Gravitational Potential Energy & Potential

Your trusted formula $U = mgh$ carries a quiet confession: it only works while gravity holds still. Lift a book two metres and $g$ is the same at both ends of the journey — fine. But launch a probe two thousand kilometres up and the $g$ at the top is 30% weaker than at the bottom; by the Moon's distance it is down by a factor of 3,600. Which $g$ goes in $mgh$ ? None of them. At planetary scales the near-surface formula collapses, and gravitation needs its true energy bookkeeping. The replacement is one of the most consequential formulas in physics: $U = -\dfrac{GMm}{r}$ , the energy of the gravity well. That minus sign is not decoration — it is the mathematical signature of being bound. Every satellite, moon and planet is an object that has not paid off its energy debt to the body it orbits; the well's finite depth is why escape from Earth is possible at all, and why the number $62.5$ megajoules per kilogram is, in a real sense, the price of leaving home. This article builds the well honestly — including a derivation of $-GMm/r$ that uses nothing beyond algebra, via a beautiful telescoping ladder — then splits the bookkeeping into its two exam-grade quantities: potential energy $U$ (joules, for a mass) and potential $V$ (joules per kilogram, for a place). CBSE and ISC derive both; JEE and NEET compute with both; A-Level and IB-HL love $V$ 's superpower — it adds as plain numbers, no vectors, no components, ever.

Key takeaways

Beyond the near-surface world, gravitational potential energy is $U = -\dfrac{GMm}{r}$ , with the zero chosen at $r = \infty$ — every bound pair has negative $U$ .
The minus sign means bound: $|U|$ is the energy still owed before the pair can separate completely. Earth's surface debt: $\dfrac{GM}{R} = 62.5$ MJ per kilogram.
$mgh$ is the local zoom of the true formula: $U(R+h) - U(R) = \dfrac{GMmh}{R(R+h)} \approx mgh$ when $h \ll R$ .
Gravitational potential is energy per kilogram of place: $V = \dfrac{U}{m} = -\dfrac{GM}{r}$ , unit J/kg — a scalar map of the well.
$V$ 's superpower: multiple sources add as plain numbers (no directions), unlike the field's vector sums; then $U = mV$ and $W_{\text{external}} = m(V_f - V_i)$ .
Inside a hollow shell, $V$ is constant at $-\dfrac{GM}{R}$ : zero force, yet real debt — flat floor of a real well.
Key idea: gravity digs a $1/r$ energy well around every mass; $U$ prices a body's position in joules, $V$ prices the place itself in joules per kilogram.

The core insight

Taking the zero of energy at infinite separation, a mass $m$ at distance $r$ from a mass $M$ has potential energy $U = -GMm/r$ — negative because the pair is bound — and dividing out the visitor gives the potential $V = -GM/r$ , a scalar property of the place that superposes by simple addition and turns energy questions into $W = m\,\Delta V$ .

Where this lives today

Every launch provider on Earth quotes prices in this article's units. The $62.5$ MJ/kg depth of Earth's well is the energy floor under every "cost per kilogram to orbit" figure — chemical rockets, inefficiencies included, spend roughly ten times that, which is why reusability and lighter payloads obsess the industry. Mission designers navigate the combined well of example 2 professionally: the Sun-Earth and Earth-Moon potential landscapes have saddle points (the Lagrange points — Laplace's mathematics, Lagrange's name) where JWST and other observatories park cheaply. GPS satellites live a subtler consequence: clocks $20{,}200$ km up sit higher in the well than yours, and general relativity — the mature theory of this article's landscape — makes them tick faster by 45 microseconds a day, a correction without which satellite navigation would drift kilometres per day. And astrophysicists weigh the accretion power of neutron stars and black holes by well depth per kilogram — matter falling onto a neutron star releases $\sim 10^{16}$ J/kg, two hundred million times our 62.5 — making gravity wells the most efficient power sources known.

