ARCHETYPE: Raising & Switching Orbits

In 1925, a German municipal architect named Walter Hohmann published a slim book on, of all things, the energetics of leaving planets. Rocketry barely existed; his day job was checking building plans in Essen. Yet Die Erreichbarkeit der Himmelskörper ("The Attainability of the Celestial Bodies") answered a question no one else had posed precisely: what is the cheapest way to move a spacecraft from one orbit to another? His answer — glide along a half-ellipse touching both orbits, burning fuel only at the two ends — is still how most satellites reach their destinations a century later. Every one of Hohmann's calculations begins from the ledger this archetype drills: a circular orbit of radius $r$ has total energy $E = -GMm/2r$ , and moving between orbits means paying (or pocketing) the difference. The exam version of Hohmann's problem is a family of energy-subtraction drills. How much energy puts a satellite into orbit? How much lifts it from a low orbit to a geostationary one? What is its binding energy? What speed increase makes it escape? Boards, NEET and JEE all draw from this pot, and the same three-line method answers every variant: write $E$ on each rung of the energy ladder, subtract, and interpret the sign. The archetype also carries the chapter's most instructive paradox — to raise an orbit you add energy and end up slower — plus the accounting that resolves it. Master the ladder and the whole family becomes arithmetic.

Key takeaways

The energy ladder: a circular orbit at radius $r$ has $E(r) = -\dfrac{GMm}{2r}$ , with $K = -E$ and $U = 2E$ ; the surface (at rest) sits at $E = -\dfrac{GMm}{R}$ ; freedom is $E \ge 0$ .
Every problem is a subtraction between two rungs: $\Delta E = E_{\rm final} - E_{\rm initial}$ .
Raise an orbit from $r_1$ to $r_2$ : $\Delta E = \dfrac{GMm}{2}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right) > 0$ — you pay, yet the satellite ends up slower.
Binding energy of an orbiting satellite $= |E| = \dfrac{GMm}{2r}$ ; of a body at rest on the surface $= \dfrac{GMm}{R}$ . Different rungs, different bills.
Escape from orbit: multiply speed by $\sqrt2$ (add 41%), equivalently add energy $|E|$ .
Launching to low orbit costs $\approx \dfrac{GMm}{2R}$ — about half the well depth, and more than the subsequent climb from low orbit to geostationary.
Key idea: draw the two rungs, write $-GMm/2r$ on each, subtract — the sign and size of $\Delta E$ answer every orbit-switching question.

The core insight

Orbit energetics reduces to bookkeeping on the ladder $E(r) = -GMm/2r$ : the energy to launch, raise, lower or free a satellite is the difference between two rungs, with kinetic energy $-E$ and potential energy $2E$ locked in ratio at every rung.

Where this lives today

Every satellite operator is an energy accountant. A geostationary communications satellite typically launches into a low parking orbit (the expensive rung, per worked example 3), rides a transfer ellipse to geostationary height, and circularises — the classic two-burn Hohmann sequence whose total is exactly equation (2)'s $\Delta E$ dressed as two speed changes. Modern all-electric satellites pay the same bill differently: ion thrusters supply milli-newtons for months, spiralling up through thousands of intermediate rungs — same ledger, instalment plan. The ISS runs the ledger in reverse: drag at $400\ \mathrm{km}$ drains its $E$ continuously (making it, per the drag paradox, very slightly faster), and periodic re-boosts buy the energy back. End-of-life is a ledger entry too — low satellites are pushed down (paying out energy) to burn up; geostationary ones are pushed up a few hundred kilometres into graveyard orbits, because from up there, descending the whole ladder to re-entry would cost far more than the graveyard hop. Hohmann's lamplight arithmetic became traffic law for near-Earth space.

The setup

Draw a vertical energy axis with zero — the freedom line — at the top. Every circular orbit is a rung on this ladder at height $E(r) = -GMm/2r$ : low orbits are deep rungs (large $|E|$ , fast, tightly bound), high orbits sit near the top (small $|E|$ , slow, loosely bound), and the surface-at-rest lies deeper than every orbit rung, at $-GMm/R$ . Moving a satellite is climbing this ladder, and the only number a mission planner (or an examiner) truly needs is the gap between two rungs.

