Gravitational Potential Energy and Escape Velocity

How hard do you have to throw something so that it never comes back? Not "very high," not "into orbit" — gone, coasting away from Earth forever. The question sounds like it needs the whole trajectory solved, drag and all; in fact it needs one line of energy bookkeeping, and the answer is a speed: $11.2$ km/s, about thirty-four times the speed of sound. Below it, what goes up comes down. At it, the projectile limps to rest infinitely far away. The number does not care about the direction of the throw, and it does not care what is being thrown — a pebble and a battleship share the same escape speed. That one-line calculation only works after an upgrade this article performs carefully. The $U = mgh$ you have used since the energy chapter is a local approximation, valid while the gravitational field looks uniform; it cannot answer any question in which the field weakens along the way. The honest potential energy of gravitation is $U = -GMm/r$ — negative everywhere, climbing toward zero at infinite separation — and once you own it, a whole class of formerly hard problems collapses: maximum altitude of a fast projectile, the speed a comet picks up falling from deep space, and the single number $E = K + U$ whose sign tells you whether any object is bound to its planet or free. You arrive with the universal law and conservation of energy; this article welds them. The payoff runs deep into the track ahead: the energy view built here is exactly how the orbit chapters classify ellipses, parabolas, and hyperbolas.

Key takeaways

Gravitational potential energy with the honest force law is $U(r) = -\dfrac{GMm}{r}$ , with the zero chosen at infinite separation — so $U$ is negative everywhere, and less negative means higher.
The familiar $mgh$ is the local expansion of this formula for height changes small compared to the planet's radius.
Escape speed comes from pure energy accounting: $\tfrac12 mv_{\rm esc}^2 - \dfrac{GMm}{R} = 0$ gives $v_{\rm esc} = \sqrt{2GM/R} = \sqrt{2gR}$ — $11.2$ km/s for Earth, independent of the projectile's mass and of launch direction.
The total energy $E = \tfrac12 mv^2 - GMm/r$ is conserved, and its sign is destiny: $E < 0$ bound, $E = 0$ barely escapes, $E > 0$ escapes with speed to spare.
A circular orbit has $E = -\dfrac{GMm}{2r}$ : kinetic energy equals half the depth of the well, and escape from orbit needs only $\sqrt2$ times the orbital speed.
Key idea: gravity digs a $-GMm/r$ energy well around every mass; whether you are trapped in it or free is decided by the sign of one conserved number.

The core insight

Because gravity weakens as $1/r^2$ , the work needed to climb from $r$ to infinity is finite — $GMm/r$ — so each position sits at potential energy $U = -GMm/r$ ; conservation of $E = \tfrac12 mv^2 - GMm/r$ then decides everything, with $E \ge 0$ (reached at speed $v_{\rm esc} = \sqrt{2GM/R}$ ) the condition for escape.

Where this lives today

Mission design is energy bookkeeping in this article's currency. Trajectory engineers quote a mission's cost as $C_3 = v_\infty^2$ — twice the specific orbital energy, literally the $E > 0$ leftover of equation (3) — and launch-vehicle brochures list payload versus $C_3$ curves. The $\sqrt2$ of Step 5 sets the architecture of everything beyond Earth: from low orbit, 3.2 km/s buys escape, which is why missions assemble speed in orbit rather than fighting the whole well in one ascent. Sample-return spacecraft exploit the other end of the scale — asteroid Bennu's escape speed is about 20 centimetres per second, so OSIRIS-REx could tap the surface and drift away without meaningfully burning fuel. The escape-speed lens also organizes the sky itself. Whether a galaxy keeps its hot supernova-blown gas, whether a planet keeps its atmosphere over billions of years (Mars's thin air is partly the slow victory of thermal molecular speeds over a 5.0 km/s escape speed), and where a black hole's point of no return sits — all are the same comparison, $v$ versus $\sqrt{2GM/R}$ , run at wildly different scales. And when a satellite operator watches drag pull a spacecraft down while its speed rises, they are watching equation (5)'s 2-to-1 virial lock operate in real time.