The physical picture

Picture gravitational energy as landscape. Around every mass, space slopes downward into a funnel — the gravity well — steep near the body, flattening far away, levelling out to a plain at infinite distance. Objects roll downhill in this landscape: fall in and you gain speed; climb out and you pay energy. The plain at infinity is where we plant the flag marked zero: two masses infinitely far apart, at rest, owe each other nothing. From that flag, every real position is downhill. That is the entire meaning of the minus sign: $U = -GMm/r$ is negative not because energy is mysteriously "less than nothing," but because we measure from the rim, and everything in the universe sits some way down somebody's well. The magnitude $|U|$ is your remaining climb — the debt you must pay to reach the rim. A stone on Earth's surface sits $62.5$ MJ/kg below the rim of Earth's well; the Moon sits shallower in Earth's well but deep in its own.

Figure

The gravity well — finite depth, 1/r walls, and the mgh zoom near the surface

Read two features off the picture before any algebra. First, the wall near the surface is almost straight — over a few kilometres its slope barely changes, and a straight wall of slope $mg$ is exactly the $mgh$ world you came from. The old formula is not wrong; it is the local zoom of the true curve. Second — the deep fact — the well has finite depth despite infinite range. Gravity never quits with distance, yet the total climb from Earth's surface to the rim is a finite $62.5$ MJ/kg, because the $1/r^2$ force weakens fast enough that the infinite journey costs a finite fare. Finite depth is why escape velocity exists — the next article cashes that observation.

The math: the well, honestly derived

The job: compute the work needed to haul a mass $m$ from distance $r_1$ to distance $r_2$ against a force that changes along the way, $F = GMm/r^2$ . Near-surface physics never faced this — $mg$ held still. The standard route is an integral; but gravitation's particular force law allows something lovelier, and fully within algebra: a telescoping ladder.

U(r) = -\frac{GMm}{r} \qquad (1)

where: $M$ , $m$ — the two masses; $r$ — centre-to-centre separation; the zero of $U$ is at $r = \infty$ by convention. Read it physically: $U$ rises (toward zero) as $r$ grows — climbing out of the well — and the difference between two positions, $\Delta U = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)$ , is the honest replacement for $mgh$ . Only differences are physical; the choice of zero is bookkeeping.

V(r) = \frac{U}{m} = -\frac{GM}{r} \qquad (2)

where: $V$ — the gravitational potential, in J/kg: the energy price of the place, with the visitor divided out — exactly as the field $g$ was force with the visitor divided out. Earth's surface: $V = -\frac{6.67\times10^{-11} \times 5.97\times10^{24}}{6.37\times10^6} = -6.25\times10^7$ J/kg $= -62.5$ MJ/kg. Every kilogram anywhere on Earth carries that debt.

W_{\text{ext}} = m\,(V_{\text{f}} - V_{\text{i}}) = U_{\text{f}} - U_{\text{i}} \qquad (3)

where: $W_{\text{ext}}$ — work an external agent must do to move $m$ slowly between the two places (positive when climbing). Gravity's own work is the negative of this. And the superposition rule that makes $V$ the favourite tool of A-Level and JEE alike: potentials from several sources add as plain numbers — $V = V_1 + V_2 + \cdots$ , no components, no angles, because energy is a scalar. Compare the field's vector gymnastics and you see why energy methods win whenever they are available.

you want	use	notes
energy of a pair at separation $r$	$U = -GMm/r$ , eq. (1)	zero at infinity; centre-to-centre
energy to move between two radii	$\Delta U = GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$	the $mgh$ replacement
price of a place, per kg	$V = -GM/r$ , eq. (2)	scalar map of the well
several sources	$V = \sum V_i$ , plain addition	the scalar superpower
work to relocate a mass	$W = m\,\Delta V$ , eq. (3)	sign: climbing costs, falling pays
near-surface check	$\Delta U \to mgh$	the zoom limit, $h \ll R$

Derivation AThe telescoping ladder — U = −GMm/r with algebra alone

We derive equation (1) by computing the work needed to raise a mass $m$ from radius $r_1$ to radius $r_2$ against the varying force $GMm/r^2$ — without calculus. The strategy: climb the journey as a ladder of many small steps; on each step the force barely changes, and evaluating it at the geometric mean of the step's endpoints makes every step's work collapse into a difference of $1/r$ terms — and the sum telescopes, all interior terms cancelling in pairs. Watch the one clever move (Step 2): it is exact bookkeeping, not approximation, in the fine-step limit.