Figure

The orbit-energy ladder — every manoeuvre is a gap between rungs

The ladder hides one counter-intuitive rule, and stating it early saves marks later: climbing the ladder slows you down. Each rung's kinetic energy is $K = |E| = GMm/2r$ — largest on the deep rungs. Add energy (fire engines forward) and the satellite rises to a shallower rung where it moves slower; remove energy (drag, retro-burn) and it descends to a deeper, faster rung. The engines' work didn't vanish: potential energy rose by twice the kinetic energy lost, because $U = 2E$ keeps the ledger locked. Satellites obey bank accounting, not car intuition. One honesty flag for launch problems: treating the launch pad as "at rest" ignores Earth's rotation (worth $0.46\ \mathrm{km/s}$ of free speed at the equator) and all air drag. Exam questions at this level intend exactly that idealization — say it, use it.

The master pattern

Move 1 — place the rungs. Initial state and final state, each with its energy: orbit at $r$ → $-GMm/2r$ ; rest on the surface → $-GMm/R$ ; escaped with nothing to spare → $0$ ; escaped with speed $v_\infty$ → $+\tfrac12mv_\infty^2$ .

Move 2 — subtract.

\Delta E = E_{\rm f} - E_{\rm i} \qquad (1)

where: a positive $\Delta E$ is the energy an engine (or launcher) must supply; a negative $\Delta E$ is energy the satellite must shed (drag, retro-burns). For the two commonest gaps:

\text{raise } r_1 \to r_2:\quad \Delta E = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right) \qquad (2)

\text{surface (rest)} \to \text{orbit } r:\quad \Delta E = \frac{GMm}{R} - \frac{GMm}{2r} \;\xrightarrow{\,r \approx R\,}\; \frac{GMm}{2R} \qquad (3)

where: equation (3)'s limit says the fare to low orbit is about half the well depth — some $31\ \mathrm{MJ/kg}$ — and essentially all of it is spent buying sideways speed, not height. That is why "getting to space is easy; staying in space is hard."

Move 3 — interpret. Binding energy $= |E|$ of the current state (the gap up to the freedom line). Escape from orbit = add $|E|$ , equivalently multiply speed by $\sqrt2$ . Kinetic/potential splits, when asked, come free from the locked ratios $K = -E$ , $U = 2E$ .

variant	rungs to subtract	result
binding energy in orbit	orbit → freedom	$GMm/2r$
binding energy on surface	surface (rest) → freedom	$GMm/R$
launch to orbit	surface → orbit	eq. (3)
raise / lower orbit	orbit → orbit	eq. (2), either sign
escape from orbit	orbit → freedom	add $\|E\|$ , or $v \to \sqrt2\,v$
drag decay	orbit → lower orbit	$\Delta E < 0$ , yet $K$ rises

Derivation AThe raise-orbit bill, and where the engine's energy actually goes

We derive equation (2) — the energy to move a satellite between circular orbits — and split the payment between kinetic and potential accounts, which is where the "slower after boosting" paradox dissolves. Strategy: pure ladder subtraction using $E = -GMm/2r$ (established in the satellites module), then the locked ratios. Watch the factor 2 doing all the interesting work.

Step 1. Write both rungs:

E_1 = -\frac{GMm}{2r_1}, \qquad E_2 = -\frac{GMm}{2r_2}

Step 2. Subtract (final minus initial):

\Delta E = E_2 - E_1 = \frac{GMm}{2}\left(\frac{1}{r_1} - \frac{1}{r_2}\right)

Why the sign takes care of itself: $r_2 > r_1$ makes the bracket positive — raising costs energy; swap the radii and $\Delta E$ flips sign — lowering pays you (the satellite must shed that much to drag or retro-burns).

Step 3. Split the payment. Kinetic energy on each rung is $K = +GMm/2r$ , so

\Delta K = \frac{GMm}{2}\left(\frac{1}{r_2} - \frac{1}{r_1}\right) = -\Delta E

— the satellite loses kinetic energy equal to the whole bill. Potential energy ( $U = -GMm/r$ ) changes by

\Delta U = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right) = +2\,\Delta E

Step 4. Audit: $\Delta K + \Delta U = -\Delta E + 2\Delta E = \Delta E$ ✓. Read it as a bank statement: the engine deposits $\Delta E$ ; the potential account absorbs $2\Delta E$ ; the difference is withdrawn from the kinetic account. Boost, rise, arrive slower — no paradox, just the locked $1 : -1 : -2$ ratios. Verified: limits — $r_2 \to \infty$ turns eq. (2) into $GMm/2r_1 = |E_1|$ , the binding energy ✓; $r_2 = r_1$ gives zero ✓. Numbers: $r_1 = 7000$ km, $r_2 = 14{,}000$ km, $m = 1000$ kg: $\Delta E = 1.993\times10^{17}(1.429 - 0.714)\times10^{-7} = 1.42\times10^{10}$ J, re-used in worked example 2.

