Vis-viva equation v² = GM(2 - r − 1 - a) — derivation: Visual walkthrough

Step 1 — The picture: one small body, one big body

Before any formula, look at the setup.

Figure — Vis-viva equation v² = GM(2 - r − 1 - a) — derivation

The big orange dot is the central body, mass $M$ (think Sun or Earth). We treat it as fixed.
The small magenta dot is the orbiting body, mass $m$ (a satellite or planet), so small it doesn't tug the big one around.
The purple arrow labelled $r$ points from the centre of $M$ to $m$ . Its length $r$ is your current distance.
The magenta arrow labelled $v$ is the velocity — which way, and how fast, $m$ is moving right now. Its length $v$ is the speed.

WHY start here? Every symbol below is either $r$ , $v$ , or built from this picture. If we can measure a distance and a speed, we have everything.

Step 2 — The shape of the loop: an ellipse and its size $a$

The orbit is not a circle in general — it's a squashed circle called an ellipse. We need names for its parts.

The central body sits at one focus (the orange dot), not the middle. This is Kepler's first law (see Kepler's Laws).
The point where $m$ is closest is the periapsis, at distance $r_{p}$ (magenta).
The point farthest away is the apoapsis, at distance $r_{a}$ (violet).
The long axis of the ellipse (periapsis straight through to apoapsis) has length $r_{p} + r_{a}$ . Half of it is the ==semi-major axis $a$ ==.

WHY define $a$ this way? Because — as we will prove in Step 7 — the total energy of the orbit turns out to depend on this one number and nothing else. $a$ is the single knob that sets how much "oomph" the whole orbit has.

Step 3 — The energy balance: kinetic plus potential is constant

Now the first of two conservation laws. See Conservation of Energy.

Gravity is a conservative force: it never secretly adds or drains energy. So the sum below is the same number at every point of the orbit.

In the figure the two coloured bars swap height as $m$ moves: near periapsis the kinetic bar is tall (fast), the potential bar is deep (deep in the well); near apoapsis they swap. Their total (the dashed line) never moves. That flat dashed line is $ε$ .

WHY this tool? Energy conservation gives us one equation that is true everywhere at once. That's exactly what we need to link a point far out to a point close in.

Step 4 — The clever trick: read energy at the two ends

Energy is the same everywhere, so we're free to evaluate it wherever the algebra is easiest. The two easiest spots are periapsis and apoapsis.

WHY the ends? At periapsis and apoapsis the velocity arrow is exactly perpendicular to $r$ — the planet is momentarily neither climbing nor falling, just sweeping sideways. So the speed is purely the sideways speed, with no confusing radial part. Everywhere else the velocity is slanted and messier.

Because $ε$ is equal at both ends:

One equation, two unknown speeds. We need one more relation — that's the next step.

Step 5 — The second balance: angular momentum ties the speeds

See Conservation of Angular Momentum. Angular momentum measures "spinning-around-ness": how much sideways sweep the body carries. Gravity pulls straight toward the centre, so it can't change this spin — it's conserved too.

At the two ends, where $v ⊥ r$ , the full speed is the sideways speed, so:

WHY do we need this? Equation (1) had two unknowns. Equation (2) expresses $v_{a}$ using $v_{p}$ . Two equations, now really only one unknown ( $v_{p}$ ). We can solve.

Step 6 — The algebra collapses to a clean speed

Substitute (2) into (1) — replace $v_{a}$ by $v_{p} (r_{p} / r_{a})$ — and watch it simplify.

$\frac{1}{2} v_{p}^{2} - \frac{μ}{r _{p}} = \frac{1}{2} v_{p}^{2} \frac{r _{p}^{2}}{r _{a}^{2}} - \frac{μ}{r _{a}}$

Gather the $v_{p}^{2}$ pieces on the left, the $μ$ pieces on the right:

$\frac{1}{2} v_{p}^{2} difference of two squares (1 - \frac{r _{p}^{2}}{r _{a}^{2}}) = μ (\frac{1}{r _{p}} - \frac{1}{r _{a}})$

The bracket $1 - r_{p}^{2} / r_{a}^{2}$ factors as $\frac{( r _{a} - r _{p} ) ( r _{a} + r _{p} )}{r _{a}^{2}}$ , and the right side over a common denominator is $\frac{r _{a} - r _{p}}{r _{p} r _{a}}$ . The factor $(r_{a} - r_{p})$ appears on both sides and cancels:

The figure highlights the twin $(r_{a} - r_{p})$ brackets glowing on each side, with an arrow showing them cancel — that cancellation is the whole magic that makes it collapse.

