Visual walkthrough — Conservation of energy and angular momentum in gravitational field

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Step 1 — Two arrows and a dot: setting the stage

WHAT. Put a heavy body (the Sun, mass $M$) at a fixed point. Put a small body (a planet, mass $m$) somewhere out at distance $r$. Draw two arrows from the planet:

• $\vec{r}$ — the position arrow, pointing from the Sun to the planet. Its length is $r$.
• $\vec{v}$ — the velocity arrow, showing which way the planet is sliding right now, and how fast (longer = faster).

WHY. Every single thing we prove is a relationship between these two arrows. Before using any symbol we must be able to see it. $r$ is "how far out", $\vec{v}$ is "where it's headed". That's all we own so far.

PICTURE. Look at the figure. $M$ sits at the origin. The cyan arrow $\vec{r}$ reaches out to the planet; the amber arrow $\vec{v}$ leaves the planet at an angle. The angle between them matters — hold that thought.

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Step 2 — The cross product: measuring "twist"

WHAT. We invent one number that measures how much the velocity is sideways to the position rather than along it. That number is the magnitude of the cross product $\vec{r}\times \vec{v}$.

WHY this tool? We need a quantity that is zero when two arrows are parallel and largest when they're perpendicular — because "orbiting" is exactly the sideways part of the motion. The dot product would do the opposite (max when parallel). So the cross product is the right instrument for "twist".

PICTURE. The figure shows the parallelogram built on $\vec{r}$ and $\vec{v}$. Its area equals $|\vec{r}||\vec{v}|\sin \varphi$, where $\varphi$ is the angle between the arrows. The little curved arrow shows the sense of rotation; by the right-hand rule the twist-vector sticks straight out of the page.

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Step 3 — Why the twist is frozen: torque is zero

WHAT. We show $\vec{L}$ never changes. The thing that would change it is torque $\vec{\tau }=\vec{r}\times \vec{F}$ — the twisting effort of the force.

WHY. From basic mechanics, the rate of change of angular momentum equals torque: $\frac{d\vec{L}}{dt}=\vec{\tau }$. So if $\vec{\tau }=\vec{0}$ always, then $\vec{L}$ can never budge.

PICTURE. In the figure, $\vec{F}$ (amber) lies exactly along $\vec{r}$ (cyan) — they're the same line. The cross product of two arrows on the same line has zero parallelogram area. So the twist of the force is nothing.

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Step 4 — Splitting velocity into "in-out" and "around"

WHAT. Inside the flat plane, chop the velocity into two clean pieces:

• $\dot{r}$ — the radial speed: how fast $r$ is growing or shrinking (in-out motion).
• $r\dot{\theta }$ — the tangential speed: how fast the planet swings around, where $\dot{\theta }$ is the angle's turning rate.

WHY. These two directions are perpendicular, so by Pythagoras the total speed splits with no cross-term. Splitting like this lets us feed the "around" part straight into $L$.

PICTURE. The figure shows the velocity arrow resolved into a cyan radial leg (along $\vec{r}$) and an amber tangential leg (perpendicular to $\vec{r}$). The right angle between them is marked.

Now feed the tangential part into $L$. Because the tangential direction is perpendicular to $\vec{r}$ ($\varphi =90^{\circ }$, $\sin \varphi =1$): $L=mr(r\dot{\theta })=m{r}^2\dot{\theta }\Rightarrow \dot{\theta }=\frac{L}{m{r}^2}.$

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Step 5 — Building the potential well by walking in from infinity

WHAT. Define the potential energy $U(r)$ — the stored "positional" energy — by counting the work gravity does as the planet drifts in from infinitely far away (where we call $U=0$).

WHY this tool? Because gravity is conservative (see Conservative Forces and Potential Energy): the work depends only on start and end radius, never the path. That is precisely the condition that lets a single number $U(r)$ exist. Without it, no energy bookkeeping is possible.

PICTURE. The figure plots $U(r)=-GMm/r$: a curve that is $0$ far out and plunges into a deep pit near the Sun — a potential well. To escape, you must climb out of the pit.