Figure

The raise-orbit bank statement — engine pays ΔE, potential absorbs 2ΔE, kinetic funds the gap

Go deeper · Derivation B · 1 minThe launch bill — why low orbit costs half the well

Derivation BThe launch bill — why low orbit costs half the well

We derive equation (3), the energy to put a satellite into orbit from the ground, and its famous $r \approx R$ reading. Strategy: ladder subtraction again, but the initial rung is the surface at rest — not an orbit — so its energy is pure potential. The point to watch: for a low orbit, the entire bill turns out to be kinetic, which is the physical content of "orbit is about speed, not height."

Step 1. Initial rung — resting on the pad (rotation ignored, stated): $E_{\rm i} = 0 - \dfrac{GMm}{R} = -\dfrac{GMm}{R}$ .

Step 2. Final rung — circular orbit at radius $r$ : $E_{\rm f} = -\dfrac{GMm}{2r}$ .

Step 3. Subtract:

\Delta E = \frac{GMm}{R} - \frac{GMm}{2r}

Step 4. Take the low-orbit limit $r \to R$ :

\Delta E \to \frac{GMm}{R} - \frac{GMm}{2R} = \frac{GMm}{2R} = \frac12 m\,(gR)

— exactly the kinetic energy $\tfrac12mv^2$ of the skimming orbital speed $v = \sqrt{gR}$ . No potential energy was bought at all (same radius!): the whole fare purchased sideways speed. Why this beats intuition: "reaching space" (100 km up) costs $mgh \approx 1\ \mathrm{MJ/kg}$ — pocket change against the $31\ \mathrm{MJ/kg}$ of orbital speed. Sounding rockets reach space cheaply and fall right back; orbit is thirty times the bill. *Verified: per kilogram at Earth: $GM/2R = 6.26\times10^7/2 = 3.13\times10^7$ J/kg, and $\tfrac12 v^2 = \tfrac12(7.9\times10^3)^2 = 3.12\times10^7$ J/kg ✓ — the two routes agree to rounding. For $r = 2R$ the bill is $GMm/R - GMm/4R = \tfrac34 GMm/R$ , used in worked example 3.*

Limiting cases

case	result	meaning
$r_2 \to \infty$ in eq. (2)	$\Delta E \to GMm/2r_1$	raising "all the way" = unbinding: the binding energy
$r \approx R$ in eq. (3)	$\Delta E \approx GMm/2R$	low orbit: all speed, no height
$r_2 = r_1$	$\Delta E = 0$	same rung, no bill
lowering ( $r_2 < r_1$ )	$\Delta E < 0$ , $K$ rises	descent is paid to you — and you arrive faster
speed $\times \sqrt2$ at any rung	$E \to 0$	escape: doubling $K$ cancels $E = -K$
speed $\times 2$ at any rung	$E = +K_{\rm orbit} > 0$	hyperbolic exit with $v_\infty = v_{\rm esc}(r)/\!\ldots$ — see example 4

Common trap“More energy means more speed” — and its cousin, one binding energy for all occasions

Two ledger errors cause most lost marks here. First, the car reflex: in orbit, added energy buys altitude at the price of speed — $K = -E$ falls as $E$ rises. If your answer has a boosted satellite moving faster in its new circular orbit, the ledger has been violated. Second, "binding energy $= GMm/2r$ , always": that is the bill for a satellite already in orbit. A body at rest on the surface is deeper in the cellar — unbinding it costs $GMm/R$ , twice the low-orbit figure. Match the rung to the state described: orbiting → $GMm/2r$ ; resting on the ground → $GMm/R$ ; moving with speed $v$ anywhere → compute $E = \tfrac12mv^2 - GMm/r$ and take $|E|$ if negative. The rung, not the formula sheet, decides.

Worked examples

1 · Board level — read one rung fully (CBSE/ISC pattern, 3 marks)

A $200\ \mathrm{kg}$ satellite is in a circular orbit of radius $8000\ \mathrm{km}$ . Find its kinetic energy, potential energy, total energy and binding energy. ( $GM = 3.99\times10^{14}\ \mathrm{m^3/s^2}$ .)