The physical picture

Picture gravity's grip on you as a well — a funnel-shaped depression in an energy landscape, deepest at the planet's surface, flattening out toward zero far away. Climbing out of a uniform-gravity well would cost $mgh$ without limit: every metre of height costs the same, forever, and no finite throw escapes. But gravity is not uniform. Each successive metre of altitude costs slightly less than the one before, because the field is $1/r^2$ -weaker up there. Sum the infinite staircase of ever-cheaper steps and the total is finite: $GMm/R$ from the surface all the way to infinity. A well of finite depth can be jumped out of — that is the entire secret of escape velocity. The well picture also explains the strange minus signs. If we agree to call the energy of two masses infinitely far apart, at rest zero — the least arbitrary choice available — then every actual configuration, being lower in the well, has negative potential energy. Negative $U$ is not owing anyone energy; it is simply "below the rim." Moving up means becoming less negative. And the sign of the total energy $E = K + U$ becomes a verdict: negative, and you lack the energy to reach the rim — you are bound, forever orbiting or falling back; zero or positive, and you can coast out of the well with speed left over.

Figure

The gravity well — finite depth is the secret of escape

One more preview the picture hands us for free. On the energy diagram a thrown object rides a horizontal line (its conserved $E$ ); where that line intersects the $U(r)$ curve, the kinetic energy has been squeezed to zero — a turning point. Bound trajectories live between such intersections; the escape case $E = 0$ is the line grazing the rim at infinity. The orbit chapters will add angular momentum to this diagram (the effective potential), but the reading habit — horizontal energy line meets potential curve — starts here.

The math: from work integral to escape

First, the honest potential energy. The near-surface formula $U = mgh$ was built from a constant force $mg$ ; now the force is $GMm/r^2$ and weakens as you climb, so the work must be integrated rather than multiplied.

Derivation APotential energy from the work integral

We are deriving $U(r) = -GMm/r$ , the potential energy of two masses a distance $r$ apart, from the definition of potential energy as minus the work done by the conservative force. The strategy is the standard one from the energy chapter — compute $W_{\rm grav}$ along a radial path, use path-independence to claim generality, then make the one free choice (where $U = 0$ ) wisely. Watch two signs: the force points inward while our path steps outward, and the choice $U(\infty) = 0$ is what plants the whole curve below zero.

Step 1. Set up the work integral along a radial path. Moving a mass $m$ from $r_1$ outward to $r_2$ , the gravitational force on it is $\vec F = -\dfrac{GMm}{r^2}\hat r$ (inward), so

W_{\rm grav} = \int_{r_1}^{r_2} \left(-\frac{GMm}{r^2}\right) dr = GMm\left[\frac{1}{r}\right]_{r_1}^{r_2} = GMm\left(\frac{1}{r_2} - \frac{1}{r_1}\right) \qquad (1)

Why: for outward motion ( $r_2 > r_1$ ) this is negative — gravity fights the climb — exactly as it should be.

Step 2. Convert work to potential energy. By definition $\Delta U = -W_{\rm grav}$ , so $U(r_2) - U(r_1) = GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$ . Why: gravity is conservative — the central-force work integral depends only on endpoints (any path decomposes into radial steps that count and angular steps that don't, since the force is radial) — so a potential energy exists at all.

Step 3. Choose the zero. Let $r_1 \to \infty$ and demand $U(\infty) = 0$ :

U(r) = -\frac{GMm}{r} \qquad (2)

Why this choice: infinity is the one configuration all problems share — two masses so far apart they no longer interact. Any other zero would work but would smuggle a planet-specific constant into every formula.

