Visual walkthrough — Potential energy — definition, gravitational (mgh and −GMm - r), elastic (½kx²)

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Step 0 — The one idea behind all three

Before any formula, fix the single sentence that generates everything:

The one master rule (see Work done by a force):

$\underset{\text{change in stored energy}}{\underbrace{\Delta U}}=-\underset{\text{work the force did}}{\underbrace{W}}=-\underset{\text{sum of tiny works from }i\text{ to }f}{\underbrace{{\int }_i^f\vec{F}\cdot d\vec{r}}}$

• $\Delta U$ — how much the stored energy changed (final minus initial).
• $W$ — the total work the force itself did.
• The minus sign — if the force helps the motion (positive $W$), it is spending the store, so $U$ goes down. That is the whole reason for the minus.
• $\int \dots d\vec{r}$ — the symbol $\int$ ("integral") means "add up an infinite number of tiny slices." We use it because the object moves through many tiny steps and we want the total.

Why an integral and not just $F\times d$? Because in two of our three cases the force changes as the object moves. When force is constant you may multiply; when it grows or shrinks you must slice-and-sum. That is exactly what an integral does — and geometrically, an integral is the area under the force-vs-position graph. Hold onto that picture; it carries this entire page.

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Step 1 — Constant force: the flat graph gives $mgh$

WHAT. Lift an object of mass $m$ straight up, from height $y=0$ to $y=h$, near the ground.

WHY start here. Near Earth's surface gravity barely changes, so the force is constant. A constant force is the simplest possible case — its graph is a flat horizontal line, and the "area under it" is just a rectangle. No calculus muscle needed yet.

PICTURE. In the figure, gravity is the downward arrow of size $mg$:

• $m$ — the mass (kilograms), "how much stuff."
• $g$ — the strength of gravity near the ground, about $9.8$ (metres per second squared).
• $mg$ — their product, the actual downward pull (in newtons).

The force graph is flat at value $mg$. The object travels a distance $h$ upward.

$\text{Area of rectangle}=\underset{\text{height of graph}}{\underbrace{mg}}\times \underset{\text{width of graph}}{\underbrace{h}}=mgh$

Because gravity points down while the lift goes up, the force does negative work, $W=-mgh$. Apply the master rule:

$\Delta U=-W=-(-mgh)=mgh$

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Step 2 — Varying force: the triangle graph gives $\frac{1}{2}k{x}^2$

WHAT. Stretch a spring slowly from its natural length ($x=0$) out to a stretch $x$.

WHY here. Now the force is not constant — the more you stretch, the harder the spring pulls back (Hooke's Law). So the graph is no longer flat; it is a sloped line. That is our first real "area under a slanted graph" — a triangle.

PICTURE. Hooke's law says the spring force is

$F=-kx$

• $x$ — how far the spring is stretched from natural length (metres).
• $k$ — the stiffness (newtons per metre): a big $k$ is a stiff spring.
• the minus — the force points back toward $x=0$, opposing your stretch. It "restores."

Plot the size of your applied force $kx$ against $x$: a straight ramp starting at $0$ and rising to $kx$ at stretch $x$. The area under a ramp is a triangle:

$\text{Area of triangle}=\frac{1}{2}\times \underset{\text{base}}{\underbrace{x}}\times \underset{\text{height}}{\underbrace{kx}}=\frac{1}{2}k{x}^2$

The spring's own force does $W=-\frac{1}{2}k{x}^2$ (it opposes the stretch), so:

$\Delta U=-W=\frac{1}{2}k{x}^2$

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Step 3 — Why stretch and squeeze cost the same

WHAT. Check the case $x<0$ (compression) as well as $x>0$ (stretch).

WHY. The contract says cover every sign. What happens when you push the spring in?

PICTURE. Because $U=\frac{1}{2}k{x}^2$ and squaring kills the sign, $U(-x)=U(+x)$. The energy curve is a symmetric parabola — a valley with its lowest point at $x=0$. Compressing $0.1\text{m}$ stores exactly what stretching $0.1\text{m}$ stores.

At $x=0$ (natural length) the stored energy is zero — the bottom of the valley. This is the degenerate case: no stretch, no store.

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Step 4 — A force that dies with distance: setting up $-\frac{GMm}{r}$

WHAT. Move a mass $m$ toward a planet of mass $M$, and ask how the stored energy behaves as the separation $r$ changes.

WHY a new tool. The spring's force grew linearly. Gravity in deep space does the opposite — it fades as you move away, following Gravitation — Newton's law:

$F=-\frac{GMm}{r^2}$

• $r$ — distance between the two centres (metres).
• $M,m$ — the two masses; both appear because gravity is a mutual pull.
• $G$ — the universal gravitational constant, a fixed tiny number.
• $\frac{1}{r^2}$ — the inverse-square shape: double the distance, quarter the force.
• the minus — the pull is inward, toward the planet.

PICTURE. The force graph is now a steep curve that plunges near the planet and flattens toward zero far away. Its area is no longer a rectangle or triangle — it is the area under a curved $1/r^2$ shape. That is why we need the full integral.

Why put the zero at infinity? For $mgh$ we set $U=0$ at the floor. Here there is no natural floor — but infinitely far away the pull is zero, so it is the cleanest place to declare "no stored energy." We measure everything relative to "escaped to infinity."

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Step 5 — Doing the curved-area sum: the formula appears

WHAT. Add up the tiny works as $m$ falls from infinity in to distance $r$.

WHY. This is the payoff of Step 4's graph: the integral is exactly the shaded curved area.

