Escape velocity — derivation

WHAT are we finding?

WHY does it not depend on $m$? Both the kinetic energy you supply ($\propto m$) and the gravitational energy you must pay ($\propto m$) scale with $m$, so $m$ cancels. A pebble and a spaceship need the same speed.

HOW to derive it — energy method (first principles)

Step 1 — Gravitational potential energy (WHY this form). Newton's gravity force on $m$ at distance $r$ from centre of $M$: $F(r) = \frac{GMm}{r^2}\quad(\text{attractive, points inward})$

PE is the work done against gravity to bring $m$ from infinity to $r$. Choose $U(\infty)=0$ (natural zero: no interaction when infinitely apart). Then

$U(r) = -\int_{\infty}^{r} F(r')\,(-dr') \;=\; -\int_{\infty}^{r}\frac{GMm}{r'^2}\,dr'$

Why this step? Moving inward, gravity does positive work, so the system loses PE → $U$ becomes negative. Evaluate:

$U(r) = -GMm\left[-\frac{1}{r'}\right]_{\infty}^{r} = -\frac{GMm}{r}$

Step 2 — Conserve total mechanical energy. No air, no engine after launch ⇒ only gravity acts ⇒ mechanical energy is conserved: $E = K + U = \tfrac{1}{2}mv^2 - \frac{GMm}{r} = \text{constant}$

Step 3 — Apply the boundary conditions. At launch ($r=R$, speed $v_e$): $\;E_i = \tfrac{1}{2}mv_e^2 - \dfrac{GMm}{R}$ At the just-escaping limit ($r\to\infty$, speed $\to 0$): $\;E_f = 0 - 0 = 0$

Why set final speed = 0? "Minimum" speed means no wasted energy — you barely make it.

Step 4 — Equate and solve. $\tfrac{1}{2}mv_e^2 - \frac{GMm}{R} = 0 \;\Rightarrow\; \tfrac{1}{2}v_e^2 = \frac{GM}{R}$

Worked examples

Active recall

Connections

• Gravitational potential energy
• Newton's law of universal gravitation
• Conservation of mechanical energy
• Orbital velocity & circular motion
• Black holes — Schwarzschild radius (set $v_e=c$)
• Why the Moon has no atmosphere

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Hinglish (regional understanding)