Step 4. Recover the old friend. For a small climb $h$ above a surface of radius $R$ : $U(R+h) - U(R) = GMm\left(\dfrac1R - \dfrac1{R+h}\right) = \dfrac{GMm\,h}{R(R+h)} \approx \dfrac{GM}{R^2}\,mh = mgh$ . Why: for $h \ll R$ the denominator is nearly $R^2$ , and $GM/R^2$ is precisely the surface field $g$ — the flat-Earth formula is the first term of the honest one. *As a check: differentiating back gives $F = -dU/dr = -GMm/r^2$ (attractive) ✓, and the $mgh$ limit follows from the Taylor expansion derived in Step 4.*

With (2) in hand, the total mechanical energy of anything coasting in a gravitational field is

E = \frac12 mv^2 - \frac{GMm}{r} \qquad (3)

where: $v$ and $r$ are the speed and centre-distance at any single instant — any instant, because with gravity the only force acting, $E$ is conserved. Read the two terms as a competition: kinetic energy is always positive, potential always negative, and their sum's sign is the object's fate. If $E < 0$ there is a radius where $U(r)$ alone uses up all of $E$ — a turning point the object cannot pass: bound. If $E \ge 0$ the object can reach $r = \infty$ , arriving with kinetic energy $E$ to spare: free. This single-number classification is the engine of every calculation below.

Go deeper · Derivation B · 1 minEscape velocity

Derivation BEscape velocity

The goal is the minimum launch speed from the surface of a body of mass $M$ and radius $R$ that never falls back: $v_{\rm esc} = \sqrt{2GM/R}$ . The strategy is conservation of energy between two snapshots — launch, and the marginal case "barely makes it": at rest, infinitely far away. No trajectory is solved and none is needed; that is the power of the energy method. The subtlety to watch is what "minimum" buys: the marginal case has zero of both energies at infinity, which pins $E = 0$ at launch too.

Step 1. Write the energy at launch: $E = \tfrac12 mv_{\rm esc}^2 - \dfrac{GMm}{R}$ . Why: speed $v_{\rm esc}$ at centre-distance $R$ , straight into equation (3).

Step 2. Write the energy of the marginal escape: at $r \to \infty$ , $U \to 0$ , and "barely" means arriving with nothing left, $K \to 0$ . So $E = 0$ . Why: any leftover speed at infinity means the launch was faster than necessary; any turning point at finite $r$ means it was slower.

Step 3. Equate and solve:

\frac12 mv_{\rm esc}^2 = \frac{GMm}{R} \quad\Longrightarrow\quad v_{\rm esc} = \sqrt{\frac{2GM}{R}} = \sqrt{2gR} \qquad (4)

Why: the projectile's mass $m$ cancels — pebble and battleship escape alike — and the last form uses $g = GM/R^2$ to write it from surface gravity alone.

Step 4. Put in Earth's numbers: $v_{\rm esc} = \sqrt{2 \times 9.81 \times 6.371 \times 10^6} = \sqrt{1.250 \times 10^8} = 1.12 \times 10^4\ \mathrm{m/s} \approx 11.2\ \mathrm{km/s}$ . Why the direction never appeared: energy is a scalar — (3) contains speed, not velocity — so any launch direction that avoids hitting the planet gives the same verdict.

Step 5. Read the anatomy: $v_{\rm esc} = \sqrt2\, v_{\rm circ}$ , where $v_{\rm circ} = \sqrt{GM/R}$ is the circular-orbit speed at the same radius. Why: escape needs kinetic energy equal to the full well depth $GMm/R$ ; a circular orbit carries exactly half that. The factor $\sqrt2$ between orbiting and escaping recurs everywhere in astronautics. *As a check: $2 \times 9.81 \times 6.371 \times 10^6 = 1.250 \times 10^8$ , whose square root is $11{,}180\ \mathrm{m/s}$ ✓; the Moon's escape speed is confirmed in worked example 1.*

Go deeper · Derivation C · 1 minThe energy of an orbit

Derivation CThe energy of an orbit

We are deriving the most quoted line in orbital mechanics: a circular orbit of radius $r$ has total energy $E = -GMm/2r$ — negative (bound, as an orbit must be), and exactly half the potential energy. The strategy is to let gravity's centripetal role fix the kinetic energy, then add. The punchline to watch: kinetic and potential energies in orbit are locked in the 2-to-1 ratio $K = -E = -U/2$ (the virial relation for $1/r^2$ forces), which produces the paradox that losing energy makes a satellite speed up.

Step 1. Let gravity supply the centripetal demand: $\dfrac{GMm}{r^2} = \dfrac{mv^2}{r}$ , so $v^2 = \dfrac{GM}{r}$ . Why: on a circular orbit gravity is the only force and points at the centre — the circular-motion article's balance, reused.

