Escape speed

Throwing something away for good

Throw a ball up and it comes back. Throw it faster and it goes higher. Is there a speed at which it never comes back? Yes: if its kinetic energy is enough to climb all the way out of the Earth's potential well.

The escape speed

From the surface, the climb to zero potential energy needs $\dfrac{GMm}{R}$. So

For the Earth, $v_e = \sqrt{2 \times 9.8\ \text{m/s}^2 \times 6.4 \times 10^{6}\ \text{m}} = 11.2\ \text{km/s}$.

• The escaping body's mass cancels: a pebble and a rocket need the same speed.
• Energy has no direction: (ignoring air) any direction that clears the ground will do.

Below escape speed

A body launched straight up at $v < v_e$ rises to

which grows without limit as $v$ approaches $v_e$.

You should find the real height far above $\dfrac{v^2}{2g}$ for fast launches, and running off the graph as the speed nears $11.2\ \text{km/s}$.

Watch out. Escape speed is not needed to reach orbit. A low orbit needs only $\sqrt{gR} \approx 7.9\ \text{km/s}$, sideways.

A worked example

What is the escape speed from Mars ($M = 6.42 \times 10^{23}\ \text{kg}$, $R = 3.39 \times 10^{6}\ \text{m}$), and from a point $300\ \text{km}$ above the Earth?

1. Mars. $GM = 6.67 \times 10^{-11}\ \text{N m}^2/\text{kg}^2 \times 6.42 \times 10^{23}\ \text{kg} = 4.28 \times 10^{13}\ \text{m}^3/\text{s}^2$, so $v_e = \sqrt{\dfrac{2 \times 4.28 \times 10^{13}\ \text{m}^3/\text{s}^2}{3.39 \times 10^{6}\ \text{m}}} = 5.03\ \text{km/s}$.
2. Above the Earth. $r = 6.70 \times 10^{6}\ \text{m}$, so $v_e = \sqrt{\dfrac{2GM}{r}} = \sqrt{\dfrac{2 \times 4.01 \times 10^{14}\ \text{m}^3/\text{s}^2}{6.70 \times 10^{6}\ \text{m}}} = 10.9\ \text{km/s}$.
3. Note. Starting higher, less speed is needed.

Think first. Escape is about energy, not force: compare the kinetic energy with the depth of the potential well.

Practise with new numbers

Try it in Python

The program finds the escape speeds of four worlds.

Remember these

• $v_e = \sqrt{\dfrac{2GM}{R}} = \sqrt{2gR}$; $11.2\ \text{km/s}$ for the Earth.
• It does not depend on the body's mass or direction.
• Straight up at $v < v_e$: $h = \dfrac{Rv^2}{2gR - v^2}$.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. A planet has the Earth's radius and $9$ times its mass. What is its escape speed?

2. Does a heavier rocket need a higher escape speed?

3. A ball is thrown up at $1.0\ \text{km/s}$. Is $\dfrac{v^2}{2g}$ a good estimate of its height?

4. Why is it easier to escape from the Moon than from the Earth?