Gravitational potential energy — U = −GMm - r (not mgh): Visual walkthrough

Step 1 — Two masses, one pull

WHAT. Put a big mass $M$ (think: a planet) at a fixed point we call the origin — the centre of our picture. Put a small mass $m$ (think: a ball) a distance $r$ away.

WHY. Before we can talk about energy, we must know the force. Energy is "stored-up work," and work comes from force. So we start with the pull.

PICTURE. Look at the figure. The straight line from $M$ to $m$ is the distance $r$ . The little pale-yellow arrow $\overset{r}{^}$ ("r-hat") is a unit vector: an arrow of length exactly $1$ that only tells us a direction — it points away from $M$ , outward. The pink arrow is the actual gravitational force $F$ , and it points the other way, inward, toward $M$ .

Figure — Gravitational potential energy — U = −GMm - r (not mgh)

Step 2 — Write the force with its minus sign

WHAT. Newton's Law of Universal Gravitation tells us the size of the pull is $\frac{GM m}{r ^{2}}$ , and the direction is inward. Put together: $F = inward - how strong \frac{GM m}{r ^{2}} which way (outward) \overset{r}{^} .$

WHY the minus? $\overset{r}{^}$ points outward, but gravity pulls inward — the exact opposite. Multiplying by $- 1$ flips the outward arrow around to point inward. The minus sign is literally "reverse the direction of $\overset{r}{^}$ ."

WHY $1/ r^{2}$ ? Double the distance and the pull drops to a quarter. This "inverse-square" fall-off is the fingerprint of gravity, and it is what makes the final energy formula look the way it does.

PICTURE. The figure shows three snapshots of $m$ at $r$ , $2 r$ , $3 r$ . The pink force arrow shrinks fast — at $2 r$ it is $1/4$ as long, at $3 r$ it is $1/9$ as long. That shrinking is $1/ r^{2}$ made visible.

Step 3 — What "potential energy" even means

WHAT. We define the change in potential energy as minus the work the force does: $Δ U = U_{B} - U_{A} = - work done by gravity \int_{A}^{B} F \cdot d r .$

WHY this definition? Because gravity is a conservative force — the work it does depends only on the start and end points, never the wiggly path. That lets us "bottle" the work into a single number $U$ that depends on position alone. The minus sign is a bookkeeping choice: when gravity does positive work (pulling $m$ in), we want the stored energy to go down, like a spent battery.

WHY the $\int$ (integral) and not just force $\times$ distance? Because the force changes as $m$ moves (Step 2 showed it shrinking). You cannot multiply a changing force by a distance. The integral is the tool that adds up force-times-tiny-step over a path where the force keeps changing — it is a running total of little pushes.

WHAT is $F \cdot d r$ ? The dot ( $\cdot$ ) is the dot product: it takes only the part of the force that lies along the direction of motion. If you push sideways to the motion, no work is done. Here we will move straight along $\overset{r}{^}$ , so the dot product will simplify nicely in the next step.

PICTURE. The figure shows a curvy path from $A$ to $B$ chopped into tiny straight steps $d r$ . At each step we grab the piece of $F$ pointing along that step (the shadow of the pink arrow on the blue step) and add them all up.

Step 4 — Move straight out, so the dot product collapses

WHAT. Choose the simplest possible path: slide $m$ straight outward along the line through $M$ . Then every tiny step is $d r = d r \overset{r}{^}$ — a step of length $d r$ in the outward direction. Feed this into the force: $F \cdot d r = (- \frac{GM m}{r ^{2}} \overset{r}{^}) \cdot (d r \overset{r}{^}) = - \frac{GM m}{r ^{2}} = 1 (\overset{r}{^} \cdot \overset{r}{^}) d r = - \frac{GM m}{r ^{2}} d r .$

WHY are we allowed to pick this path? Because gravity is conservative (Step 3), the answer does not depend on the path. So we choose the laziest one — a straight radial line — and the messy dot product turns into ordinary numbers.

WHY does $\overset{r}{^} \cdot \overset{r}{^} = 1$ ? The dot product of an arrow with itself is (length) $^{2}$ . Here the length is $1$ , so $\overset{r}{^} \cdot \overset{r}{^} = 1$ . The two unit vectors "cancel," leaving a plain scalar.

PICTURE. The figure shows the force arrow $F$ (inward, pink) and the step $d r$ (outward, blue) lying on the same line but pointing opposite ways. Same line ⇒ the dot product is just the product of the signed lengths; opposite ways ⇒ the result is negative.

Step 5 — Pin the zero of energy at infinity

WHAT. Potential energy only ever appears as a difference ( $Δ U$ ), so we get to choose where " $U = 0$ " lives. We choose: $U (\infty) = 0.$ Then the energy at any distance $r$ is $U (r) = U (r) - = 0 U (\infty) = - \int_{\infty}^{r} F \cdot d r .$

WHY infinity? For an inverse-square force, infinity is the only natural landmark — it's the one place where the pull has fully faded to nothing. Any other choice would be arbitrary and would spoil the clean formula. "Free of the pull ⇒ zero energy" is the cleanest possible reference.

PICTURE. The figure is a horizontal energy line. Far to the right (huge $r$ ) the level sits at $0$ . As we walk left toward $M$ we will be walking downhill, into a valley — that valley is what the next step computes.