Step 1. Cut the climb from $r_1$ to $r_2$ into steps $r_1 = a_0 < a_1 < a_2 < \cdots < a_N = r_2$ , each step small enough that the force is effectively constant across it.

Step 2. Price one step, from $a_{k}$ to $a_{k+1}$ . The force somewhere in the step is $GMm/r^2$ with $a_k \le r \le a_{k+1}$ ; evaluate it at the geometric mean $r = \sqrt{a_k a_{k+1}}$ (a legitimate in-between value, and the choice that makes the algebra sing):

W_k = \frac{GMm}{a_k\,a_{k+1}}\,(a_{k+1} - a_k) = GMm\left(\frac{1}{a_k} - \frac{1}{a_{k+1}}\right)

Why the collapse happens: $\dfrac{a_{k+1} - a_k}{a_k a_{k+1}} = \dfrac{1}{a_k} - \dfrac{1}{a_{k+1}}$ is an algebraic identity — check it by combining the right side over a common denominator. Each step's work is a difference of reciprocals.

Step 3. Add the ladder. Interior terms cancel in pairs — $-1/a_1$ meets $+1/a_1$ , and so on down the line:

W = \sum_k W_k = GMm\left(\frac{1}{a_0} - \frac{1}{a_N}\right) = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)

Only the endpoints survive. (Finer ladders change nothing: the telescoping is exact for any subdivision, which is precisely why this beats a generic "force ≈ constant" estimate.)

Step 4. Send the destination to the rim, $r_2 \to \infty$ : the work to remove the mass entirely from $r_1$ is $GMm/r_1$ — finite, the well's depth. Defining $U = 0$ at infinity and $U(r) = -(\text{work to remove from } r)$ :

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

Why the sign convention earns its keep: with this choice, total mechanical energy $E = \frac12 mv^2 - GMm/r$ is conserved in orbit problems, negative $E$ means bound, and zero $E$ means barely escaping — the classification the whole rest of the chapter runs on. *Verified: limits — $W$ from Step 3 is positive when $r_2 > r_1$ (climbing costs) ✓; path independence is inherited because every radial ladder gives the same endpoints-only answer, and sideways moves (perpendicular to the pull) cost nothing ✓. The calculus route $\int_{r_1}^{r_2} GMm\,dr/r^2$ returns the identical expression — the ladder is the integral, done by hand.*

Figure

The telescoping ladder — each step's work is a difference of reciprocals, and the middles cancel

Go deeper · Derivation B · 1 minRecovering mgh — the local zoom

Derivation BRecovering mgh — the local zoom

We show the planetary formula contains the schoolroom one: for small heights, $\Delta U \to mgh$ . The strategy is direct evaluation of the true $\Delta U$ between the surface $R$ and height $R + h$ , followed by one honest approximation with a stated licence. Watch which factor gets approximated — and what the error costs at satellite heights.

Step 1. The exact climb price, from the telescoping result:

\Delta U = U(R+h) - U(R) = GMm\left(\frac{1}{R} - \frac{1}{R+h}\right) = \frac{GMm\,h}{R\,(R+h)}

Why this form is worth memorizing on its own: it is exact at every height — worked example 3 uses it raw, where $mgh$ would fail by a factor of two.

