Escape Velocity and Gravitational Potential Energy
Escape Velocity and Gravitational Potential Energy
Gravitational potential energy is the energy stored in the gravitational field between two masses. Unlike the simplified formula mgh used near Earth's surface, the general expression for gravitational potential energy is U equals negative G M m over r. The negative sign reflects the convention that potential energy is zero at infinite separation — you must add energy to move an object farther from the gravitating body.
Escape velocity is the minimum speed needed to escape a gravitational field entirely — to reach infinite distance with zero residual velocity. Setting the total energy (kinetic plus potential) to zero gives v-escape equals the square root of 2 G M over r. Notice that escape velocity is exactly the square root of 2 times the circular orbital velocity at the same radius. A satellite in circular orbit at any altitude need only increase its speed by about 41 percent to escape.
Earth's escape velocity at the surface is about 11.2 kilometers per second. For the Moon it is only 2.4 kilometers per second, which is why the Apollo lunar modules needed much less fuel to leave the Moon than to leave Earth. Jupiter's escape velocity is a staggering 59.5 kilometers per second, and the Sun's surface escape velocity is 618 kilometers per second.
The concept of escape velocity also explains black holes: if the escape velocity at some radius exceeds the speed of light, nothing — not even light — can escape. That radius is the Schwarzschild radius, and the boundary it defines is the event horizon.
$$U = -\frac{GMm}{r}$$
$U < 0$ always (bound state). $U \to 0$ as $r \to \infty$.
$$v_{\text{esc}} = \sqrt{\frac{2GM}{r}} = \sqrt{\frac{2\mu}{r}} = \sqrt{2} \, v_{\text{circ}}$$
At Earth's surface: $v_{\text{esc}} = \sqrt{2 \times 9.81 \times 6.371 \times 10^6} \approx 11{,}186$ m/s $\approx 11.2$ km/s.
$$E = \frac{1}{2}mv^2 - \frac{GMm}{r} = -\frac{GMm}{2a}$$
$E < 0$: bound orbit (ellipse/circle). $E = 0$: parabolic escape. $E > 0$: hyperbolic trajectory.
Calculate Earth's surface escape velocity.
Step 1: Apply the formula.
$$v_{\text{esc}} = \sqrt{\frac{2\mu}{R_E}} = \sqrt{\frac{2 \times 3.986 \times 10^{14}}{6.371 \times 10^6}}$$
Step 2: Evaluate.
$$v_{\text{esc}} = \sqrt{1.251 \times 10^8} = 11{,}186 \text{ m/s} \approx 11.2 \text{ km/s}$$
That's about 40,270 km/h — 33 times the speed of sound.
A 2000 kg satellite is in circular orbit at 500 km altitude. What is its total orbital energy?
Step 1: $r = 6371 + 500 = 6871$ km $= 6.871 \times 10^6$ m.
Step 2: $E = -GMm/(2r) = -(3.986 \times 10^{14})(2000)/(2 \times 6.871 \times 10^6)$
$$E = -5.81 \times 10^{10} \text{ J} = -58.1 \text{ GJ}$$
The negative sign confirms the satellite is gravitationally bound.
Find the escape velocity from the surface of Mars. ($M = 6.417 \times 10^{23}$ kg, $R = 3390$ km)
Step 1: $v_{\text{esc}} = \sqrt{2GM/R} = \sqrt{2(6.674 \times 10^{-11})(6.417 \times 10^{23})/(3.39 \times 10^6)}$
$$= \sqrt{2.525 \times 10^7} = 5024 \text{ m/s} \approx 5.0 \text{ km/s}$$
Mars's escape velocity is less than half of Earth's, which is why Mars has lost most of its atmosphere over billions of years.
Practice Problems
Show Answer Key
1. $r = 6779$ km. $v_{\text{esc}} = \sqrt{2\mu/r} = \sqrt{2 \times 3.986 \times 10^{14}/6.779 \times 10^6} = 10{,}841$ m/s.
2. $v_{\text{esc}} = \sqrt{2(6.674 \times 10^{-11})(7.342 \times 10^{22})/1.737 \times 10^6} = 2376$ m/s $= 2.38$ km/s.
3. $v_{\text{circ}} = 7726$ m/s. $v_{\text{esc}} = \sqrt{2} \times 7726 = 10926$ m/s. $\Delta v = 10926 - 7726 = 3200$ m/s.
4. $r = 7171$ km. $E = -\mu m/(2r) = -(3.986 \times 10^{14})(1000)/(2 \times 7.171 \times 10^6) = -2.78 \times 10^{10}$ J.
5. $8000 = \sqrt{2\mu/r} \Rightarrow r = 2\mu/8000^2 = 1.245 \times 10^7$ m. Alt $= 12450 - 6371 = 6079$ km.
6. $\mu = v^2 R/2 = (59500)^2 \times 6.991 \times 10^7/2 = 1.267 \times 10^{17}$ m³/s².
7. $v = \sqrt{2\mu/r}$ at every point along the trajectory — the speed exactly equals the local escape velocity.
8. Need $v_{\text{heliocentric}} = 42.1$ km/s. Earth gives 29.8 km/s for free. Extra: $42.1 - 29.8 = 12.3$ km/s relative to Earth (plus overcoming Earth's gravity well).
9. $R_s = 2GM/c^2 = 2(3.986 \times 10^{14})/(3 \times 10^8)^2 = 8.87 \times 10^{-3}$ m $\approx 8.9$ mm.
10. Surface: $U = -\mu m/R_E = -6.26 \times 10^9$ J. ISS: $U = -\mu m/r_{ISS} = -5.88 \times 10^9$ J. Difference: $3.8 \times 10^8$ J.