is frozen forever. This is why the orbit stays in one flat plane and why equal areas sweep in equal times.

PICTURE. Two snapshots of the orbit, early and late. The parallelogram tilts and stretches as the satellite moves, but its area stays identical and the purple $\vec{h}$ arrow never budges.

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Step 4 — Rewrite $h$ in polar coordinates: $h=r^2\dot{\theta }$

WHAT. Describe the satellite by its distance $r$ from $O$ and its angle $\theta$ (the true anomaly) measured from a fixed reference direction. The sideways speed is $r\dot{\theta }$ (distance $\times$ angular rate), and the outward speed is $\dot{r}$.

WHY. We showed $h$ is the sideways part of the motion times $r$. In polar language the sideways velocity is exactly $r\dot{\theta }$, so

$$h=r\cdot (\text{sideways speed})=r\cdot r\dot{\theta }=r^2\dot{\theta }.$$

The dot means "rate of change per second": $\dot{\theta }=d\theta /dt$ is how fast the angle grows.

PICTURE. The velocity arrow $\vec{v}$ is split into a white radial piece $\dot{r}$ (along $\vec{r}$) and an amber tangential piece $r\dot{\theta }$ (perpendicular). Only the amber piece feeds the spin.

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Step 5 — The clever swap: let $u=1/r$

WHAT. We want the orbit's shape $r(\theta )$, not $r(t)$. So we trade time $t$ for angle $\theta$ using $\dot{\theta }=h/r^2$, and we replace $r$ by its reciprocal $u=1/r$.

WHY $u=1/r$? The equation of motion has $1/r^2$ terms (from gravity) and $r$ terms (from the curving). Written in $r$ they fight each other and stay nonlinear. Written in $u=1/r$, magic happens: the gravity term becomes a plain constant, and everything collapses into the simplest equation in physics — a spring equation. The tool is chosen because it linearises the problem.

Step-by-step (each line is one substitution):

$$\dot{r}=\frac{dr}{d\theta }\dot{\theta }=\left(\underset{dr/d\theta ,\text{since }r=1/u}{\underbrace{-\frac{1}{u^2}\frac{du}{d\theta }}}\right)\left(\underset{\dot{\theta }=h/r^2}{\underbrace{h{u}^2}}\right)=-h\frac{du}{d\theta }$$ $$\ddot{r}=\frac{d}{dt}\left(-h\frac{du}{d\theta }\right)=-h\frac{d^{2}u}{d{\theta }^2}\dot{\theta }=-h^2{u}^2\frac{d^{2}u}{d{\theta }^2}$$

PICTURE. Two side-by-side plots of the same orbit: on the left the real ellipse $r(\theta )$ (curvy, awkward); on the right $u(\theta )=1/r$, which traces a clean cosine wave. The swap literally straightens the problem into a wave.

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Step 6 — Binet's equation: the orbit is a spring

WHAT. Plug $\ddot{r}$ and $r{\dot{\theta }}^2$ into the radial equation of motion

$$\ddot{r}-r{\dot{\theta }}^2=-\frac{GM}{r^2}.$$

Using $r{\dot{\theta }}^2=\frac{1}{u}(h{u}^2{)}^2=h^2{u}^3$ and $\ddot{r}=-h^2{u}^2{u}^{\prime \prime }$ (where $u^{\prime \prime }=d^{2}u/d{\theta }^2$):

$$-h^2{u}^2{u}^{\prime \prime }-h^2{u}^3=-GM{u}^2.$$

Divide every term by $-h^2{u}^2$:

$$\boxed{u^{\prime \prime }+u=\frac{GM}{h^2}}$$

WHY this is beautiful. This is Binet's equation, and it is identical in form to a mass on a spring being pushed by a steady force: $x^{\prime \prime }+x=\text{const}$. We already know its solution — a constant offset plus a cosine wave.

$$u(\theta )=\underset{\text{constant offset (the push)}}{\underbrace{\frac{GM}{h^2}}}+\underset{\text{free oscillation}}{\underbrace{A\cos (\theta -{\theta }_0)}}$$

Choose the reference angle so ${\theta }_0=0$ and write $A=\frac{GM}{h^2}e$ (this just names the wave's height in terms of a number $e\ge 0$):

$$u=\frac{GM}{h^2}(1+e\cos \theta ).$$

PICTURE. The straight offset line $GM/h^2$ with a cosine riding on top of it — the sum is $u(\theta )$. The amplitude of the ripple is what we will call eccentricity $e$.

