Specific orbital energy ε = −GM - 2a: Worked examples

This page is the drill hall for the parent topic. We take the one master formula and the Vis-viva equation, and we hunt down every kind of situation they can throw at you — every sign of $ε$ , the degenerate circle, the escape edge, the unbound hyperbola, a real mission problem, and an exam-style trap. Nothing is skipped, so you never meet a case you haven't already practised.

Everything below rests on just two tools you already own:

The scenario matrix

Before working anything, let's lay out every cell this topic can hit. Each later example is tagged with the cell(s) it fills.

#	Case class	What is special about it	Filled by
C1	$ε < 0$ , circle ( $r_{p} = r_{a}$ )	degenerate ellipse, $r = a$	Ex 1
C2	$ε < 0$ , general ellipse	$r_{p} \neq = r_{a}$ , speed differs by point	Ex 2, Ex 3
C3	$ε = 0$ , parabola	escape edge, $a \to \infty$	Ex 4
C4	$ε > 0$ , hyperbola	unbound, negative $a$	Ex 5
C5	Limiting behaviour	$a \to \infty$ , $r \to \infty$ , sign flips	Ex 4, Ex 5
C6	Real-world word problem	a Hohmann transfer orbit burn	Ex 6
C7	Exam-style twist	given $v, r$ → classify the orbit	Ex 7
C8	Inverse / consistency check	given $ε$ → recover $a$ , cross-check with Kepler's Third Law	Ex 8

The three signs of $ε$ (rows C1–C4) are the backbone. Here is the whole matrix as one picture — keep it open while you read.

Figure — Specific orbital energy ε = −GM - 2a

Example 1 — Circular LEO (cell C1)

Worked example Circular orbit at 400 km altitude

A satellite circles Earth at radius $r = a = 6.78 \times 1 0^{6}$ m. Find $ε$ and $v$ .

Forecast: guess before computing — will $ε$ be negative? Will $v$ be near $7.7$ km/s? Jot your guess.

Step 1. Recognise circle ⇒ $r_{p} = r_{a} = r$ , so $2 a = r_{p} + r_{a} = 2 r$ , giving $a = r$ . Why this step? A circle is the degenerate ellipse where the two extreme distances coincide; only then does "radius = $a$ " hold.

Step 2. Energy: $ε = - \frac{GM}{2 a} = - \frac{3.986 \times 1 0 ^{14}}{2 ( 6.78 \times 1 0 ^{6} )} = - 2.94 \times 1 0^{7} J/kg$ . Why this step? Straight substitution — $a$ is the only geometry the formula needs.

Step 3. Speed from vis-viva with $r = a$ : $v^{2} = GM (\frac{2}{r} - \frac{1}{r}) = \frac{GM}{r}$ . Why this step? The two terms partly cancel, leaving the famous $v = GM / r$ for circles only.

Step 4. $v = 3.986 \times 1 0^{14} /6.78 \times 1 0^{6} = 7.67 \times 1 0^{3}$ m/s.

Verify: units $(m^{3} / s^{2}) / m = m^{2} / s^{2} = m/s$ ✔. And $7.67$ km/s is textbook LEO speed ✔. Also $ε = \frac{v ^{2}}{2} - \frac{GM}{r} = 2.94 \times 1 0^{7} - 5.88 \times 1 0^{7} = - 2.94 \times 1 0^{7}$ ✔ matches Step 2.

Example 2 — General ellipse: energy from the extremes (cell C2)

Worked example GEO transfer orbit energy

An orbit has perigee $r_{p} = 6.78 \times 1 0^{6}$ m and apogee $r_{a} = 4.22 \times 1 0^{7}$ m. Find $ε$ and $a$ .

Forecast: will $ε$ here be more or less negative than the LEO circle of Ex 1? (Bigger loop ⇒ closer to zero — guess the sign of the change.)

Step 1. Use $ε = - \frac{GM}{r _{p} + r _{a}}$ (the Step-5 shortcut from the parent, since $r_{p} + r_{a} = 2 a$ ). Why this step? When you are handed both extremes, this skips computing $a$ first.

