5.7 Escape Velocity

Introduction

Escape velocity is the minimum initial speed an object needs to completely break free from the gravitational pull of a massive body, like a planet or a star, without any further propulsion. An object launched with this speed will travel infinitely far away, eventually slowing down but never falling back. It is a crucial concept in rocketry and space exploration.

The Energy Balance

The concept of escape velocity is rooted in the conservation of energy. To escape a planet's gravitational field, an object must be given enough initial kinetic energy (K.E.) to overcome its absolute gravitational potential energy.

• Kinetic Energy: The energy of motion, given by $K.E.=\frac{1}{2}m{v}^2$.
• Gravitational Potential Energy: The energy an object has due to its position in a gravitational field. Using the universal definition, the G.P.E. of an object of mass $m$ at a distance $r$ from the center of a planet of mass $M$ is: $U_g=-\frac{GMm}{r}$ (The potential energy is negative and becomes zero at an infinite distance).

For an object to just barely escape, its total mechanical energy (K.E. + G.P.E.) must be zero. This means it will arrive at an infinite distance with zero kinetic energy.

Derivation of the Escape Velocity Formula

Let's find the escape velocity ($v_{esc}$) for an object of mass $m$ launched from the surface of a planet of mass $M$ and radius $R$.

1. Set up the energy conservation equation: $E_{initial}=E_{final}$ $(K.E.+G.P.E.{)}_{surface}=(K.E.+G.P.E.{)}_{infinity}$

2. Define the initial and final states:

   • Initial State (at the surface):
     • $K.E{.}_{initial}=\frac{1}{2}m{v}_{esc}^2$
     • $G.P.E{.}_{initial}=-\frac{GMm}{R}$
   • Final State (at infinity):
     • $K.E{.}_{final}=0$ (the object has just enough energy to arrive at infinity with zero speed).
     • $G.P.E{.}_{final}=0$ (by definition).

3. Substitute into the conservation equation: $\frac{1}{2}m{v}_{esc}^2-\frac{GMm}{R}=0+0$

4. Solve for $v_{esc}$: $\frac{1}{2}m{v}_{esc}^2=\frac{GMm}{R}$ The mass of the object, $m$, cancels out from both sides. $v_{esc}^2=\frac{2GM}{R}$ Taking the square root gives the final formula: $v_{esc}=\sqrt{\frac{2GM}{R}}$

Alternative Formula using Surface Gravity (g)

We can express the escape velocity in terms of the acceleration due to gravity, $g$, at the planet's surface. We know that $g=\frac{GM}{R^2}$, which can be rearranged to $GM=g{R}^2$.

Substituting $GM=g{R}^2$ into the escape velocity formula: $v_{esc}=\sqrt{\frac{2(g{R}^2)}{R}}$ This simplifies to: $v_{esc}=\sqrt{2gR}$ This is an equally valid and often more convenient formula.

Numerical Value for Earth

Let's calculate the escape velocity from the surface of the Earth.

• $g\approx 9.8{\text{ m/s}}^2$
• $R\approx 6.4\times 10^6\text{ m}$

Using the formula $v_{esc}=\sqrt{2gR}$: $v_{esc}=\sqrt{2\times (9.8{\text{ m/s}}^2)\times (6.4\times 10^6\text{ m})}$ $v_{esc}\approx \sqrt{125.44\times 10^6{\text{ m}}^2/{\text{s}}^2}\approx 11,200\text{ m/s}$ Converting to kilometers per second: $v_{esc}\approx 11.2\text{ km/s}$

Possible Questions/Answer

• Q: What is the difference between orbital velocity and escape velocity? A: Orbital velocity is the speed needed to maintain a stable orbit around a planet. Escape velocity is the higher speed needed to break free from the planet's gravity completely. In fact, the escape velocity is exactly $\sqrt{2}$ times the orbital velocity at the same radius ($v_{esc}=\sqrt{2}\times v_{orbit}$).

• Q: Does an object need to be pointed straight up to escape a planet? A: No. The direction of launch does not matter (as long as it is above the horizontal, so it doesn't immediately hit the ground). Escape velocity is a speed, not a velocity. As long as the initial speed is at least 11.2 km/s, and the object doesn't pass through the planet, it will eventually escape the gravitational pull.