Gravitational Potential Energy Far from Earth gave the potential energy $U = -GMm/r$, and Potential Energy from Force derived it by integrating Newton's gravity from infinity. Now use it to answer an old question: how fast must something be launched so that it never falls back? Build and fly a rocket in the Rocket Builder lab to see the numbers at work.
Why it works
Ignore the air. After launch, only gravity acts, so $K + U$ stays constant: $$\tfrac{1}{2}mv^2 - \dfrac{GMm}{r} = E.$$ As the object climbs, $U$ rises toward 0 and $K$ falls. If $E < 0$, then $K$ would have to become negative before $r$ reaches infinity, which can't happen: the object stops and falls back. If $E \ge 0$, it can reach any distance with $K \ge 0$ to spare. The boundary is $E = 0$: $$\tfrac{1}{2}mv_{\text{esc}}^2 = \dfrac{GMm}{R}.$$
Key idea
From a distance $r$ from a body's center, $$v_{\text{esc}} = \sqrt{\dfrac{2GM}{r}},$$ independent of the mass of the object launched. From Earth's surface it is about $11.2\ \text{km/s}$. Launched faster, the object coasts away with $v_\infty = \sqrt{v_0^2 - v_{\text{esc}}^2}$.
Worked example: Escape from Earth
Find the escape speed from Earth's surface. ($M = 5.972 \times 10^{24}\ \text{kg}$, $R = 6.371 \times 10^{6}\ \text{m}$, $G = 6.674 \times 10^{-11}\ \text{N·m}^2/\text{kg}^2$.)
- $$\dfrac{2GM}{R} = \dfrac{2(6.674 \times 10^{-11})(5.972 \times 10^{24})}{6.371 \times 10^{6}} = 1.251 \times 10^{8}\ \text{m}^2/\text{s}^2$$
- $v_{\text{esc}} = 1.12 \times 10^{4}\ \text{m/s} = 11.2\ \text{km/s}$.
Answer: About $11.2\ \text{km/s}$.
Worked example: Faster than escape
A probe leaves Earth (ignore the air and the Sun) at $15\ \text{km/s}$. How fast is it moving far away?
- $\tfrac{1}{2}mv_0^2 - \dfrac{GMm}{R} = \tfrac{1}{2}mv_\infty^2$, so $v_\infty^2 = v_0^2 - v_{\text{esc}}^2$.
- $$v_\infty = \sqrt{15^2 - 11.19^2}\ \text{km/s} = \sqrt{99.9} \approx 10.0\ \text{km/s}$$
- Most of the launch speed goes into the climb, even for a fast start.
Answer: About $10.0\ \text{km/s}$.
Common mistake
Thinking a heavier object needs a higher escape speed. It needs more energy, since both $K$ and $U$ are proportional to $m$, but the mass cancels from the speed. A grain of sand and a spacecraft need the same 11.2 km/s from Earth's surface. Check: the formula $\sqrt{2GM/r}$ contains the planet's mass, not the projectile's.
Real rockets don't fire once at escape speed from the ground; they burn for minutes, climb out of the thick air, and usually go into orbit first. Escape speed is still the energy budget they must reach: $E \ge 0$.