§ 13.3 — Escape Speed and Orbital Speed

Escape Speed: Breaking Free from Gravity

Escape speed is the minimum velocity an object must have at a given location to escape to infinity without further propulsion. It's a fundamental concept in celestial mechanics, applicable to planets, moons, black holes, and any gravitating body.

To derive escape speed, we use energy conservation. An object at the surface (or any radius \( r \)) with velocity \( v_{\text{esc}} \) has total energy:

\[ E = \frac{1}{2}mv_{\text{esc}}^2 - G\frac{Mm}{r} \]

For escape to infinity, the object should just barely reach infinite distance with zero velocity. At infinity, both kinetic and potential energy vanish, so:

\[ E_{\text{final}} = 0 \]

By conservation of energy:

\[ \frac{1}{2}mv_{\text{esc}}^2 - G\frac{Mm}{r} = 0 \]

\[ v_{\text{esc}} = \sqrt{\frac{2GM}{r}} \]

For Earth's surface (\( M = 5.972 \times 10^{24} \, \text{kg} \), \( r = 6.371 \times 10^6 \, \text{m} \)):

\[ v_{\text{esc}} = \sqrt{\frac{2(6.674 \times 10^{-11})(5.972 \times 10^{24})}{6.371 \times 10^6}} \approx 11.2 \, \text{km/s} \]

Note that escape speed is independent of the mass of the escaping object—it depends only on the gravitating mass and distance. A feather and a boulder require the same speed to escape Earth.

Orbital Speed: Stable Circular Motion

An object in a circular orbit moves in a path where gravitational force provides the centripetal acceleration needed to maintain circular motion:

\[ G\frac{Mm}{r^2} = m\frac{v_{\text{orbit}}^2}{r} \]

Solving for orbital velocity:

\[ v_{\text{orbit}} = \sqrt{\frac{GM}{r}} \]

The relationship between orbital and escape speeds is elegant:

\[ v_{\text{esc}} = \sqrt{2} \cdot v_{\text{orbit}} \]

This means escape speed is always \( \sqrt{2} \approx 1.414 \) times the orbital speed at the same radius. For a circular orbit just above Earth's surface:

\[ v_{\text{orbit}} = \sqrt{\frac{GM_E}{R_E}} \approx 7.9 \, \text{km/s} \]

Orbital Period and Kepler's Third Law

For a circular orbit, the orbital period \( T \) relates to the radius through energy considerations and Newton's laws. Combining the centripetal force requirement with gravitational force:

\[ \frac{v_{\text{orbit}}}{r} = \frac{2\pi}{T} \]

\[ \sqrt{\frac{GM}{r}} = \frac{2\pi r}{T} \]

Squaring and rearranging:

\[ T^2 = \frac{4\pi^2 r^3}{GM} \]

This is Kepler's Third Law, which we'll explore in detail in the next section. It shows that orbital period increases with the 3/2 power of orbital radius.

A Parallel Problem: Multi-Stage Rocket to Escape

A three-stage rocket on Earth's surface initially carries fuel with a total exhaust velocity \( v_e = 4.5 \, \text{km/s} \) (relative to the rocket). The rocket's dry mass (structure + payload) is \( M_{\text{dry}} = 5000 \, \text{kg} \), and the initial fuel mass is \( M_{\text{fuel}} = 95,000 \, \text{kg} \). The rocket operates in three stages, each expelling 1/3 of the initial fuel mass.

Determine the final velocity of the payload after all three stages have fired, and verify whether it achieves escape velocity.

Solution:

Using the Tsiolkovsky rocket equation for each stage:

\[ \Delta v = v_e \ln\left(\frac{M_{\text{initial}}}{M_{\text{final}}}\right) \]

First stage: Initial total mass = 100,000 kg, final mass = 5000 + 66,667 = 71,667 kg

\[ \Delta v_1 = 4500 \ln\left(\frac{100,000}{71,667}\right) = 4500 \ln(1.395) = 4500(0.333) = 1499 \, \text{m/s} \]

Second stage: Initial mass = 71,667 kg, final mass = 5000 + 33,333 = 38,333 kg

\[ \Delta v_2 = 4500 \ln\left(\frac{71,667}{38,333}\right) = 4500 \ln(1.869) = 4500(0.626) = 2817 \, \text{m/s} \]

Third stage: Initial mass = 38,333 kg, final mass = 5000 kg

\[ \Delta v_3 = 4500 \ln\left(\frac{38,333}{5000}\right) = 4500 \ln(7.667) = 4500(2.037) = 9167 \, \text{m/s} \]

Total velocity gain:

\[ v_{\text{total}} = \Delta v_1 + \Delta v_2 + \Delta v_3 = 1499 + 2817 + 9167 = 13,483 \, \text{m/s} \]

Since escape velocity from Earth's surface is \( v_{\text{esc}} = 11,200 \, \text{m/s} \) and the rocket achieves \( 13,483 \, \text{m/s} \), the payload successfully escapes Earth's gravity. The excess kinetic energy corresponds to a final speed at infinity of:

\[ \frac{1}{2}v_{\infty}^2 = \frac{1}{2}v_{\text{total}}^2 - G\frac{M_E}{R_E} \]

\[ v_{\infty} = \sqrt{(13,483)^2 - 2(6.674 \times 10^{-11})(5.972 \times 10^{24})/(6.371 \times 10^6)} \]

\[ v_{\infty} = \sqrt{181.8 \times 10^6 - 125.4 \times 10^6} = \sqrt{56.4 \times 10^6} \approx 7.5 \, \text{km/s} \]

Real-World Applications

Escape speed determines the feasibility of space launch from any planetary body. It's why rockets need enormous energy to escape Earth, but why a person jumping could escape from a small asteroid. Black holes have escape speeds exceeding the speed of light, making them true prisons of spacetime. Satellite communication networks rely on understanding orbital speeds—geostationary satellites orbit at a specific radius where their period matches Earth's rotation period (24 hours).

⇧ Back to chapter