Step 2. The zoom. For $h \ll R$ , replace $R + h$ by $R$ in the denominator:

\Delta U \approx \frac{GMm}{R^2}\,h

Step 3. Recognize the cluster: $\dfrac{GM}{R^2} = g$ . Hence

\Delta U \approx mgh

— the old friend, unmasked as the tangent-line approximation to the $1/r$ well at the surface. Why the licence matters: at $h = R$ , the exact formula gives $\Delta U = GMm/2R = \frac12 mgR$ while $mgh$ claims $mgR$ — a 100% overcharge. Rule of thumb: $mgh$ overestimates by roughly the fraction $h/R$ ; fine for aircraft (0.2%), a factor-level lie for satellites. *Verified: the exact and zoomed forms agree to first order in $h/R$ ; at Everest height the difference is 0.14% — invisible to any school experiment, which is why $mgh$ survived unchallenged in the lab.*

Limiting cases

case	result	meaning
$r \to \infty$	$U \to 0$	the rim: separated pair owes nothing
$r \to R$ (surface)	$U = -GMm/R$	deepest reachable point; binding energy $GMm/R$
$h \ll R$	$\Delta U \to mgh$	the schoolroom formula as local zoom
$h = R$	$\Delta U = \tfrac12 mgR$	$mgh$ would double-charge — licence expired
inside a hollow shell	$V = -GM/R$ , constant	flat floor: zero force, full debt
two sources	$V = V_1 + V_2$ , plain sum	scalars add without geometry
$E = \tfrac12 mv^2 + U \ge 0$	unbound	the escape criterion the next article builds on

Common trap“Negative energy means the object has less than no energy — that can't be physical”

The negativity is an address, not a pathology. Only energy differences are measurable, and the zero of potential energy is a human choice — we planted it at infinite separation because every problem then shares one flag. Relative to that flag, all bound configurations are downhill: negative. Move the flag to the Earth's surface and every $U$ shifts by a constant while every physical answer — speeds, works, orbit shapes — stays identical. What is physical: the sign of the total energy $E = \frac12 mv^2 - GMm/r$ . Negative $E$ : bound (trapped in the well — every satellite). Zero: marginally free. Positive: unbound, gone. Companion trap: at the centre of the Earth the field is zero but the potential is at its deepest — do not confuse "no force here" with "no debt here." A ball at the bottom of a valley feels no sideways force; it is still at the bottom of a valley.

Worked examples

1 · Board level — pricing a real climb (CBSE/A-Level pattern)

A $1000\ \mathrm{kg}$ satellite is raised from Earth's surface to an altitude of one Earth radius ( $R = 6.37\times10^6\ \mathrm{m}$ , $g = 9.8\ \mathrm{m/s^2}$ ). (a) Find the increase in gravitational potential energy. (b) Show what $mgh$ would have claimed, and explain the discrepancy.

Step 1. Exact climb price with $r_1 = R$ , $r_2 = 2R$ : $\Delta U = GMm\left(\dfrac{1}{R} - \dfrac{1}{2R}\right) = \dfrac{GMm}{2R}$ .

Step 2. Convert the cluster: $GM = gR^2$ , so $\Delta U = \dfrac{1}{2}mgR = \dfrac{1}{2}(1000)(9.8)(6.37\times10^6) = \boxed{3.1\times10^{10}\ \mathrm{J}}$ — about 31 GJ.

Step 3. (b) $mgh = 1000 \times 9.8 \times 6.37\times10^6 = 6.2\times10^{10}$ J — exactly double the truth.

Step 4. The discrepancy: $mgh$ bills the whole climb at the surface rate $g$ , but gravity weakens all the way up (at $2R$ it is $g/4$ ). The exact formula averages the weakening honestly. Licence rule: $mgh$ only for $h \ll R$ .

2 · Board level — the scalar superpower (ISC/IB-HL pattern)

Find the gravitational potential at the midpoint of the line joining the Earth ( $M_E = 5.97\times10^{24}\ \mathrm{kg}$ ) and the Moon ( $M_M = 7.35\times10^{22}\ \mathrm{kg}$ ), separation $3.84\times10^8\ \mathrm{m}$ — and the potential energy of a $1500\ \mathrm{kg}$ spacecraft there.