WHY does this feel like a miracle? The difference-of-squares was hiding a common factor with the right-hand side. Nature arranged it so a messy quadratic reduces to one tidy fraction.

Step 7 — The big reveal: energy depends ONLY on $a$

Now use Step 2's fact: $r_{a} + r_{p} = 2 a$ . Substitute it into the $v_{p}^{2}$ result:

$v_{p}^{2} = \frac{2 μ r _{a}}{r _{p} \cdot 2 a} = \frac{μ r _{a}}{a r _{p}}$

Put this back into the energy expression at periapsis and use $r_{a} = 2 a - r_{p}$ :

$ε = \frac{1}{2} v_{p}^{2} - \frac{μ}{r _{p}} = \frac{μ r _{a}}{2 a r _{p}} - \frac{μ}{r _{p}} = \frac{μ}{r _{p}} (\frac{2 a - r _{p}}{2 a} - 1) = \frac{μ}{r _{p}} (- \frac{r _{p}}{2 a})$

The $r_{p}$ cancels and we land on:

The figure shows three different-shaped ellipses that share the same $a$ ; a bar on the side shows they all sit at the same energy level. Different eccentricities, identical energy.

Step 8 — Read off vis-viva at any point

Energy is constant, so at any point (distance $r$ , speed $v$ ) the balance equals the value we just found:

$\frac{1}{2} v^{2} - \frac{μ}{r} = - \frac{μ}{2 a}$

Multiply through by 2 and rearrange:

Step 9 — All the cases in one picture

The single formula secretly covers four different kinds of path depending on $a$ . We must check them all.

Circle ( $r = a$ ): $v^{2} = μ (\frac{2}{r} - \frac{1}{r}) = \frac{μ}{r}$ . Constant speed — the loop never dips or climbs. (Magenta path.)
Ellipse ( $a$ finite, $r$ varies): speed rises near periapsis, falls near apoapsis. (Violet path.)
Parabola / escape ( $a \to \infty$ , so $\frac{1}{a} \to 0$ ): $v^{2} = \frac{2 μ}{r}$ . This is the boundary $ε = 0$ — just barely free. See Escape Velocity. (Orange path.)
Hyperbola ( $a < 0$ , so $- \frac{1}{a} > 0$ ): $v^{2} > \frac{2 μ}{r}$ . Extra "leftover" speed even at infinity — a flyby that never returns. (Navy dashed path.)

The one-picture summary

Two conservation laws (energy + angular momentum), evaluated at the two easy ends of the orbit, collapse by a difference-of-squares cancellation into $ε = - μ /2 a$ — energy set by size alone — which unrolls into vis-viva.

Recall Feynman: the whole walk in plain words

Picture a ball whirling around you on a stretchy string in an oval loop. Two things about it never change no matter where it swings: its total energy (motion-energy plus how-high-up energy) and its spinning-around-ness. I look at the two extreme moments — closest and farthest — because there the ball moves purely sideways, so its speed is clean and simple. Writing "energy at closest = energy at farthest" and "spin at closest = spin at farthest" gives me two facts. I mash them together, a big pile of terms cancels, and out pops a shockingly simple truth: the total energy depends only on how big the oval is — half its longest width, which we call $a$ . Not its shape, not where the ball is right now — just its size. Once I know the energy is fixed by $a$ , I can stand anywhere on the loop, plug in my current distance $r$ , and immediately know my speed. That recipe is the vis-viva equation.

Connections

Conservation of Energy — the flat dashed line in Step 3.
Conservation of Angular Momentum — the equal-sweep rule in Step 5.
Kepler's Laws — Sun sits at a focus (Step 2).
Escape Velocity — the $a \to \infty$ case in Step 9.
Hohmann Transfer Orbit — applies this twice to change orbits.
Orbital Elements — formal definitions of $a$ , $e$ , $r_{p}$ , $r_{a}$ .

Recall Quick self-check

Why can we evaluate energy just at periapsis and apoapsis? ::: Energy is the same everywhere, and at the apsides $v ⊥ r$ , so the speed is purely sideways — the algebra is cleanest there. What single quantity sets the total orbital energy? ::: The semi-major axis $a$ , via $ε = - μ /2 a$ . What makes the hyperbola faster than escape in the formula? ::: A negative semi-major axis, $a < 0$ , which makes $- 1/ a > 0$ and pushes $v^{2}$ above $2 μ / r$ . Define the specific angular momentum $h$ . ::: $h = ∣ r \times v ∣$ — distance times the sideways part of the velocity; constant because gravity applies no twist.