How fast would you have to fire a cannonball into the sky for it to leave the world behind and fly into space? The speed necessary for this to happen is called the escape velocity.
Escape velocity is a concept for any field that exerts an attractive force on an object. Essentially, it is the speed needed for the object to escape the influence of the attractive field starting from a certain spatial point.
In this explanation, we specifically focus on the gravitational field created by the Earth.
A rocket must have an initial speed greater than the escape velocity of the Earth in order to travel into space
Energetic considerations and the escape velocity formula
To derive the formula for the escape velocity, we start by writing the formula for the total energy of a body in the presence of a gravitational field described by Newton’s law of gravitation:
\[E = U + E_k = -G \frac{M\cdot m}{r} + \frac{1}{2} \cdot m \cdot v^2\]
U is the gravitational potential energy due to the gravitational field, and K is the kinetic energy due to the state of movement of the body measured in joules. G is the gravitational constant (with an approximate value of 6.67 · 10-11m3/kg·s2), M is the mass of Earth (approximately 5.97 · 1024kg), m is the mass of the body under study measured in kg, r is the radial distance between them measured in m (whose presence indicates spherical symmetry), and v is the speed of the body measured in m/s.
You might also use K as the symbol for kinetic energy.
Possible trajectories under gravitational attraction
Consider an object on the surface of the Earth, like a ball or a space rocket. If the object is static (with zero velocity), its energy is just given by the potential energy. However, if we move the object, its velocity will determine the growth in kinetic energy (and total energy).
Now consider a point at an infinite radial distance from the Earth. Since the radial distance is in the denominator in the formula for the potential energy, the gravitational potential energy heads towards zero if the radial distance becomes infinite. This means that a static body at an infinite radial distance has zero energy. If, however, the object has a certain speed, it will have kinetic energy and, therefore, total energy.
The conservation of energy
Recall the principle of conservation of energy: energy is conserved for isolated systems. Energy is conserved since we aren’t considering any other elements besides the Earth and a certain body. This implies that the energy of a body on the surface of Earth must be the same as the energy of the body at an infinite distance.
With this in mind, we can classify the possible trajectories of a body:
- If the total energy of the body is positive on the surface of Earth, it will be positive at an infinite radial distance, which means that it will reach infinity with non-zero velocity.
- If the total energy of the body is zero on the surface of Earth, it will be zero at an infinite radial distance, which means that it will reach infinity with zero velocity.
- If the total energy of the body is negative on the surface of Earth, it should be negative at an infinite radial distance. However, since it cannot have negative energy at an infinite radial distance because kinetic energy is always positive or zero, we find that bodies with total negative energy never reach infinite radial distances. They either fall back to the surface of the Earth or go into orbit.
Escape velocity on the surface of the Earth
With this classification, we can see that for a body to escape to an infinite radial distance, it needs to have zero energy. If we are on the surface of the Earth, we can find the escape velocity ve by equating the energy to zero:
\[E = -G \cdot \frac{M \cdot m}{r_E} + \frac{1}{2} \cdot m \cdot v^2_e\]
\[\frac{1}{2} \cdot m \cdot v_e^2 = G \cdot \frac{M \cdot}{r_E}\]
\[v^2_e = G \cdot \frac{M \cdot m}{r_E} \cdot \frac{2}{m} = \frac{2 \cdot G \cdot M}{r_E}\]
\[v_e = \sqrt{\frac{2 \cdot G \cdot M}{r_E}}\]
Here, rE is the radius of the Earth (approximately 6371km). Using the actual values, we find that the escape velocity of a body on the surface of the Earth is independent of its mass and has an approximate value of 11.2km/s.
Escape velocity example
Let’s calculate the escape velocity for a different planet.
What is the escape velocity for Mars, where Mars has a mass of 6.39 · 1023kg and a radius of 3.34 · 106m? The gravitational constant G equals 6.67 · 10-11m3/kg·s2.
The escape velocity is given by the equation:
\[v = \sqrt{\frac{2 \cdot G \cdot M}{r}}\]
By plugging in the values given in the question, we get v = 5.05 · 103m/s as the escape velocity for Mars.
How can satellites orbit the Earth?
What happens if an object is fast enough to make it into the atmosphere but not fast enough to escape? In this case, the body is trapped under the influence of the Earth’s gravitational field and will follow (closed) orbits around Earth.
Closed orbits for the interaction determined by Newton’s gravitational law are usually elliptical. Ellipses can take many forms that are determined by specific parameters. For a unique combination of these parameters, we get a circle, which is simply a very special case of an ellipse with singular properties. While studying circular orbits is a huge simplification, it helps us understand the concept of escape velocity. Note that most of the planets in our Solar System follow elliptical orbits, which are almost circular. You can check out our explanation on Planetary Orbits for more information on this!
For circular orbits, there is a relationship between the orbital speed v0 of the object and the radial distance r from the centre of the Earth. We can derive this by using the equation for centripetal force, as this is provided by the gravitational attraction to the object in orbit:
\[\frac{mv_0^2}{r} = \frac{G \cdot M \cdot m}{r^2}\]
\[v_0^2 \cdot r = G \cdot M\]
\[v_0 = \sqrt{\frac{G \cdot M}{r}}\]
This implies that depending on the orbit’s altitude, bodies will orbit at a certain speed. The orbital speed should not be confused with escape velocity, as they differ by a factor of √2.
\[v_e = \sqrt{\frac{2 \cdot G \cdot M}{r}} = \sqrt{2} \cdot \sqrt{\frac{G \cdot M}{r}} = \sqrt{2} \cdot v_0 \approx 1.4142 \cdot v_0\]
This means any satellite with any stable orbit around the Earth would have to increase its velocity by 41.42% to escape Earth’s gravity.
If you look at the image below, you can see the different trajectories of an object based on its initial velocity.
- In A, the object falls back towards Earth.
- In B, the object enters a stationary orbit with an orbital velocity.
- In C, the object’s velocity is too high for a stationary orbit but too low to leave the Earth’s gravitational field. Therefore, it will fall into an orbit at a higher altitude.
- In D, the object leaves the Earth’s gravitational field with the escape velocity.
Different trajectories of an object based on its initial velocity, commons.wikimedia.org
The orbital speed should not be confused with escape velocity, as they differ by a factor of √2.
Escape Velocity - Key takeaways
The gravitational field causes bodies to describe certain trajectories whose characteristics can be studied according to their energy.
The escape velocity is the velocity at which a body must travel to escape the influence of an attractive field from a certain point.
The escape velocity of a body on the surface of the Earth is independent of its mass.
If an object is fast enough to make it into the atmosphere but not fast enough to escape it, it will follow (closed) orbits around the Earth.
The orbital velocity of a satellite depends on its radius from the object it is orbiting.
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