Step 6 — Do the integral

WHAT. Substitute Step 4's result into Step 5 and integrate. (I rename the running distance $r^{'}$ so it doesn't clash with the endpoint $r$ .) $U (r) = - \int_{\infty}^{r} (- \frac{GM m}{r ^{'2}}) d r^{'} = GM m \int_{\infty}^{r} \frac{d r ^{'}}{r ^{'2}} .$

The two minus signs (one from the definition of $U$ , one from the inward force) cancel, leaving a clean positive $GM m$ out front. Now use the fact that the antiderivative of $\frac{1}{r ^{'2}}$ is $- \frac{1}{r ^{'}}$ : $U (r) = GM m [- \frac{1}{r ^{'}}]_{\infty}^{r} = GM m at r^{'} = r - \frac{1}{r} - = 0 (- \frac{1}{\infty}) = - \frac{GM m}{r} .$

WHY does the $\infty$ term vanish? $- 1/\infty$ is $- 1/ (huge) = 0$ . This is exactly why infinity was the smart reference: it kills one whole term.

PICTURE. The figure plots $U (r) = - GM m / r$ . It is a curve that plunges toward $- \infty$ as $r \to 0$ (deep in the well, right next to $M$ ) and rises gently toward the $0$ line as $r$ grows. The whole curve lives below zero.

Step 7 — Edge case: what happens as $r \to 0$ and $r \to \infty$ ?

WHAT. Check the two extremes of the curve so no reader is ever surprised.

As $r \to \infty$ : $U = - \frac{GM m}{r} \to 0^{-}$ (approaches zero from below). The masses are essentially free.
As $r \to 0^{+}$ : $U = - \frac{GM m}{r} \to - \infty$ . The well is bottomless in the ideal point-mass model.

WHY does $U \to - \infty$ not break physics? Real bodies have a finite radius; you cannot actually reach $r = 0$ because you hit the planet's surface first. The bottomless well is an idealisation for point masses — physically you stop at $r = R$ (the surface), a finite, sensible number.

WHY care about the sign of the approach? Because a common error (see below) is thinking bigger $r$ means "more negative." The figure kills that: as $r$ grows, the curve climbs toward zero — $U$ increases.

PICTURE. Same curve as Step 6, now with the two limits flagged: a pink arrow diving to $- \infty$ on the left, a blue arrow flattening onto the $0$ line on the right, and a shaded band marking the physically reachable region $r \geq R$ .

Step 8 — Sanity check: the well is $m g h$ up close

WHAT. Zoom into a tiny sliver of the curve near the surface, $r = R + h$ with $h ≪ R$ . The change in $U$ from surface to height $h$ : $Δ U = - \frac{GM m}{R + h} - (- \frac{GM m}{R}) = GM m \cdot \frac{h}{R ( R + h )} \approx GM m \cdot \frac{h}{R ^{2}} = m = g \frac{GM}{R ^{2}} h = m g h .$

WHY does this matter for a "derivation in pictures"? It shows the schoolbook $m g h$ is not a rival law — it is the straight-line tangent to our curve near $r = R$ . See Acceleration due to gravity g for $g = GM / R^{2}$ .

PICTURE. The figure zooms the well near $r = R$ . Over a short span the curve looks like a straight ramp; its constant slope is $m g$ , so climbing a height $h$ costs $m g h$ . Far from the surface the ramp bends back into the true curve — where $m g h$ fails and you must use $- GM m / r$ .

The one-picture summary

WHAT. One figure holding the whole story: the inward force (pink), the outward path to infinity (blue), the shaded work being integrated, the reference $U (\infty) = 0$ , and the resulting energy well $U = - GM m / r$ with its $m g h$ tangent near the surface.

Recall Feynman: retell the whole walkthrough to a 12-year-old

A planet is like the mouth of a deep round pit dug into space. A ball near the planet has fallen partway into that pit. To describe "how deep in the pit" we first ask: how hard does the pit pull? Answer — it pulls inward, and the pull gets weaker fast the farther out you go (twice as far, one-quarter the pull). To find the "depth energy" we imagine dragging the ball all the way out of the pit to infinitely far away, adding up every little tug of the pull along the way — that grand total is the energy. We decide that being fully out of the pit (infinitely far) counts as zero depth. Since anywhere inside the pit is below that, the number comes out negative: $U = - GM m / r$ . Right at the top rim, if you only hop up a little, the pit wall looks like a straight ramp, and hopping up a height $h$ costs $m g h$ — that is the flat-Earth shortcut. But for rockets leaving the whole pit, you need the real curved-well formula.

Connections

Newton's Law of Universal Gravitation — the force we integrated in Steps 2–4.
Conservative Forces and Potential Energy — why picking a lazy straight path (Step 4) is legal.
Work-Energy Theorem — the work we bottled into $U$ .
Escape Velocity — set total energy to $0$ using this well.
Kepler's Laws & Orbital Energy — reads $E = - GM m /2 r$ off this same curve.
Acceleration due to gravity g — the tangent slope of the well in Step 8.

Visual walkthrough — Gravitational potential energy — U = −GMm - r (not mgh)