Lessons · gravitation

Gravitational energy: why escape velocity is finite

Regime 2 — calculus does what algebra cannot. The integrand isn't constant, so no single algebra product gives the answer; the accumulated quantity is the area under the curve — and SymPy proves that area is exactly the closed-form result.

Regime 2 · calculus does more Math machine-derived & checked by SymPy 3 modeling assumptions (author-asserted)

Lifting a mass away from a planet ( $\mu = GM = 4\times10^{14}$ m³/s², surface radius $R = 6.4\times10^{6}$ m): the gravitational pull weakens as $F = GMm/r^2$ . How much energy does it take to climb — and why is the energy to escape finite? Drag the cursor outward and watch the potential energy approach a ceiling as the area under $1/r^2$ converges.

PE gained reaching

r=2R

31250000 J

Escape energy

GMm/R

(area to

\infty

)

62500000 J

Surface weight

F(R)=GMm/R^2=mg

9.766 N

Potential energy is the area under the force–distance curve

\Delta U = \int_R^r F\,dr' = \int_R^r \frac{GMm}{r'^2}\,dr'

The work to lift a mass against gravity is the integral of the force over distance — the area under the $F$ – $r$ curve. For a constant force this is a rectangle ( $mgh$ ); for the inverse-square force it is the area under a falling curve.

check $\tfrac{d}{dr}\Delta U = F$ ; check $\Delta U = \int_R^r F\,dr'$ ; check $\lim_{r\to\infty}\Delta U = GMm/R$ ; collapse the near-surface case to $mgh$

✓ The PE's slope is the force: $U'(r) = F(r)$ — the area's rate of growth is the curve's height. [structural]
✓ The accumulated PE is the area: $\Delta U(r) = \int_R^r F\,dr'$ . [structural]
✓ The area to infinity converges: the escape energy $\to GMm/R$ is finite as $r\to\infty$ . [structural]
✓ Near the surface $F\approx mg$ is constant, so $\Delta U \approx mgh$ — the area is a rectangle (the quadrature). [structural]

Dimensional homogeneity: checked by SymPy (holds).

✗ Common misconception: “It takes infinite energy to fully escape a planet's gravity, since gravity reaches out forever.”

Gravity does reach out forever, but it weakens as $1/r^2$ — and the area under $1/r^2$ from the surface to infinity is finite. That finite area is the escape energy $GMm/R$ ; divide by the mass and you get the escape speed $\sqrt{2GM/R}$ . Drag the cursor out: the potential energy rises toward a ceiling, not without bound.

Modeling assumptions — author-asserted, disclosed not discharged

Spherically symmetric planet, so outside the surface the field is $GMm/r^2$ (point-mass equivalent).
Only gravity acts; the lift is quasi-static (no kinetic energy bookkeeping — this is the potential energy alone).
$\mu = GM$ is used so the force is unit-clean; energies are shown per kilogram ( $m = 1$ kg).

The F–r graph, fully annotated

A static rendering (Matplotlib): the shaded area under F is the accumulated integral ΔU, and the slope of ΔU is F. The interactive version with a draggable cursor is in the Graph tab above.

Formulas used

Hover a formula to preview its reference entry; click to open it in the reference (or the concept graph):