KAHIBARO
Table of Contents
Leaving a Gravitational Body
Escape velocity is the minimum speed an object must have, at a given distance from a massive body, so that it can move away forever without any further propulsion, assuming no air resistance or other forces act on it. It is called a velocity, but in this context what matters is the speed, not the direction.
The key idea is energy. A launched object starts with kinetic energy and gravitational potential energy. If it has just enough total mechanical energy to reach infinitely far away and come to rest there, then it has the escape velocity. If it has less than this, it will eventually fall back. If it has more, it will still escape, and it will keep some speed even very far away.
Energy Argument
For an object of mass $m$ at a distance $r$ from the center of a planet or star of mass $M$, the gravitational potential energy is
$$
U(r) = -\frac{GMm}{r}
$$
and the kinetic energy is
$$
K = \frac{1}{2}mv^2
$$
To just escape, the final speed at infinity is zero, so the final total energy is zero. Therefore the initial total energy must also be zero:
$$
\frac{1}{2}mv_e^2 - \frac{GMm}{r} = 0
$$
Solving for the escape velocity $v_e$ gives
$$
v_e = \sqrt{\frac{2GM}{r}}
$$
Important escape velocity formula:
$$
v_e = \sqrt{\frac{2GM}{r}}
$$
At the surface of a spherical body of radius $R$:
$$
v_e = \sqrt{\frac{2GM}{R}}
$$
Escape velocity depends on the mass of the central body and the distance from its center. It does not depend on the mass of the escaping object.
Meaning of the Formula
The formula shows several important facts. A more massive planet has a larger escape velocity. A larger launch distance from the center gives a smaller escape velocity. This is why it is easier to escape from high altitude than from the surface.
It is also important to notice that the mass $m$ of the object cancels out. A small rock and a large spacecraft need the same escape speed from the same location, if we ignore air resistance and propulsion details.
Escape from Earth
For Earth, using mass $M_E$ and radius $R_E$, the surface escape velocity is approximately
$$
v_e \approx 11.2 \text{ km/s}
$$
This is about $11200 \text{ m/s}$.
That does not mean a rocket must instantly be fired to exactly this speed in one moment. Real rockets can reach space by accelerating over time. Escape velocity is best understood as an energy threshold, not as a requirement for an instantaneous launch.
Comparison with Circular Orbital Speed
Escape velocity is larger than the speed needed for a circular orbit at the same radius. The circular orbital speed is
$$
v_{\text{orb}} = \sqrt{\frac{GM}{r}}
$$
so
$$
v_e = \sqrt{2}\,v_{\text{orb}}
$$
This means escape speed is $\sqrt{2}$ times the circular orbital speed at the same distance.
Relationship between orbital speed and escape speed:
$$
v_{\text{orb}} = \sqrt{\frac{GM}{r}}, \qquad
v_e = \sqrt{\frac{2GM}{r}} = \sqrt{2}\,v_{\text{orb}}
$$
An object in circular orbit is bound gravitationally. An object with escape velocity has just enough energy to become unbound.
Physical Interpretation
If the total mechanical energy is negative, the object is gravitationally bound. If the total mechanical energy is zero, it is at the threshold of escape. If the total mechanical energy is positive, it escapes with leftover kinetic energy.
This gives a useful way to classify motion.
| Total energy | Result |
|---|---|
| $E < 0$ | Bound motion |
| $E = 0$ | Just escapes |
| $E > 0$ | Escapes with extra speed |
Launch from Different Heights
If an object is launched from height $h$ above the surface of a planet of radius $R$, then the starting distance from the center is
$$
r = R + h
$$
So the escape velocity becomes
$$
v_e = \sqrt{\frac{2GM}{R+h}}
$$
As $h$ increases, the required escape speed decreases.
Visualizing Escape
A useful picture is to think of the object climbing out of a gravitational well. The deeper the well, the more speed it needs at the start.
Common Misunderstanding
A common mistake is to think that if a projectile is launched upward slower than escape velocity, gravity will stop it immediately. In reality, gravity gradually reduces its speed as it rises. If the initial speed is below escape velocity, it reaches a maximum distance and then returns. If the initial speed is exactly the escape velocity, it keeps slowing down and approaches zero speed only infinitely far away.
Typical Escape Velocities
Different bodies have very different escape velocities.
| Body | Escape velocity |
|---|---|
| Moon | $\approx 2.38 \text{ km/s}$ |
| Earth | $\approx 11.2 \text{ km/s}$ |
| Jupiter | $\approx 59.5 \text{ km/s}$ |
| Sun | $\approx 617 \text{ km/s}$ |
These values help explain why it is much easier to leave the Moon than Earth, and much harder to leave the Sun.
Final Rule
To escape from a spherical body, neglecting air resistance and thrust after launch, the object must have at least enough initial kinetic energy to overcome the magnitude of its gravitational potential energy:
$$
\frac{1}{2}mv^2 \ge \frac{GMm}{r}
$$
Equivalently,
$$
v \ge \sqrt{\frac{2GM}{r}}
$$
Escape velocity is one of the clearest examples of how conservation of energy determines motion in gravity.