Step 2. Convert to kinetic energy: $K = \tfrac12 mv^2 = \dfrac{GMm}{2r} = -\dfrac{U}{2}$ . Why: the orbit's speed is not free — the radius fixes it, and with it the 2-to-1 energy ratio.

Step 3. Add the potential energy:

E = K + U = \frac{GMm}{2r} - \frac{GMm}{r} = -\frac{GMm}{2r} \qquad (5)

Why: negative, so bound ✓; half the well depth at that radius, so escape from orbit needs a kinetic top-up equal to $|E|$ — one more factor that makes low orbit "halfway to anywhere."

Step 4. Extract the paradox. Drag (or a retro-burn) makes $E$ more negative, so by (5) $r$ shrinks — and by Step 1, $v = \sqrt{GM/r}$ grows. A satellite losing energy to thin air spirals down and speeds up. Why: the shrinking orbit releases potential energy twice as fast as the drag drains the total; the surplus becomes speed. *For elliptical orbits the same formula holds with the semi-major axis $a$ in place of $r$ : $E = -GMm/2a$ . The energy ratios $K = |E|$ and $U = 2E$ are consistent with the virial theorem for inverse-square forces.*

Which equation answers which question

you want	use	notes
energy to climb between radii	$\Delta U = GMm\left(\tfrac{1}{r_1} - \tfrac{1}{r_2}\right)$	reduces to $mgh$ only when $h \ll R$
bound or free?	sign of $E = \tfrac12 mv^2 - \tfrac{GMm}{r}$	one number, computed at any instant
escape from a surface	$v_{\rm esc} = \sqrt{2GM/R} = \sqrt{2gR}$ , eq. (4)	mass-free, direction-free
escape from an orbit	$v_{\rm esc}(r) = \sqrt2\,v_{\rm circ}(r)$	the ubiquitous $\sqrt2$
speed elsewhere on the path	solve (3) between two snapshots	no trajectory needed
orbit energy / de-orbit budget	$E = -GMm/2r$ , eq. (5)	drag makes satellites speed up
leftover speed after escape	$v_\infty^2 = v^2 - v_{\rm esc}^2$	from $E > 0$ in (3)

Limiting cases

Case	Result	Meaning
$h \ll R$	$\Delta U \to mgh$	the flat-Earth formula is the first Taylor term of the honest one
$v < v_{\rm esc}$ (radial throw)	turning point at finite $r$ , falls back	$E < 0$ : trapped below the rim
$v = v_{\rm esc}$	arrives at infinity with $v = 0$	$E = 0$ : the marginal, parabolic case
$v > v_{\rm esc}$	$v_\infty = \sqrt{v^2 - v_{\rm esc}^2}$	$E > 0$ : hyperbolic, speed to spare
$r \to \infty$	$U \to 0^-$	the well flattens to its rim
circular orbit	$K = -E = -U/2$	the 2-to-1 virial lock behind the drag paradox
$v_{\rm esc} \to c$	$R \to 2GM/c^2$	Michell's "dark star" — numerically the black-hole radius

Common trap"A rocket must reach 11.2 km/s to leave Earth"

Escape velocity is the speed a ballistic projectile — thrown once, then coasting — needs at Earth's surface. A rocket under continuous thrust is not ballistic and can leave Earth at walking pace, given enough fuel: it climbs a metre, holds, climbs another. What (4) really says is subtler and more useful: it fixes the total energy bill, $GMm/R$ per kilogram lifted out of the well, however you choose to pay it. Real missions pay in one early burst precisely because of energy accounting (burning fuel while deep in the well and moving fast extracts the most benefit per kilogram of propellant), so their burnout speeds do approach the ballistic number. Two companion misreadings to dodge: escape speed is not a speed you must maintain (past the burn you coast, slowing forever without ever stopping), and escaping is not "leaving Earth's gravity" — the field has infinite range; you merely outrun its finite energy bill.