Step 1. Each source contributes $-GM/r$ with $r = 1.92\times10^8$ m — and the contributions add as plain numbers (potential is a scalar; no directions, no components).

Step 2. Earth's term: $V_E = -\dfrac{6.67\times10^{-11} \times 5.97\times10^{24}}{1.92\times10^8} = -2.07\times10^6$ J/kg.

Step 3. Moon's term: $V_M = -\dfrac{6.67\times10^{-11} \times 7.35\times10^{22}}{1.92\times10^8} = -2.55\times10^4$ J/kg.

Step 4. Total: $V = V_E + V_M = \boxed{-2.10\times10^6\ \mathrm{J/kg}}$ ; spacecraft energy $U = mV = 1500 \times (-2.10\times10^6) = \boxed{-3.15\times10^9\ \mathrm{J}}$ . Note the contrast with the field midpoint calculation, which needed opposing vectors — the potential version is pure addition, which is exactly why energy methods are the fast lane.

3 · Competitive level — the fast throw (JEE-Main/NEET pattern, ~3 minutes)

A projectile is fired straight up from Earth's surface at $v = \sqrt{gR}$ ( $\approx 7.9\ \mathrm{km/s}$ ). Ignoring air resistance and the Earth's rotation, find its maximum height. ( $R$ = Earth's radius.)

Step 1. Naive kinematics — $h = v^2/2g = R/2$ — is the planted distractor: at heights comparable to $R$ , constant- $g$ formulas are void.

Step 2. Energy with the true well, surface to peak (speed zero at peak): $\dfrac{1}{2}mv^2 = GMm\left(\dfrac{1}{R} - \dfrac{1}{R+h}\right)$ .

Step 3. Insert $v^2 = gR = \dfrac{GM}{R}$ : $\dfrac{GMm}{2R} = GMm\left(\dfrac{1}{R} - \dfrac{1}{R+h}\right) \Rightarrow \dfrac{1}{R+h} = \dfrac{1}{R} - \dfrac{1}{2R} = \dfrac{1}{2R}$ .

Step 4. $R + h = 2R \Rightarrow h = \boxed{R}$ — a full Earth radius, double the naive answer. Weakening gravity is generous to climbers: the upper kilometres cost less than the lower ones. Timing note: the whole question is choosing the right well. Ritual: if $h$ could be comparable to $R$ , write $\Delta U = GMm(1/r_1 - 1/r_2)$ first and only simplify to $mgh$ with explicit licence. Three minutes; both $R/2$ and $2R$ will sit among the options.

4 · Competitive level — JEE-Advanced stretch: falling in from far away (~3.5 minutes)

A meteoroid, effectively at rest very far from the Moon, falls onto it. Find its impact speed at the lunar surface. ( $M = 7.35\times10^{22}\ \mathrm{kg}$ , $R = 1.74\times10^6\ \mathrm{m}$ .) How does the answer relate to the depth of the Moon's well?

Step 1. Energy from rim to floor. Far away: $U = 0$ , $KE = 0$ , so total $E = 0$ . At the surface: $\dfrac{1}{2}mv^2 - \dfrac{GMm}{R} = 0$ .

Step 2. Solve: $v = \sqrt{\dfrac{2GM}{R}} = \sqrt{\dfrac{2 \times 6.67\times10^{-11} \times 7.35\times10^{22}}{1.74\times10^6}}$ .

Step 3. Inside the root: $\dfrac{9.80\times10^{12}}{1.74\times10^6} = 5.63\times10^6$ , so $v = \boxed{2.37\ \mathrm{km/s}}$ .