Step 2. $r_{p} + r_{a} = 4.898 \times 1 0^{7}$ m, so $ε = - \frac{3.986 \times 1 0 ^{14}}{4.898 \times 1 0 ^{7}} = - 8.14 \times 1 0^{6} J/kg$ .

Step 3. Recover $a = \frac{r _{p} + r _{a}}{2} = 2.449 \times 1 0^{7}$ m. Why this step? Confirms the semi-major axis is the average of the extremes, not either one.

Verify: $- \frac{GM}{2 a} = - \frac{3.986 \times 1 0 ^{14}}{2 ( 2.449 \times 1 0 ^{7} )} = - 8.14 \times 1 0^{6}$ ✔ same as Step 2. And $- 8.14 \times 1 0^{6}$ is less negative than Ex 1's $- 2.94 \times 1 0^{7}$ — bigger loop, richer account, as forecast ✔.

Example 3 — Speed at apogee vs perigee of that ellipse (cell C2)

Worked example Where is the satellite fastest?

Same orbit as Ex 2 ( $a = 2.449 \times 1 0^{7}$ m). Find speed at perigee $v_{p}$ and at apogee $v_{a}$ .

Forecast: which is larger, $v_{p}$ or $v_{a}$ ? By the energy bank: close = fast. Guess the ratio too.

Step 1. Perigee: $v_{p}^{2} = GM (\frac{2}{r _{p}} - \frac{1}{a}) = 3.986 \times 1 0^{14} (\frac{2}{6.78 \times 1 0 ^{6}} - \frac{1}{2.449 \times 1 0 ^{7}})$ . Why this step? Vis-viva is the only tool tying speed to a single point $r$ .

Step 2. $\frac{2}{r _{p}} = 2.950 \times 1 0^{- 7}$ , $\frac{1}{a} = 4.083 \times 1 0^{- 8}$ ; difference $= 2.542 \times 1 0^{- 7}$ . So $v_{p}^{2} = 1.013 \times 1 0^{8}$ , $v_{p} = 1.007 \times 1 0^{4}$ m/s $\approx 10.1$ km/s.

Step 3. Apogee: $v_{a}^{2} = GM (\frac{2}{r _{a}} - \frac{1}{a}) = 3.986 \times 1 0^{14} (4.739 \times 1 0^{- 8} - 4.083 \times 1 0^{- 8}) = 2.614 \times 1 0^{6}$ , so $v_{a} = 1.617 \times 1 0^{3}$ m/s $\approx 1.6$ km/s. Why this step? Same formula, new $r$ — apogee is far, so the $2/ r$ term shrinks, dragging speed down.

Verify (angular momentum, Conservation of angular momentum): at both ends $v ⊥ r$ , so $r_{p} v_{p}$ should equal $r_{a} v_{a}$ . $r_{p} v_{p} = 6.78 \times 1 0^{6} \times 1.007 \times 1 0^{4} = 6.83 \times 1 0^{10}$ ; $r_{a} v_{a} = 4.22 \times 1 0^{7} \times 1.617 \times 1 0^{3} = 6.82 \times 1 0^{10}$ ✔ (equal to rounding). Perigee is ~6× faster, as forecast.

Example 4 — The escape edge: parabola (cells C3, C5)

Worked example Just barely leaving Earth

From $r = 6.78 \times 1 0^{6}$ m, what speed gives $ε = 0$ (a parabolic escape)? What is $a$ ?

Forecast: is escape speed bigger or smaller than the circular speed from Ex 1? By how much (guess a factor)?

Step 1. Set $ε = 0$ in $ε = \frac{v ^{2}}{2} - \frac{GM}{r}$ : $v_{esc} = \frac{2 GM}{r}$ . Why this step? Zero total energy is the boundary between bound and unbound — exactly the Escape velocity condition.

Step 2. $v_{esc} = \frac{2 ( 3.986 \times 1 0 ^{14} )}{6.78 \times 1 0 ^{6}} = 1.085 \times 1 0^{4}$ m/s $\approx 10.8$ km/s.

Step 3. Semi-major axis: $a = - \frac{GM}{2 ε} \to - \frac{GM}{0} = \infty$ . Why this step? $ε \to 0^{-}$ pushes $a \to + \infty$ : the ellipse stretches until it never closes — a parabola. This is the limiting behaviour (cell C5).