Worked examples

1 · Escape from the Moon (confidence builder)

The Moon has $M = 7.35 \times 10^{22}\ \mathrm{kg}$ and $R = 1.74 \times 10^6\ \mathrm{m}$ . Find its escape speed, and compare the energy cost per kilogram of escaping the Moon versus escaping the Earth.

Step 1. Compute $GM$ : $(6.674 \times 10^{-11})(7.35 \times 10^{22}) = 4.91 \times 10^{12}\ \mathrm{m^3/s^2}$ .

Step 2. Apply (4): $v_{\rm esc} = \sqrt{2 \times 4.91 \times 10^{12}/1.74 \times 10^6} = \sqrt{5.64 \times 10^6} = \boxed{2.37\ \mathrm{km/s}}$ — about a fifth of Earth's.

Step 3. Energy per kilogram is the well depth: Moon $GM/R = 2.82\ \mathrm{MJ/kg}$ ; Earth $GM_E/R_E = 3.986 \times 10^{14}/6.371 \times 10^6 = 62.6\ \mathrm{MJ/kg}$ — twenty-two times costlier.

Step 4. Read the consequence: the Apollo ascent stage left the Moon on a single modest engine, while leaving Earth took a Saturn V. The ratio of well depths, not of surface gravities (which differ only six-fold), is the honest measure of how trapped you are — the well depth carries an extra factor of $R$ .

2 · A very fast throw, and the road to interplanetary space (exam level)

(a) A projectile is launched straight up from Earth's surface at $v_0 = 5.0\ \mathrm{km/s}$ (ignore drag and Earth's rotation). Find its maximum altitude, and compare with the flat-Earth prediction $v_0^2/2g$ . (b) A spacecraft coasts in low Earth orbit at $r = 6.65 \times 10^6\ \mathrm{m}$ , where $v_{\rm circ} = 7.74\ \mathrm{km/s}$ . What extra speed must one burn supply to escape Earth? (Use $GM_E = 3.986 \times 10^{14}\ \mathrm{m^3/s^2}$ , $R_E = 6.371 \times 10^6\ \mathrm{m}$ .)

Step 1. Part (a), conservation of energy between surface and apex (where $v = 0$ ): $\tfrac12 v_0^2 - \dfrac{GM}{R} = -\dfrac{GM}{r_{\max}}$ , so $\dfrac{GM}{r_{\max}} = \dfrac{GM}{R} - \dfrac{v_0^2}{2}$ .

Step 2. Numbers: $GM/R = 3.986 \times 10^{14}/6.371 \times 10^6 = 6.257 \times 10^7$ ; $\ v_0^2/2 = 1.25 \times 10^7$ . So $GM/r_{\max} = 5.007 \times 10^7$ , giving $r_{\max} = 3.986 \times 10^{14}/5.007 \times 10^7 = 7.96 \times 10^6\ \mathrm{m}$ .

Step 3. Altitude: $h = r_{\max} - R = 7.96 \times 10^6 - 6.37 \times 10^6 = \boxed{1.59 \times 10^6\ \mathrm{m} \approx 1590\ \mathrm{km}}$ .

Step 4. The flat-Earth formula predicts $h = v_0^2/2g = 1.25 \times 10^7/9.81 = 1270\ \mathrm{km}$ — a 20% underestimate, because it keeps charging full-strength gravity for a climb on which the field actually weakens by a third. (At 5 km of altitude the two answers agree to five figures; the divergence is the breakdown of $mgh$ .)

Step 5. Part (b): escape speed at the spacecraft's radius is $v_{\rm esc}(r) = \sqrt2\,v_{\rm circ} = 1.414 \times 7.74 = 10.95\ \mathrm{km/s}$ . The burn must supply $\Delta v = 10.95 - 7.74 = \boxed{3.21\ \mathrm{km/s}}$ .

Step 6. Compare the two legs: reaching that orbit from the ground cost roughly $8$ km/s of velocity (with gravity and drag losses); leaving Earth entirely from there costs only $3.2$ more. Low orbit really is "halfway to anywhere": the deepest, most expensive part of the well is the first few hundred kilometres.