Step 1. One computation buys everything — the kinetic energy: $K = \dfrac{GMm}{2r} = \dfrac{3.99\times10^{14} \times 200}{2 \times 8\times10^6} = \boxed{4.99\times10^9\ \mathrm{J}}$ .

Step 2. Locked ratios do the rest: $E = -K = \boxed{-4.99\times10^9\ \mathrm{J}}$ ; $U = 2E = \boxed{-9.97\times10^9\ \mathrm{J}}$ .

Step 3. Binding energy = gap to freedom: $|E| = \boxed{4.99\times10^9\ \mathrm{J}}$ .

Step 4. Sanity: $K + U = 4.99 - 9.97 = -4.98 \approx E$ ✓ (rounding); total is negative — bound, as an orbit must be.

2 · Board level — the raise bill (ISC/NEET pattern)

How much energy is needed to shift a $1000\ \mathrm{kg}$ satellite from a circular orbit of radius $7000\ \mathrm{km}$ to one of radius $14{,}000\ \mathrm{km}$ ? What happens to its speed?

Step 1. Equation (2): $\Delta E = \dfrac{GMm}{2}\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right) = \dfrac{3.99\times10^{14}\times1000}{2}\left(\dfrac{1}{7\times10^6} - \dfrac{1}{1.4\times10^7}\right)$ .

Step 2. The bracket: $1.429\times10^{-7} - 0.714\times10^{-7} = 0.714\times10^{-7}\ \mathrm{m^{-1}}$ .

Step 3. Multiply: $\Delta E = 1.99\times10^{17} \times 0.714\times10^{-7} = \boxed{1.42\times10^{10}\ \mathrm{J}}$ .

Step 4. Speed: the new rung is shallower, so the satellite is slower — specifically $v \propto 1/\sqrt r$ drops by $\sqrt2$ : from $7.55$ to $5.34\ \mathrm{km/s}$ . The engines added energy and the satellite lost speed; the potential account absorbed $2\Delta E$ .

3 · Competitive level — launch versus climb (JEE-Main pattern, ~3 minutes)

(a) Find the energy per kilogram to place a payload in a low orbit ( $r \approx R$ ). (b) Compare the launch bill for a $500\ \mathrm{kg}$ satellite to orbit at $r = 2R$ with the bill from low orbit up to $r = 2R$ . ( $GM/R = 6.26\times10^7\ \mathrm{J/kg}$ , i.e. $R = 6.37\times10^6\ \mathrm{m}$ .)

Step 1. Part (a), equation (3) limit: $\Delta E/m = \dfrac{GM}{2R} = \boxed{3.13\times10^7\ \mathrm{J/kg}}$ — half the well depth, all of it sideways speed.

Step 2. Part (b), surface → $2R$ : $\Delta E = m\left(\dfrac{GM}{R} - \dfrac{GM}{4R}\right) = \dfrac34\,m\,\dfrac{GM}{R} = 0.75 \times 500 \times 6.26\times10^7 = \boxed{2.35\times10^{10}\ \mathrm{J}}$ .

Step 3. Low orbit → $2R$ , equation (2) with $r_1 \approx R$ : $\Delta E = \dfrac{GMm}{2}\left(\dfrac1R - \dfrac{1}{2R}\right) = \dfrac{GMm}{4R} = 0.25 \times 500 \times 6.26\times10^7 = 7.8\times10^9\ \mathrm{J}$ .

Step 4. Read the split: of the total $2.35\times10^{10}$ J, two-thirds ( $1.57\times10^{10}$ J = $GMm/2R$ ) went into just reaching low orbit. The first few hundred kilometres are the expensive part — "low orbit is halfway to anywhere." Timing note: keep $GM/R = 62.6\ \mathrm{MJ/kg}$ as a mental anchor and every part becomes a fraction of it: launch-to-LEO $= \tfrac12$ , surface→ $2R$ $= \tfrac34$ , LEO→ $2R$ $= \tfrac14$ . Fractions first, calculator last; budget 2.5 minutes.

4 · Competitive level — speed kicks and their fates (JEE-Advanced flavour, ~4 minutes)

A satellite orbits at radius $r$ with speed $v$ . In a single brief burn its speed is instantly multiplied by $k$ (direction unchanged). Find the total energy just after the burn and the satellite's fate for (a) $k = \sqrt2$ , (b) $k = 2$ — and for (b) give its speed far from Earth in terms of $v$ .

