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

Before we start, three plain-language reminders so no symbol is unearned:

We will always work in km and seconds, so $GM$ in $km^{3} / s^{2}$ keeps every answer in km/s. Mixing metres and km is the #1 silent error — we avoid it by picking km once and never switching.

The scenario matrix

Every vis-viva problem lives in one of these cells. The examples below are tagged with the cell they cover.

Cell	Orbit type	Sign / value of $a$	What's special	Example
C1	Circular	$a = r$ (finite $> 0$ )	$\frac{2}{r} - \frac{1}{a} = \frac{1}{r}$ ; speed constant	Ex 1
C2	Elliptical, at perigee	$a > 0$ , $r = r_{p} < a$	fastest point	Ex 2
C3	Elliptical, at apogee	$a > 0$ , $r = r_{a} > a$	slowest point	Ex 2
C4	Elliptical, general point	$a > 0$ , $r_{p} < r < r_{a}$	in-between speed	Ex 3
C5	Parabolic (escape boundary)	$a \to \infty$ , $\frac{1}{a} \to 0$	$ε = 0$ exactly	Ex 4
C6	Hyperbolic (unbound)	$a < 0$	$v^{2} > 2 GM / r$ , excess speed	Ex 5
C7	Solve backwards for $a$	given $v, r$ → find orbit size	classifies the orbit	Ex 6
C8	Real-world word problem	mission $Δ v$	Hohmann transfer	Ex 7
C9	Exam twist / degenerate	$r \to r$ at $a \to r$ limit, sign trap	catch the mistake	Ex 8

Worked Examples

Worked example Ex 1 — Circular orbit (Cell C1)

A GPS satellite circles Earth at radius $r = 26560$ km. How fast is it going?

Forecast: Guess first — will it be faster or slower than the $\sim 7.7$ km/s of a low satellite? (Higher orbit → slower.)

Step 1. Recognise circular ⇒ $a = r$ . Why this step? A circle is an orbit whose "longest width" is its diameter, so half of that, $a$ , equals the radius $r$ . This is the only case where $r$ and $a$ are the same number.

Step 2. Substitute $a = r$ into vis-viva: $v^{2} = GM (\frac{2}{r} - \frac{1}{r}) = \frac{GM}{r} .$ Why this step? The two fractions collapse because they share the same denominator — that's the algebraic fingerprint of "circular means constant speed."

Step 3. Plug numbers: $v = \frac{398600}{26560} = 15.008 \approx 3.874 km/s .$

Verify: Units: $km^{3} / s^{2} \div km = km^{2} / s^{2} = km/s$ ✓. And $3.87 < 7.7$ km/s — slower than LEO, exactly as a bigger orbit should be. ✓

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

Step 1. Find $a$ from the apsides. Why this step? Vis-viva always needs $a$ , and the two apsides sit at opposite ends of the long axis, so their distances add to the full length $2 a$ : $a = \frac{r _{p} + r _{a}}{2} = \frac{7000 + 42000}{2} = 24500 km .$

Step 2. Speed at perigee, $r = r_{p} = 7000$ km: $v_{p}^{2} = 398600 (\frac{2}{7000} - \frac{1}{24500}) = 398600 (2.8571 \times 1 0^{- 4} - 4.0816 \times 1 0^{- 5}) .$ Why this step? We just drop $r = r_{p}$ and the fixed $a$ into the formula — nothing else changes. $v_{p}^{2} = 398600 \times 2.4490 \times 1 0^{- 4} = 97.62 \Rightarrow v_{p} \approx 9.88 km/s .$

Step 3. Speed at apogee, $r = r_{a} = 42000$ km, same $a$ : $v_{a}^{2} = 398600 (\frac{2}{42000} - \frac{1}{24500}) = 398600 (4.7619 \times 1 0^{- 5} - 4.0816 \times 1 0^{- 5}) .$ $v_{a}^{2} = 398600 \times 6.803 \times 1 0^{- 6} = 2.712 \Rightarrow v_{a} \approx 1.647 km/s .$

Verify: Angular momentum must match: $r_{p} v_{p} = 7000 \times 9.88 = 6.92 \times 1 0^{4}$ and $r_{a} v_{a} = 42000 \times 1.647 = 6.92 \times 1 0^{4}$ . Equal ✓ (this is Conservation of Angular Momentum in action). And $v_{p} > v_{a}$ as forecast. ✓

Worked example Ex 4 — Parabolic escape, the

$a \to \infty$ limit (Cell C5) A probe sits at $r = 7000$ km. What speed does it need to just barely escape Earth (reach infinity with zero speed left)?