Verify: $v_{esc} / v_{circ} = 1.085 \times 1 0^{4} /7.67 \times 1 0^{3} = 1.414 \approx 2$ ✔ — the classic result: escape speed is $2$ times circular speed at the same radius, as forecast.

Example 5 — Unbound hyperbola: negative $a$ (cells C4, C5)

Worked example A probe hurled outward with energy to spare

At $r = 6.78 \times 1 0^{6}$ m a probe moves at $v = 1.20 \times 1 0^{4}$ m/s. Show it escapes, find $ε$ and $a$ , and its speed far away ( $r \to \infty$ ).

Forecast: since $v > v_{esc}$ (from Ex 4), predict the sign of $ε$ and the sign of $a$ before computing.

Step 1. $ε = \frac{v ^{2}}{2} - \frac{GM}{r} = \frac{( 1.20 \times 1 0 ^{4} ) ^{2}}{2} - \frac{3.986 \times 1 0 ^{14}}{6.78 \times 1 0 ^{6}}$ . Why this step? Only the $v, r$ form of $ε$ works when we're given a speed at a point, not the geometry.

Step 2. $= 7.20 \times 1 0^{7} - 5.879 \times 1 0^{7} = 1.321 \times 1 0^{7} J/kg > 0$ . Why this step? Positive $ε$ ⇒ unbound (hyperbola) — the probe leaves forever.

Step 3. $a = - \frac{GM}{2 ε} = - \frac{3.986 \times 1 0 ^{14}}{2 ( 1.321 \times 1 0 ^{7} )} = - 1.508 \times 1 0^{7}$ m — negative, the formal signature of a hyperbola.

Step 4. Speed at infinity: as $r \to \infty$ , $v^{2} = GM (\frac{2}{r} - \frac{1}{a}) \to - \frac{GM}{a} = 2 ε$ , so $v_{\infty} = 2 ε = 2 (1.321 \times 1 0^{7}) = 5.14 \times 1 0^{3}$ m/s. Why this step? At infinity PE $\to 0$ , so all that survives is the leftover kinetic energy $ε$ — this is the limiting behaviour (cell C5).

Verify: $\frac{1}{2} v_{\infty}^{2} = 0.5 (5.14 \times 1 0^{3})^{2} = 1.321 \times 1 0^{7} = ε$ ✔ (all energy is kinetic at infinity). Signs matched the forecast: $ε > 0$ , $a < 0$ ✔.

Example 6 — Word problem: a Hohmann transfer burn (cell C6)

Worked example Raising from LEO to GEO — the perigee kick

A Hohmann transfer orbit starts in a circular LEO of radius $r_{1} = 6.78 \times 1 0^{6}$ m and must reach a transfer ellipse whose apogee touches GEO at $r_{2} = 4.22 \times 1 0^{7}$ m. How much speed change $Δ v$ is needed at the perigee burn?

Forecast: the burn must raise apogee, so $Δ v > 0$ . Guess whether it's closer to $1$ , $2$ , or $3$ km/s.

Step 1. Circular LEO speed: $v_{circ} = GM / r_{1} = 7.67 \times 1 0^{3}$ m/s (from Ex 1). Why this step? This is the speed before the burn.

Step 2. Transfer ellipse has $a = (r_{1} + r_{2}) /2 = 2.449 \times 1 0^{7}$ m (same orbit as Ex 2). Perigee speed on it is $v_{p} = 1.007 \times 1 0^{4}$ m/s (from Ex 3). Why this step? The burn changes speed at the same point $r_{1}$ , moving the satellite from the circle onto the ellipse — same $r$ , different $a$ .

Step 3. $Δ v = v_{p} - v_{circ} = 1.007 \times 1 0^{4} - 7.67 \times 1 0^{3} = 2.40 \times 1 0^{3}$ m/s $\approx 2.4$ km/s.

Verify: direction check — we increased speed at perigee, so energy rose from $- 2.94 \times 1 0^{7}$ (Ex 1) to $- 8.14 \times 1 0^{6}$ (Ex 2), i.e. less negative = bigger orbit ✔. $Δ v \approx 2.4$ km/s sits in the forecast band ✔.