Figure

One launch site, four fates — speed decides the conic

Origins · 1 minPierre-Simon Laplace and how this was discovered

Pierre-Simon Laplace

1749–1827 · Paris

Laplace was Newtonian gravity's great accountant: his five-volume Mécanique céleste rebuilt the solar system's motions from the inverse-square law with such completeness that Napoleon reportedly asked where God appeared in it ("I had no need of that hypothesis"). In the Exposition du système du monde (1796) he ran escape-velocity logic to its limit — a star 250 times the Sun's diameter at the Sun's density would have $v_{\rm esc}$ exceeding light speed, so "the largest luminous bodies in the universe may be invisible." John Michell had reached the same "dark star" conclusion in 1783, even proposing to find such objects by the wobble of luminous companions — essentially how black holes are found today. The Newtonian argument abuses light (which does not slow like a stone), yet the radius it predicts, $R = 2GM/c^2$ , is exactly the Schwarzschild radius general relativity would derive honestly in 1916 — a coincidence that keeps the calculation in every relativity course as the world's most instructive wrong derivation.

History

1687

— Newton's Principia supplies the force law and, with it, the finite well every escape calculation climbs out of.

1783

— John Michell's letter to Cavendish: a star can be massive enough that "all light emitted from such a body would be made to return towards it" — the dark star.

1796

— Laplace's Exposition du système du monde independently publishes the dark-star argument from escape-velocity logic.

1865

— Jules Verne's From the Earth to the Moon fires a crewed shell from a giant cannon at close to escape speed — wrong on survivability, right on the number.

1959

— Luna 1 becomes the first artifact to reach escape speed, leaving Earth's well for a heliocentric orbit.

1977

— Voyagers 1 and 2 launch; gravity assists top up their energy until both exceed the Sun's local escape speed — Voyager 1 crossed into interstellar space in 2012, coasting on a permanently positive

E

Going further

Texts. Halliday, Resnick & Walker, Fundamentals of Physics, Ch. 13 · Young & Freedman, University Physics, Ch. 13 · Kleppner & Kolenkow, An Introduction to Mechanics, Ch. 9 · Taylor, Classical Mechanics, Ch. 4 and 8 for the energy method and the road to orbits. Key numerical results derived above: the work integral (1) and its $mgh$ limit; $v_{\rm esc} = \sqrt{2 \times 9.81 \times 6.371 \times 10^6} = 11.2$ km/s (Earth) and $2.37$ km/s (Moon); well depths $2.82$ vs $62.6$ MJ/kg; worked example 2 gives $r_{\max} = 7.96 \times 10^6$ m, $h \approx 1590$ km vs the flat-Earth prediction of 1270 km, and $\Delta v = (\sqrt2 - 1) \times 7.74 = 3.21$ km/s; solar escape speed at 1 AU is $42.1$ km/s against Earth's orbital speed of $29.8$ km/s.

Builds on: Newton's Law of Universal Gravitation · Conservative Forces and Potential Energy · Conservation of Mechanical Energy. Leads to: The Effective Potential · Classification of Orbits by Energy · The Kepler Problem.

How can negative energy be physical? Energy can't be less than nothing.

Only energy differences are physical, and a zero must be declared before any number means anything — declaring it is bookkeeping, not physics. For gravitation the cleanest zero is "two masses infinitely separated, at rest," because every problem shares it. Relative to that reference, any actual pair of masses sits below zero: you would have to add energy to pull them apart to the reference state. So $U = -GMm/r$ reads "this configuration is $GMm/r$ joules short of being disassembled." Nothing is owed to the universe; the minus sign just records being below the rim. The same convention gives the hydrogen atom its $-13.6$ eV ground state — bound systems have negative energy under this zero by definition of bound.

Why does escape speed not depend on the launch direction? Surely straight up is easier than sideways.

Because the criterion is an energy statement, and energy is a scalar. Equation (3) contains the speed squared, not the velocity vector; gravity's pull is radial, so it drains kinetic energy at a rate that depends only on how your distance changes, not on which way you point. A sideways launch at 11.2 km/s follows a very different path from a vertical one — a long unwinding spiral versus a straight climb — but both arrive at infinity with exactly nothing left. The direction matters only through the practical caveat: the trajectory must not intersect the planet. (Real launches do prefer particular directions — eastward, with Earth's rotation — but that is about getting free speed from the spinning launch pad, not about the physics of escape.)