Forecast: Escape speed is famously $2$ times circular speed — so guess about $1.41 \times$ the circular speed at this radius.

Step 1. "Just barely escape" means the orbit is parabolic, i.e. $a \to \infty$ . Why this step? An escaping-but-not-overshooting path has total specific energy $ε = 0$ (the exact boundary between bound and free). Since $ε = - GM / (2 a)$ , setting $ε = 0$ forces $a \to \infty$ , so $\frac{1}{a} \to 0$ .

Step 2. Drop the $\frac{1}{a}$ term: $v_{esc}^{2} = GM (\frac{2}{r} - 0) = \frac{2 GM}{r} = \frac{2 \times 398600}{7000} = 113.89.$ $v_{esc} \approx 10.67 km/s .$

Verify: Circular speed here is $398600/7000 = 56.94 = 7.546$ km/s, and $2 \times 7.546 = 10.67$ km/s ✓ — matches Escape Velocity exactly.

Worked example Ex 5 — Hyperbolic orbit, the

$a < 0$ trap (Cell C6) An interstellar object flies past the Sun on a hyperbolic orbit with $a = - 50 \times 1 0^{6}$ km (negative — it is unbound). At $r = 150 \times 1 0^{6}$ km (1 AU), how fast is it moving? Use $G M_{⊙} = 1.327 \times 1 0^{11} km^{3} / s^{2}$ .

Forecast: For a hyperbola $v^{2} > 2 GM / r$ (escape speed and then some). So expect a speed above the local escape speed.

Step 1. Keep $a$ negative — do not flip its sign. Why this step? The convention " $a < 0$ for hyperbolas" is exactly what makes $- \frac{1}{a}$ turn into $+ \frac{1}{∣ a ∣}$ , adding energy so the speed exceeds escape. If you "fix" the sign you destroy the physics.

Step 2. Substitute with $\frac{1}{a} = \frac{1}{- 50 \times 1 0 ^{6}} = - 2.0 \times 1 0^{- 8}$ : $v^{2} = 1.327 \times 1 0^{11} (\frac{2}{150 \times 1 0 ^{6}} - (- 2.0 \times 1 0^{- 8})) .$ $= 1.327 \times 1 0^{11} (1.3333 \times 1 0^{- 8} + 2.0 \times 1 0^{- 8}) = 1.327 \times 1 0^{11} \times 3.3333 \times 1 0^{- 8} .$ $v^{2} = 4423.3 \Rightarrow v \approx 66.51 km/s .$

Verify: Local escape speed is $2 GM / r = 2 \times 1.327 \times 1 0^{11} / (150 \times 1 0^{6}) = 1769.3 = 42.06$ km/s. Our $66.5 > 42.06$ ✓ — genuinely hyperbolic, as forecast.

Worked example Ex 6 — Solve backwards for

$a$ (Cell C7) A spacecraft at $r = 8000$ km is measured moving at $v = 9.0$ km/s. What is the semi-major axis $a$ , and is the orbit bound or unbound?

Forecast: $9.0$ km/s — is that above or below escape at 8000 km? Escape is $2 \times 398600/8000 = 9.98$ km/s. We're below it, so the orbit should be bound (positive $a$ ).

Step 1. Rearrange vis-viva for $\frac{1}{a}$ : $\frac{1}{a} = \frac{2}{r} - \frac{v ^{2}}{GM} .$ Why this step? We know $v$ and $r$ ; the only unknown is $a$ , so isolate it. This is the "diagnose the orbit" move.

Step 2. Plug in: $\frac{1}{a} = \frac{2}{8000} - \frac{9. 0 ^{2}}{398600} = 2.5000 \times 1 0^{- 4} - 2.0321 \times 1 0^{- 4} = 4.679 \times 1 0^{- 5} .$ $a = \frac{1}{4.679 \times 1 0 ^{- 5}} \approx 21372 km .$

Step 3. Classify: $a > 0$ and finite ⇒ bound ellipse.

Verify: Recompute $v$ from this $a$ : $v^{2} = 398600 (2/8000 - 1/21372) = 398600 (2.5 \times 1 0^{- 4} - 4.679 \times 1 0^{- 5}) = 398600 \times 2.032 \times 1 0^{- 4} = 81.00$ , so $v = 9.00$ km/s ✓ round-trips. And $a > 0$ matches the bound forecast. See Orbital Elements.

Worked example Ex 7 — Real-world word problem: Hohmann

$Δ v$ (Cell C8) A satellite is in a circular orbit at $r_{1} = 7000$ km. We want to raise it to a circular orbit at $r_{2} = 42000$ km using a Hohmann Transfer Orbit. Find the first burn $Δ v_{1}$ — the speed boost needed to leave the low circle and enter the transfer ellipse.