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Step 7 — Flip back to $r$ and read off $p$

WHAT. Turn $u=1/r$ over to recover the orbit:

$$r=\frac{1}{u}=\frac{h^2/GM}{1+e\cos \theta }.$$

Compare, term for term, with the standard conic form from the Two-body problem:

$$r=\frac{p}{1+e\cos \theta }⟹\underset{\text{numerator}}{\underbrace{p}}=\frac{h^2}{GM}.$$

WHY the last squeeze. Two fractions that are equal for every $\theta$ must have equal numerators (the denominators are already identical). So the numerator $h^2/GM$ is the semi-latus rectum $p$. Solve for $h$:

$$p=\frac{h^2}{GM}⟹h^2=GMp⟹\boxed{h=\sqrt{GMp}}.$$

We take the positive square root because $h=r^2\dot{\theta }$ is a magnitude — a length² over time, never negative.

PICTURE. The finished ellipse with $O$ at the focus. The amber segment drawn straight up from the focus (at $\theta =90^{\circ }$, where $\cos \theta =0$) has length exactly $p$ — the "perpendicular peek."

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Step 8 — Every case, checked

WHAT. We sweep through all the shapes the number $e$ can produce, so no reader ever meets an un-drawn scenario.

$e$	shape	what $p$ and $h$ do
$e=0$	circle	$p=a=r$, so $h=\sqrt{GMr}=v_{\text{circ}}r$
$0<e<1$	ellipse	$p=a(1-e^2)$, so $h=\sqrt{GMa(1-e^2)}$
$e=1$	parabola	$a\to \infty$ but $p$ finite; still $h=\sqrt{GMp}$
$e>1$	hyperbola	$p=a(e^2-1)$ (with $a<0$); still $h=\sqrt{GMp}$

Degenerate case $h=0$. If $\varphi =0$ (velocity dead-on radial), then $h=rv\sin 0=0$, so $p=h^2/GM=0$: the "orbit" collapses to a straight up-and-down line through $O$ — no going-around at all. The formula still holds, it just describes a fall.

WHY this matters. The single relation $h=\sqrt{GMp}$ survives every conic. Only the link $p\leftrightarrow a$ changes with the shape; $p$ itself always carries $h$.

PICTURE. All four conics sharing the same focus $O$, each with its own semi-latus rectum drawn as an amber vertical segment — fatter segment, faster spin.

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The one-picture summary

Everything above compressed into one diagram: the two arrows $\to$ the frozen spin $\vec{h}$ $\to$ the wave $u(\theta )$ $\to$ the conic with its half-width $p$ $\to$ the boxed formula.

Recall Feynman retelling of the whole walkthrough

Imagine a stone tied to a heavy magnet by a springy string, whirling around. The string only ever pulls straight toward the magnet — never sideways. Because it never pushes sideways, the stone's "amount of going-around" can never change. We gave that go-around amount a name, $h$, and built it as the area of the little parallelogram made by "where the stone is" and "where it's heading." Then we asked: what shape does the path make? Instead of tracking the stone in time, we watched its distance as the angle turned, and we cleverly tracked $1/\text{distance}$ instead of the distance. That flipped the ugly curve into a smooth wave — a wave whose flat baseline is the pull $GM$ divided by $h^2$, and whose ripple size is the eccentricity. Turning the wave back over gives the orbit equation, and its numerator — the width of the orbit measured straight out from the magnet — is $p$. Line up the two ways of writing the numerator, and out drops the whole story in one breath: $h=\sqrt{GMp}$. Wider peek, stronger pull $\Rightarrow$ more spin.

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Connections

• Two-body problem — supplies the standard conic form we matched against in Step 7.
• Kepler's Second Law — Step 3/4: constant $h$ is exactly equal-areas-in-equal-times.
• Flight-path angle — the angle $\varphi$ introduced in Step 1, used in $h=rv\sin \varphi$.
• Eccentricity vector — the ripple amplitude $e$ born in Step 6.
• Vis-viva equation — pairs speed with radius; complements $h$ to fix the orbit fully.
• Specific orbital energy — the other conserved number; together with $h$ it determines $a$ and $e$.