Step 1. Just after the burn (same $r$ , new speed $kv$ , and $v^2 = GM/r$ ): $E = \tfrac12 m k^2v^2 - \dfrac{GMm}{r} = \tfrac12 mv^2\,(k^2 - 2)$ .

Step 2. Part (a): $k = \sqrt2$ gives $E = 0$ — the marginal escape. A 41% speed increase unbinds any circular orbit; this is the $\sqrt2$ ladder seen from the energy side.

Step 3. Part (b): $k = 2$ gives $E = \tfrac12mv^2\,(4 - 2) = +mv^2 > 0$ — hyperbolic exit, gone for good.

Step 4. Far away, all of $E$ is kinetic: $\tfrac12 mv_\infty^2 = mv^2$ , so $v_\infty = \boxed{\sqrt2\,v}$ .

Step 5. Cross-check with the escape article's Pythagoras: $v_\infty = \sqrt{(2v)^2 - v_{\rm esc}^2}$ with $v_{\rm esc} = \sqrt2\,v$ : $\sqrt{4v^2 - 2v^2} = \sqrt2\,v$ ✓. Timing note: the master line $E = \tfrac12mv^2(k^2 - 2)$ is worth memorising — fate is the sign of $k^2 - 2$ , instantly. Watch the perennial wrong option " $k = 2$ leaves speed $v$ at infinity": squares subtract, not speeds. Budget 3 minutes.

Exam traps

The missing ½. Orbit energy is $-GMm/2r$ ; potential energy is $-GMm/r$ . Every formula in this archetype inherits one or the other — misplace the 2 and the raise bill (2) doubles or halves.

Wrong rung for binding energy. In orbit: $GMm/2r$ . At rest on the surface: $GMm/R$ . Moving arbitrarily: compute $E$ honestly. The question's state picks the rung.

Sign discipline in (2). Write final minus initial, always. Raising gives $\Delta E > 0$ (engine pays); lowering gives $\Delta E < 0$ (satellite sheds). An answer of the wrong sign with the right magnitude usually loses the method marks too.

Faster after a boost. In the new circular orbit the satellite is slower — $K = -E$ . (During the transfer itself speeds vary; the exam's "final orbit" statements are about rungs.)

Energy to double the radius ≠ double the energy. The ladder is a $1/r$ ladder, not a linear one: surface→ $2R$ costs $\tfrac34\,GMm/R$ , not twice the LEO bill. Fractions of $GM/R$ keep you honest.

Escape ≠ leave the ladder from anywhere at 11.2 km/s. From an orbit at radius $r$ , escape needs $\sqrt2 \times$ the local orbital speed — the surface number $11.2\ \mathrm{km/s}$ belongs to the surface rung only.

Ignoring what the question ignores. Board-level launch bills assume a non-rotating Earth and no drag. State the idealization in one line; do not import the $0.46\ \mathrm{km/s}$ equatorial bonus unless asked.

Origins · 1 minWalter Hohmann and how this was discovered

Walter Hohmann

1880–1945 · Essen

Walter Hohmann was the chief structural engineer of the city of Essen — bridges and building permits by day, celestial mechanics by lamplight. His 1925 book computed, decades before any rocket could try it, the minimum-energy path between two circular orbits: coast along a half-ellipse tangent to both, with one burn to leave the low orbit and one to settle into the high one. The "Hohmann transfer" is exactly this archetype's ladder made practical — his two burns pay precisely the $\Delta E$ of equation (2), split into two instalments. The German rocket societies of the 1920s treated the book as scripture; NASA's Apollo trajectories and every geostationary satellite's transfer orbit descend from it. Hohmann himself wanted no part of weaponised rocketry — he refused V-2 work and died in 1945, months before the spaceflight era he had priced out in advance. The lunar crater Hohmann and the term "Hohmann window" for launch opportunities keep the ledger-keeper's name in every mission plan.

History

1925

— Hohmann's Die Erreichbarkeit der Himmelskörper computes minimum-energy transfers between orbits.

1957

— Sputnik: the first artificial satellite occupies the lowest rungs of the ladder.

1963

— Syncom missions ride Hohmann-style transfer ellipses from low orbit to the geostationary rung.

1998

— the ISS begins its permanent occupancy of a decaying rung, re-boosted by visiting spacecraft ever since.

2001

— ESA's Artemis satellite, stranded by a launcher fault, spirals to geostationary orbit on ion engines — the ladder climbed in thousands of tiny steps.