Forecast: The transfer ellipse's perigee sits on the low circle, and its perigee speed (from Ex 2, $9.88$ km/s) is faster than the low circular speed. So $Δ v_{1}$ is a modest positive number.

Step 1. Low circular speed (Cell C1 machinery) at $r_{1} = 7000$ : $v_{c 1} = \frac{398600}{7000} = 56.943 = 7.546 km/s .$ Why this step? Before the burn the satellite moves at circular speed; that's our starting point.

Step 2. Transfer-ellipse perigee speed. The transfer ellipse has $r_{p} = 7000$ , $r_{a} = 42000$ ⇒ $a = 24500$ km — identical to Ex 2. So its perigee speed is $v_{p} = 9.883$ km/s (from Ex 2). Why this step? The burn happens at perigee of the transfer ellipse, which coincides with the low circle, so we need that ellipse's perigee speed.

Step 3. The burn just makes up the difference: $Δ v_{1} = v_{p} - v_{c 1} = 9.883 - 7.546 = 2.337 km/s .$

Verify: $Δ v_{1} > 0$ (a speed-up to climb higher) ✓, and it's a realistic few-km/s figure for such a transfer ✓. Units km/s ✓.

Worked example Ex 8 — Exam twist: the degenerate / sign trap (Cell C9)

"A student computes the speed of a satellite at $r = 30000$ km on an orbit with $a = 24500$ km and gets $v^{2} = 398600 (2/30000 - 1/24500)$ . They panic because the two fractions are close. Then a second student, given a hyperbolic orbit with $a = 24500$ km (they forgot the sign), computes the same way. What's wrong, and what's the correct speed in the first (legitimate) case?"

Forecast: The first calculation is fine (bound ellipse, $r < r_{a} = ?$ ). The second is the classic sign mistake. Guess: legit answer is a couple of km/s.

Step 1. First student — check the orbit is real. Here $a = 24500$ is the Ex 2 orbit, whose apogee is $r_{a} = 2 a - r_{p} = 2 (24500) - 7000 = 42000$ km. Since $7000 \leq 30000 \leq 42000$ , the point $r = 30000$ actually exists on the orbit. Good. Why this step? Vis-viva silently returns a value even for an $r$ the orbit never reaches; you must confirm $r_{p} \leq r \leq r_{a}$ first, or the "speed" is fictitious.

Step 2. Compute: $v^{2} = 398600 (\frac{2}{30000} - \frac{1}{24500}) = 398600 (6.6667 \times 1 0^{- 5} - 4.0816 \times 1 0^{- 5}) = 398600 \times 2.585 \times 1 0^{- 5} .$ $v^{2} = 10.305 \Rightarrow v \approx 3.210 km/s .$

Step 3. Second student's error: for a hyperbola $a$ must be negative. Using $a = + 24500$ describes a bound ellipse, not the unbound path they meant — the sign is not cosmetic, it flips $- \frac{1}{a}$ from adding energy to removing it. The fix is $a = - 24500$ , which would give a much larger $v^{2} = 398600 (2/30000 + 1/24500)$ .

Verify: First case: $3.21$ km/s sits between perigee $9.88$ and apogee $1.65$ ? No — wait, $3.21 < 1.65$ is false; $1.65 < 3.21 < 9.88$ ✓ (it's between, closer to apogee since $30000$ is near the far side). Consistent ✓.

Recall Which cell am I in? (quick reflex)

Given an orbit, first name its type ::: circle→ $a = r$ ; ellipse→ $0 < a$ finite; parabola→ $a \to \infty$ ; hyperbola→ $a < 0$ . Given $v$ and $r$ , how do I find the type ::: compute $1/ a = 2/ r - v^{2} / GM$ ; sign of $a$ tells you (+ bound, ∞ parabolic, − hyperbolic). Before trusting a computed $v$ on an ellipse, what must I check ::: that $r_{p} \leq r \leq r_{a}$ , else the point isn't on the orbit.

Connections

Vis-viva equation — derivation — the formula every example uses.
Conservation of Energy — why $a$ alone sets the total energy in each cell.
Conservation of Angular Momentum — the perigee/apogee cross-check in Ex 2.
Kepler's Laws — third law links $a$ to the period once you know $a$ .
Hohmann Transfer Orbit — Ex 7 is one burn of it.
Escape Velocity — the parabolic Cell C5 limit of Ex 4.
Orbital Elements — $a$ , $e$ , $r_{p}$ , $r_{a}$ definitions used throughout.