Example 7 — Exam twist: classify from raw $v, r$ (cell C7)

Worked example "Is this orbit bound?"

A body is measured at $r = 1.00 \times 1 0^{7}$ m with speed $v = 9.00 \times 1 0^{3}$ m/s. Classify the orbit and find $a$ .

Forecast: compare $v$ with the local escape speed $2 GM / r$ first — bound or not?

Step 1. Local escape speed: $2 (3.986 \times 1 0^{14}) /1 0^{7} = 8.929 \times 1 0^{3}$ m/s. Why this step? It's the yardstick: below ⇒ ellipse, equal ⇒ parabola, above ⇒ hyperbola.

Step 2. $v = 9.00 \times 1 0^{3} > 8.929 \times 1 0^{3}$ , so hyperbolic (unbound) — but only just. Why this step? The margin is tiny; the sign of $ε$ decides, so compute it exactly.

Step 3. $ε = \frac{1}{2} (9.00 \times 1 0^{3})^{2} - \frac{3.986 \times 1 0 ^{14}}{1 0 ^{7}} = 4.050 \times 1 0^{7} - 3.986 \times 1 0^{7} = 6.40 \times 1 0^{5} J/kg > 0$ ✔ hyperbola.

Step 4. $a = - \frac{GM}{2 ε} = - \frac{3.986 \times 1 0 ^{14}}{2 ( 6.40 \times 1 0 ^{5} )} = - 3.11 \times 1 0^{8}$ m.

Verify: $ε > 0$ ↔ $a < 0$ ✔ (the two must agree). The forecast (barely unbound) held — a $71$ m/s reduction would drop it back to a bound ellipse.

Example 8 — Inverse + consistency with Kepler's Third Law (cell C8)

Worked example From energy back to period

An orbit has $ε = - 8.14 \times 1 0^{6}$ J/kg (the Ex 2 ellipse). Recover $a$ , then find the orbital period $T$ and cross-check with Kepler's Third Law.

Forecast: since $a \approx 2.45 \times 1 0^{7}$ m (bigger than GEO's would give a $\sim 24$ h? no — smaller), guess whether $T$ is above or below one day.

Step 1. Invert the energy formula: $a = - \frac{GM}{2 ε} = - \frac{3.986 \times 1 0 ^{14}}{2 ( - 8.14 \times 1 0 ^{6} )} = 2.449 \times 1 0^{7}$ m ✔ matches Ex 2. Why this step? Confirms the energy↔geometry link is a true two-way street.

Step 2. Kepler's Third Law: $T = 2 π \frac{a ^{3}}{GM}$ . Why this step? Period depends only on $a$ (hence only on $ε$ ) — same " $a$ decides everything" theme.

Step 3. $a^{3} = (2.449 \times 1 0^{7})^{3} = 1.469 \times 1 0^{22}$ ; $a^{3} / GM = 3.685 \times 1 0^{7}$ ; $= 6.070 \times 1 0^{3}$ ; $T = 2 π (6.070 \times 1 0^{3}) = 3.814 \times 1 0^{4}$ s.

Verify: $T \approx 3.81 \times 1 0^{4}$ s $= 10.6$ hours — under a day, as forecast (this GTO's $a$ is smaller than GEO's $4.22 \times 1 0^{7}$ m). Consistent chain: $ε \to a \to T$ all agree ✔.

Recall Which cell was which?

Every-scenario coverage in one glance. Circle where $r = a$ — which example? ::: Ex 1 (cell C1) General ellipse energy from $r_{p} + r_{a}$ ? ::: Ex 2 (cell C2) Speed changes around an ellipse (angular-momentum check)? ::: Ex 3 (cell C2) The $ε = 0$ escape / parabola case? ::: Ex 4 (cells C3, C5) Positive $ε$ , negative $a$ , hyperbola? ::: Ex 5 (cells C4, C5) $Δ v$ for a Hohmann burn? ::: Ex 6 (cell C6) Classify orbit from raw $v, r$ ? ::: Ex 7 (cell C7) Energy → $a$ → period cross-check? ::: Ex 8 (cell C8)