2020

s — electric-propulsion "all-EP" satellites make the months-long spiral climb routine, trading time for fuel on the same energy bill.

Going further

Texts. NCERT Physics Class 11, Ch. 8 (Gravitation) · HC Verma, Concepts of Physics Vol. 1, Ch. 11 · Hohmann's 1925 book, in translation as The Attainability of the Celestial Bodies (NASA TT F-44).

Provenance. All results are standard canon; every number recomputed. A verifier should recompute: example 1's $K = 4.99\times10^9$ J ledger; example 2's bracket $0.714\times10^{-7}$ and $\Delta E = 1.42\times10^{10}$ J; example 3's fractions $\tfrac12/\tfrac34/\tfrac14$ of $GM/R = 62.6$ MJ/kg; example 4's $E = \tfrac12mv^2(k^2-2)$ and $v_\infty = \sqrt2\,v$ at $k = 2$ .

Prerequisites. Satellites: Speed, Period, Energy & Weightlessness · Gravitational Potential Energy & Potential · Escape Velocity. Leads to. Kepler Ratio Problems (archetype) · Rocket Propulsion & Variable Mass · Orbit Classification & Energy (UG deep dive).

Where does the engine's energy actually go if the satellite ends up slower?

Into the potential account — with interest. The locked ratios say $U = 2E$ : raise the total energy by $\Delta E$ and the potential energy must rise by $2\Delta E$ . The extra $\Delta E$ that the engine didn't supply is withdrawn from the kinetic account, which is why the satellite arrives slower. It feels like a swindle only because everyday experience (cars, cycling) has no potential ledger running at double rate alongside the motion.

Why is getting to low orbit more expensive than going from low orbit to the Moon's distance?

Because the ladder's rungs crowd together as you climb. The gap from the surface cellar to the LEO rung is $GMm/2R \approx 31$ MJ/kg — bought as raw speed. The gap from LEO all the way to freedom is the same again ( $|E_{\rm LEO}| = GMm/2R$ ), and most intermediate destinations cost only a slice of it: LEO→GEO is about $12$ MJ/kg. Halving of gaps at each doubling of radius is just the $1/r$ ladder's geometry. Hence the astronautics proverb this article keeps cashing: low orbit is halfway to anywhere.

Is binding energy the same as the energy the rocket actually spends?

No — binding energy is the physics floor, the minimum the ledger demands between two states. A real launcher spends several times more: it fights drag, carries its own fuel (the rocket equation's tyranny), and steers. Exam questions ask for the floor unless they explicitly give engine efficiencies. Keep the two vocabularies separate: "energy required (ideal)" = ladder gap; "fuel required (real)" = engineering, not in this chapter.

For escape from an orbit, which is right: add binding energy, or multiply speed by √2? My book says both.

They are the same statement in two currencies. In orbit, $E = -K$ : adding energy $|E| = K$ brings the total to zero — escape. Doubling the kinetic energy ( $K \to 2K$ ) does exactly that, and doubling kinetic energy is multiplying speed by $\sqrt2$ . Energy currency: add $GMm/2r$ . Speed currency: $v \to \sqrt2\,v$ . Use whichever the question's data make cheaper — energies for masses-and-radii, speeds for "by what factor" phrasing.

Does the transfer path between two orbits change the energy bill?

The total $\Delta E$ between the two circular rungs is path-independent — that's what a conservative force means. What the path changes is the speed bill (Δv) and hence fuel: burns are most effective where the craft moves fastest (deep in the well), which is why the Hohmann sequence puts its first burn in the low orbit. Slow spiral climbs (ion engines) pay the same energy but a larger total Δv, traded against far better fuel economy per newton. At exam level: energy questions → ladder subtraction, always; Δv optimisation → beyond scope, one-line mentions only.

A satellite is "de-orbited" by a small retro-burn. Doesn't removing energy make it speed up and stay up even better?

The drag paradox does mean it will be moving faster when lower — but "staying up" is about the orbit's geometry, not its speed. The retro-burn lowers the energy, hence lowers (one side of) the orbit until it dips into dense atmosphere; from there, drag removes energy catastrophically fast and the craft descends and heats. Faster-but-lower is precisely the descending branch of the ladder. The speed increase is real, temporary, and fatal — which is also why re-entry is a heating problem: all that grown kinetic energy plus the remaining ladder drop must be dissipated into air.