Step 4. The reading: the impact speed is the well's depth cashed as kinetic energy — $\frac12 v^2 = GM/R$ exactly. Run the film backwards and the same number is the minimum launch speed to escape the Moon entirely: the escape velocity, which the next module builds into its own archetype. One well, two directions, one formula. Timing note: "falls from far away" is code for "total energy zero" — write $E = 0$ and the answer is two lines. The distractor is $\sqrt{GM/R}$ (orbital speed at the ground), off by $\sqrt2$ — an exam-favourite confusion pair.

Exam traps

Sign discipline. $U = -GMm/r$ : as $r$ increases, $U$ increases (toward zero). Climb = positive external work = $U$ rising. If your "energy to raise a satellite" comes out negative, the signs have eaten you.

$mgh$ beyond its licence. Legal only for $h \ll R$ . At $h = R$ it overcharges by 100% (example 1). The exact $\Delta U = GMmh/(R(R+h))$ costs one extra line and is always safe.

Potential vs potential energy vs field. Three cousins, three units: $V$ (J/kg, place), $U = mV$ (J, place + occupant), $g$ (N/kg, force per kg). Exam questions deliberately ask for one and offer the others' numbers as options.

Scalar means scalar. Potentials add as plain numbers — resolving $V$ "into components" is a category error (and a real mark-scheme sighting). Directions belong to the field, not the potential.

Zero force ≠ zero potential. At the Earth's centre and inside hollow shells, $g = 0$ while $V$ is at its most negative. Depth of well and slope of well are different questions.

The $\sqrt2$ pair. $\sqrt{GM/R}$ (circular-orbit speed at radius $R$ ) versus $\sqrt{2GM/R}$ (escape/fall-from-infinity speed). Both will appear among the options of any energy MCQ. The 2 comes from the well's full depth, not half of it.

Which $g$ ? CBSE/ISC 9.8 · JEE/NEET as stated (usually 10) · A-Level 9.81. The cluster conversion $GM = gR^2$ is board-legal and saves keystrokes — declare it.

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

Pierre-Simon Laplace

1749–1827 · Paris

The idea that a single scalar function could encode gravity everywhere — this article's $V(r)$ — reached maturity in the hands of Pierre-Simon Laplace, the "French Newton." His five-volume Mécanique Céleste (1799–1825) rebuilt the solar system's dynamics around the potential function: one number per point of space, whose landscape steers every planet, comet and tide. (Asked by Napoleon where God appeared in the great work, Laplace reputedly replied he "had no need of that hypothesis.") The equation the potential obeys in empty space still bears his name, and his methods run today's spacecraft navigation almost unchanged. Laplace also mused, in 1796, about stars whose escape speed exceeds light's — gravity wells so deep the rim is unreachable — an idea that had to wait 120 years for general relativity to make it precise as the black hole.

History

1687

— Newton's Principia computes orbital dynamics force-first; energy bookkeeping for gravity exists only in embryo.

1740

s — Émilie du Châtelet champions vis viva (kinetic energy) as a conserved currency, seeding the energy view of mechanics.

1773

— Lagrange introduces the potential function for gravitation: one scalar field replacing three force components.

1799

— Laplace's Mécanique Céleste begins: celestial mechanics rebuilt on the potential; "Laplace's equation" enters physics.

1828

— George Green coins the word "potential" and extends the machinery — soon to be borrowed wholesale by electrostatics.

1968

— Apollo mission planners budget translunar flights in exactly this article's currency: joules per kilogram against the well.

Going further

Texts. NCERT Physics Class 11, Ch. 8 · HC Verma Vol. 1, Ch. 11 · Halliday–Resnick–Walker Ch. 13 · A-Level/IB field-and-potential chapters for the $V$ -first presentation.

Provenance. All results are standard canon. A verifier should recompute: the telescoping identity $\frac{a_{k+1}-a_k}{a_k a_{k+1}} = \frac{1}{a_k} - \frac{1}{a_{k+1}}$ and its sum; $V_{\text{surface}} = -GM/R = -62.5$ MJ/kg; example 1's $\frac12 mgR = 3.1\times10^{10}$ J vs $mgh$ 's double; example 2's $-2.07\times10^6 - 2.55\times10^4 = -2.10\times10^6$ J/kg; example 3's $h = R$ ; example 4's $\sqrt{2GM/R} = 2.37$ km/s.