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Step 6 — Freezing the total energy

WHAT. Add moving-energy and stored-energy. The sum never changes.

WHY. The work-energy theorem says the force feeds kinetic energy exactly at the rate it drains potential energy — they trade one-for-one, so their sum has zero rate of change.

PICTURE. The figure shows two "savings accounts", KE (amber bar) and $U$ (cyan bar), at perihelion and aphelion. The bars change height individually, but their combined total line stays flat.

Now substitute the split speed and $\dot{\theta }=L/(m{r}^2)$ from Step 4: $E=\frac{1}{2}m{\dot{r}}^2+\underset{U_{\text{eff}}(r)}{\underbrace{\frac{L^2}{2m{r}^2}-\frac{GMm}{r}}}.$

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Step 7 — Kepler's 2nd law falls out (equal areas)

WHAT. The frozen $L$ instantly gives Kepler's second law: the line Sun→planet sweeps equal areas in equal times.

WHY. A thin triangular sliver swept in time $dt$ has area $dA=\frac{1}{2}r\cdot (rd\theta )$ (base along the arc, height $r$). Divide by $dt$ and use $L=m{r}^2\dot{\theta }$.

PICTURE. The figure shows two shaded slivers — a fat short one near perihelion and a thin long one near aphelion — with equal area, showing the planet must move fast when close and slow when far.

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Step 8 — Edge and degenerate cases (never leave a gap)

WHAT / WHY / PICTURE — all four corner cases in one figure:

• $L=0$ (degenerate). $\sin \varphi =0$: velocity is purely radial. No orbit — the planet plunges straight into the Sun and back out along a line. The centrifugal barrier vanishes, so nothing stops the fall.
• $\dot{r}=0$ (turning point). All speed is tangential, $v=L/(mr)$. These are perihelion (closest, fastest) and aphelion (farthest, slowest).
• $r\to \infty$ with $E=0$. Speed drops to zero exactly at infinity — the boundary between trapped and free: this defines escape, $v_{\text{esc}}=\sqrt{2GM/r}$ (see Escape Velocity and Hohmann Transfers).

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Step 9 — Two boxes become vis-viva

WHAT. Evaluate $E$ at both turning points and the $e$'s and $L$'s cancel, leaving energy that depends on $a$ only. Then eliminate $E$ to get the master speed formula.

WHY. At perihelion $r_p=a(1-e)$ and aphelion $r_a=a(1+e)$ with $r_p+r_a=2a$. Writing $E$ at both and adding collapses eccentricity $e$.

PICTURE. The figure marks $r_p$, $r_a$, and $a$ on one ellipse and shows the two energy boxes merging into the single line $E=-GMm/2a$.

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The one-picture summary

This single frame stitches every step: the flat plane (Step 3), the split velocity (Step 4), the potential well and effective barrier (Steps 5–6), the equal-area sweep (Step 7), and the shape dial set by the sign of $E$ (Step 8), all leading to the two boxed master results (Step 9).

Recall Feynman retelling — the whole walkthrough in plain words

Picture a planet as a bead on a rubber sheet with a heavy ball (the Sun) in the middle. First we noticed the pull always points straight at the ball, so it can't give the bead any twist — that "twist number" (angular momentum) is locked forever, which flattens the whole dance into one sheet and means the bead sweeps equal fan-shaped areas in equal time (fast when close, slow when far). Then we split the bead's motion into "in-and-out" and "around", and noticed the "around" part is really just the frozen twist in disguise, acting like an invisible wall that keeps the bead from falling all the way in. Next we counted the energy: moving-energy plus well-depth-energy, and found their sum never changes — one account fills as the other empties. Finally, evaluating that energy at the closest and farthest points made the orbit's wobbliness cancel out, leaving a beautifully simple truth: the total energy depends only on the orbit's width, not its squish. Turn one dial — the sign of the energy — and you slide smoothly from a tidy circle, to an oval, to a just-barely-free parabola, to a runaway hyperbola.

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