If Earth's gravity extends forever, how does anything truly "escape" it?

It doesn't — and that is the beauty of the calculation. Escape is not outrunning the field (impossible; it has infinite range) but outrunning its energy bill, which is finite: $GMm/R$ , the integral of an ever-weakening force. A projectile at escape speed is decelerated by Earth's pull at every instant of its infinite journey, forever — but the deceleration shrinks so fast with distance that the total speed lost converges to exactly 11.2 km/s. It always moves outward, merely ever more slowly. Compare a field that fell off as $1/r$ : its energy bill would diverge, no finite speed would escape, and the concept would not exist. Escape velocity is a gift of the inverse square.

Where exactly does U = mgh stop being valid, and how wrong is it?

The honest climb cost from radius $R$ to $R + h$ is $GMm\,h/[R(R+h)]$ , and $mgh$ replaces the denominator's $R(R+h)$ with $R^2$ — so the flat formula overcharges by the factor $(R+h)/R$ , about $h/R$ fractionally. That is 0.16% at airliner altitude ( $h = 10$ km), 6% at the ISS ( $h = 400$ km), and 25% for worked example 2's 1590 km throw. The rule of thumb: below about 60 km ( $h/R < 1\%$ ) the flat formula is fine for anything but precision work; above a few hundred kilometres it is materially wrong; and for any escape or orbit-energy question it is useless at every altitude, because those questions probe the well's total shape, not a local slope.

Is the "dark star" argument legitimate physics? Light doesn't obey Newtonian energy conservation.

The reasoning is illegitimate and the answer is right — which is why it endures. Michell and Laplace treated light as ballistic corpuscles that slow as they climb; really, light always moves at $c$ and instead loses energy by redshift, and near such compact objects Newtonian gravity itself fails. Yet setting $v_{\rm esc} = c$ in equation (4) yields $R = 2GM/c^2$ — precisely the Schwarzschild radius that general relativity derives correctly. The agreement is a numerical accident (two compensating factors of 2), but a productive one: it says black holes are what escape-velocity logic becomes when gravity is treated honestly, and it hands you the right scale — 3 km per solar mass — from UG-level algebra. Treat it as a memorable estimate wearing borrowed rigor, and let the relativity track do it properly.

Why does a circular orbit have exactly half its potential energy as kinetic energy? The 2-to-1 ratio seems too neat.

It is the fingerprint of the inverse square, through one line of algebra: the centripetal balance $GMm/r^2 = mv^2/r$ , multiplied through by $r/2$ , is the statement $K = GMm/2r = -U/2$ . A different force law would give a different ratio (this is the virial theorem: for $F \propto r^n$ , $2K = (n+1)\,U$ -type relations). The neatness has audible consequences: because $E = -K$ , adding energy to an orbit lowers the kinetic energy once the orbit re-circularizes — thrust forward and you end up moving slower, in a higher orbit. Rendezvous pilots genuinely have to unlearn car intuition: to catch a spacecraft ahead of you, you brake, drop lower, and let the faster inner orbit carry you forward.

What would it take to escape the Sun starting from Earth's orbit? Voyager did it — how?

At Earth's distance from the Sun ( $r = 1.50 \times 10^{11}$ m, $GM_\odot = 1.327 \times 10^{20}$ ), the solar escape speed is $\sqrt{2GM_\odot/r} = 42.1$ km/s. Earth itself orbits at 29.8 km/s, so a probe launched in Earth's direction of motion needs about 12.3 km/s more heliocentric speed — beyond any single launcher once you have also paid Earth's own 11.2. Voyager's answer was the gravity assist: swinging close behind a moving planet, the probe steals a whisper of the planet's orbital energy — an elastic collision in gravity's currency. Jupiter alone added roughly 16 km/s to Voyager 1's heliocentric speed. Every flyby is pure equation-(3) bookkeeping done in two frames, and the result is a permanently positive solar $E$ : the probes are on hyperbolas with respect to the Sun, gone for good.