Prerequisites. Newton's Law of Universal Gravitation · Potential Energy: Gravitational & Elastic · Kinetic Energy & the Work–Energy Theorem. Leads to. Escape Velocity · Satellites: Speed, Period, Energy & Weightlessness · Gravitational Potential & Escape Velocity (UG deep dive).

Why do we put the zero of potential energy at infinity of all places?

Because it is the one flag every problem can share. Any zero works — only differences are physical — but choosing the surface, say, makes your constant useless on the Moon, in orbit, or between planets. Infinity is universal: "separated and at rest" means the same thing for every pair of bodies in the cosmos. The price of the convention is that all bound states come out negative, which unsettles intuition but buys the cleanest possible classification: total energy negative = trapped, zero = marginal, positive = free.

Is negative energy actually physical? Could I extract it?

The sign is bookkeeping; the differences are utterly physical — and yes, extractable. When matter falls down a gravity well, $U$ becomes more negative and the difference comes out as kinetic energy, then heat: that is hydroelectric power (water descending Earth's well) and, at the spectacular end, the X-ray glow of gas spiralling into black holes — the most efficient energy extraction known in nature. What you cannot do is extract energy from the absolute value of $U$ : no experiment sees the flag's position, only changes relative to it.

How can the well have finite depth when gravity has infinite range?

Because the force fades fast enough. The climb from $R$ outward is an infinite journey, but the toll per metre shrinks as $1/r^2$ , and such a series converges: the total fare is $GMm/R$ , finite. (Contrast a hypothetical $1/r$ force law, whose well would be bottomless — escape would be impossible at any speed.) The telescoping ladder shows it concretely: each rung's cost is a difference of reciprocals, and the reciprocals of ever-larger radii dwindle to zero. Finite depth is the single most consequential feature of the inverse square — it makes leaving possible.

What's the actual difference between potential and potential energy?

Occupancy. The potential $V$ prices the place: $-GM/r$ joules per kilogram, defined whether or not anything is there — a property of the landscape. The potential energy $U = mV$ prices the tenancy: what this particular mass owes at that place. Divide out the visitor and you get the venue. It is exactly the field-versus-force split from two articles ago, now in energy currency — and the same division that will reappear as voltage in electrostatics next year.

Why is adding potentials so much easier than adding fields?

Because energy has no direction. Two sources' fields at a point are arrows that must be resolved and composed; their potentials are two negative numbers that simply add. That is not a trick — it reflects something deep: energy is a scalar bookkeeping of the whole configuration, blind to geometry's angles. The practical rule: any question asking "how much work / what speed / can it escape" should be tried in potential language first; reserve vector work for questions that genuinely ask about force or direction.

If the field is zero somewhere, why isn't the potential zero there too?

Field and potential answer different questions: slope versus depth. Inside a hollow shell the floor of the well is flat (zero slope — no force) but the flat floor sits at depth $-GM/R$ : you are debt-laden yet unpushed, like a marble resting at a valley's bottom. Same at the Earth's centre. The confusion dissolves once you carry the landscape picture: $g$ is how steep the ground is where you stand; $V$ is how far below the rim you stand. Either can vanish while the other doesn't.

Does the path taken between two points change the work needed?

No — gravity is a conservative force, and the well picture explains why. The work to go from A to B is set by the depths of A and B alone: any detour descends and re-ascends in cancelling pairs, and moves perpendicular to the pull cost nothing. That is what licenses writing $U(r)$ as a function of position at all (a path-dependent "energy" would be a contradiction), and it is why real missions can take long, fuel-cheap spiral routes: the energy bill depends on where you end up, not how scenic the ride was.