= -GMm\left[-\frac{1}{r'}\right]_{\infty}^{r} = -GMm\left(-\frac{1}{r}+0\right) = \frac{GMm}{r}$$ Reading it term by term: - $r'$ — a dummy stand-in for the varying distance while we sweep from $\infty$ to $r$. - $\left[-\dfrac1{r'}\right]$ — the "antiderivative": the function whose slope is $\dfrac1{r'^2}$. We recognise it because $\dfrac{d}{dr'}\!\left(-\dfrac1{r'}\right)=\dfrac1{r'^2}$. - plugging in $\infty$ gives $0$; plugging in $r$ gives $-\dfrac1r$. Gravity did **positive** work $+\dfrac{GMm}{r}$ (it *helped* the mass fall in), so by the master rule: $$U(r) = -W = -\frac{GMm}{r}$$ ![[deepdives/dd-physics-1.3.05-d2-s06.png]] > [!formula] Universal gravitational PE > $$U(r) = -\frac{GMm}{r} \qquad (U=0 \text{ at } r=\infty)$$ > The **negative** value means "below the infinity line" — the mass sits in a **well**. To climb back out to infinity you must *add* energy (that is [[Escape velocity|escape energy]]). > [!intuition] Why it *has* to be negative > Start at infinity where $U=0$. Fall inward — gravity does positive work, so $U$ must *drop*, going below zero. A deeper fall (smaller $r$) means a more negative $U$. See [[Conservative vs non-conservative forces]] for why this single-number $U$ even exists. --- ## Step 6 — The two gravity formulas are the same formula **WHAT.** Show $-\dfrac{GMm}{r}$ becomes $mgh$ when you only move a little. **WHY.** A smart reader worries: "two different formulas for gravity?" They must agree for small climbs. **PICTURE.** Zoom into the deep gravity curve right at the surface $r=R$. Over a tiny height $h$, the curve looks like a *straight ramp* — and a straight ramp is exactly the flat-force world of Step 1. $$\Delta U = -GMm\left(\frac{1}{R+h}-\frac{1}{R}\right)\approx GMm\,\frac{h}{R^2}=\left(\frac{GM}{R^2}\right)mh = mgh$$ using $g=\dfrac{GM}{R^2}$. - $R$ — Earth's radius (where you stand). - $h$ — small climb, tiny compared to $R$. - the approximation $\approx$ — valid only because $h\ll R$ flattens the curve. ![[deepdives/dd-physics-1.3.05-d2-s07.png]] So $mgh$ is just the **first tiny slice** of the big gravity well, seen so close up that the curve looks flat. --- ## Step 7 — Reading force back off the energy landscape **WHAT.** Recover force from any $U$ using the slope (see [[Force from potential — F = -dU/dx]]). **WHY.** We built $U$ *from* force; the reverse should also work — and it double-checks our answers. $$F_x = -\frac{dU}{dx}$$ - $\dfrac{dU}{dx}$ — the **slope** of the energy curve (how steeply $U$ rises as you move). - the **minus** — force points *downhill*, toward lower energy, like a ball rolling into a valley. **PICTURE.** On the spring parabola: at $x>0$ the curve slopes *up*, slope positive, so $F=-kx$ points back left. At $x<0$ it slopes down, force points right. Everywhere the arrow points toward the bottom of the valley. $$F=-\frac{d}{dx}\!\left(\tfrac12 kx^2\right)=-kx\;\checkmark$$ ![[deepdives/dd-physics-1.3.05-d2-s08.png]] The arrows on the landscape are exactly Hooke's restoring force we started from — the loop closes. This is [[Conservation of mechanical energy]] made visible: energy sloshes between the height of the curve ($U$) and speed ($KE$), never lost. --- ## The one-picture summary ![[deepdives/dd-physics-1.3.05-d2-s09.png]] Three forces, three graph shapes, three areas: | Case | Force graph | Area = $U$ | |---|---|---| | Near-Earth gravity | flat line $mg$ | rectangle → $mgh$ | | Spring | ramp $kx$ | triangle → $\tfrac12kx^2$ | | Deep-space gravity | curve $\dfrac{GMm}{r^2}$ | curved area → $-\dfrac{GMm}{r}$ | > [!recall]- Feynman retelling — the whole walkthrough in plain words > Potential energy is just "work you banked." To find how much you banked, draw a graph of the force against how far the thing moved, and measure the area underneath — because area on that graph *is* total work. A flat force (lifting near the ground) gives a plain rectangle, and its area is $mgh$. A spring pulls harder the more you stretch it, so its graph is a slanted ramp, and a ramp's area is a triangle — that triangle is $\tfrac12kx^2$, and the little $\tfrac12$ is just "half base times height." Deep-space gravity does the opposite: it fades with distance in a swooping $1/r^2$ curve, so its area is a curvy piece we add up with an integral, giving $-GMm/r$ — negative because we call "escaped to infinity" the zero point, so being trapped near a planet counts as sitting in a hole. Zoom into that hole at the surface and the curve looks straight again — that straight bit is exactly $mgh$, so the two gravity formulas were always one. Finally, to get the force back, just measure how *steeply* the energy curve rises: the force always points downhill, toward the bottom of the valley. > [!mnemonic] The shape tells the formula > **Rectangle → $mgh$. Triangle → $\tfrac12kx^2$. Curve → $-GMm/r$.** And to reverse: **slope down = force's way** ($F=-dU/dx$). ## Connections - [[1.3.05 Potential energy — definition, gravitational (mgh and −GMm - r), elastic (½kx²) (Hinglish)]] - [[Work done by a force]] - [[Hooke's Law]] - [[Gravitation — Newton's law]] - [[Force from potential — F = -dU/dx]] - [[Conservation of mechanical energy]] - [[Conservative vs non-conservative forces]] - [[Escape velocity]]