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

WHY do we even define potential energy?

HOW to derive $U=-GMm/r$ from scratch

Step 1 — Write the force. Newton's law of gravitation: a mass $M$ at the origin pulls mass $m$ at distance $r$ with $\vec F = -\frac{GMm}{r^2}\,\hat r.$ Why the minus sign? The force points inward (toward $M$ ), opposite to the outward radial unit vector $\hat r$ .

Step 2 — Move $m$ purely radially from some distance $r$ out to infinity. Then $d\vec r = dr\,\hat r$ and $\vec F \cdot d\vec r = -\frac{GMm}{r^2}\,dr.$

Step 3 — Set the reference. Choose $U(\infty)=0$ . Then $U(r) = U(r) - U(\infty) = -\int_{\infty}^{r}\vec F\cdot d\vec r = -\int_{\infty}^{r}\left(-\frac{GMm}{r'^2}\right)dr'.$

Step 4 — Do the integral. $U(r) = GMm\int_{\infty}^{r}\frac{dr'}{r'^2} = GMm\left[-\frac{1}{r'}\right]_{\infty}^{r} = GMm\left(-\frac1r + 0\right).$

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

WHY $mgh$ is just the local approximation

Near Earth's surface, $r = R + h$ with $h \ll R$ ( $R=$ Earth's radius). The change in PE from surface to height $h$ : $\Delta U = -\frac{GMm}{R+h} - \left(-\frac{GMm}{R}\right) = GMm\left(\frac1R - \frac1{R+h}\right) = GMm\cdot\frac{h}{R(R+h)}.$

Since $h\ll R$ , $R+h\approx R$ , so $\Delta U \approx GMm\cdot\frac{h}{R^2} = m\underbrace{\left(\frac{GM}{R^2}\right)}_{=\,g}h = mgh.$

Worked examples

Worked example (1) Escape speed from a planet

Q: Minimum launch speed so a mass $m$ never falls back? Energy conservation, just barely reaching $\infty$ with zero speed: $\underbrace{\tfrac12 mv_{\text{esc}}^2 + \left(-\tfrac{GMm}{R}\right)}_{\text{surface}} = \underbrace{0 + 0}_{r=\infty}.$ Why total energy = 0 at infinity? Both KE and U vanish there ("just escapes"). Solve: $\tfrac12 mv_{\text{esc}}^2 = \frac{GMm}{R}\ \Rightarrow\ \boxed{v_{\text{esc}}=\sqrt{\frac{2GM}{R}}}.$ For Earth $\approx 11.2$ km/s. Why mass-independent? $m$ cancels — a feather and a rocket need the same escape speed (ignoring air).

Common mistakes (Steel-man + fix)

Forecast-then-Verify

Recall Predict before you peek

Forecast: If you double the orbital radius $r$ , how does total orbital energy $E=-GMm/2r$ change? Verify: $E$ becomes half as negative (closer to 0), i.e. energy increases. Higher orbits are higher-energy — consistent with needing to do work to raise an orbit. ✔

Flashcards

Why is gravitational PE negative?

We set

U(\infty)=0

; every finite

r

is a bound state with less energy than the reference, so

U<0

Derive

U(r)

from the force.

U=-\int_\infty^r \vec F\cdot d\vec r=-\int_\infty^r(-GMm/r'^2)dr'=-GMm/r

Why does

mgh

work near the surface?

It's the Taylor/small-

h

limit of

-GMm/r

with

g=GM/R^2

and

R+h\approx R

Relation between

g

and

G

g=GM/R^2

— local field from the universal constant.

Escape speed formula and why mass-independent?

v_{esc}=\sqrt{2GM/R}

;

m

cancels in

\frac12 mv^2=GMm/R

Total energy of a circular orbit?

E=-GMm/2r

(and

E=\tfrac12 U=-K

r

increases, does

U

increase or decrease?

Increases (toward 0); magnitude shrinks.

What kind of force allows a potential energy to exist?

A conservative force (path-independent work).

Recall Feynman: explain to a 12-year-old

Imagine a ball stuck in a deep pit. To get it out you have to push it up the whole way — that takes effort (energy). Space works the same: planets sit in invisible "gravity pits." We say the bottom of the pit is below "zero," so it's a negative number, and the top of the pit (very far away in space) is zero. The deeper in the pit (closer to the planet), the more negative. $mgh$ is what you use when the pit is so shallow it looks like a flat ramp — handy for hills and buildings, useless for rockets going to the Moon.

Connections

Newton's Law of Universal Gravitation — the force we integrated.
Conservative Forces and Potential Energy — why $U$ exists at all.
Work-Energy Theorem — links $W$ and $\Delta KE$ to $\Delta U$ .
Escape Velocity — direct application setting $E=0$ .
Kepler's Laws & Orbital Energy — uses $E=-GMm/2r$ .
Acceleration due to gravity g — the local limit $g=GM/R^2$ .

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, gravitational potential energy ka asli formula hai $U=-GMm/r$ , na ki sirf $mgh$ . Idea simple hai: do masses ko alag karne ke liye gravity ke against kaam karna padta hai, to system energy store karta hai. Hum zero energy ka reference infinity par rakhte hain (jahan dono masses bahut door, force almost zero). Isliye jab masses paas hote hain, unki energy zero se kam, yaani negative hoti hai. Negative ka matlab "trapped in gravity well" — escape karne ke liye energy add karni padegi.

Derivation bas itni si hai: force $-GMm/r^2$ ko $\infty$ se $r$ tak integrate karo (minus sign ke saath, kyunki $U=-\int \vec F\cdot d\vec r$ ), aur seedha $-GMm/r$ aa jaata hai. Koi ratt-ne wali baat nahi.

Ab $mgh$ kahan se aaya? Jab height $h$ Earth ke radius $R$ se bahut chhoti ho, to $U=-GMm/r$ ka chhota approximation lo, aur tumhe milta hai $\Delta U \approx mgh$ jahan $g=GM/R^2$ . Matlab $mgh$ koi alag rule nahi — yeh wahi bada formula hai, bas flat-Earth (constant $g$ ) wali soch ke liye. Building ya pahaad ke liye $mgh$ theek hai, lekin satellite, rocket, ya escape velocity ke liye hamesha $-GMm/r$ use karo, warna answer galat aayega.

Yaad rakhne ka trick: "Negative kyunki phase gaye ho, zero jab azaad ho." Escape speed nikalna ho to total energy zero rakho infinity par: $\frac12 mv^2 = GMm/R$ , aur $v_{esc}=\sqrt{2GM/R}$ — mass cancel ho jaata hai, isliye feather aur rocket dono ko same escape speed chahiye.

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

WHY do we even define potential energy?

HOW to derive $U=-GMm/r$ from scratch

WHY $mgh$ is just the local approximation

Worked examples

Common mistakes (Steel-man + fix)

Forecast-then-Verify

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Newton's Laws & Dynamics

Connections