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Escape Velocity: How to Escape the Gravitational Field?

Important Formula

Formula: Second Cosmic Velocity

$$\class{purple}{v_2} ~=~ \sqrt{\frac{2G\, \class{brown}{M_{\text p}} }{R_{\text p}}}$$ $$\class{purple}{v_2} ~=~ \sqrt{\frac{2G\, \class{brown}{M_{\text p}} }{R_{\text p}}}$$ $$R_{\text p} ~=~ \frac{ 2 G\, \class{brown}{M_{\text p}} }{ {\class{purple}{v_2}}^2 }$$ $$\class{brown}{M_{\text p}} ~=~ \frac{ R_{\text p} \, {\class{purple}{v_2}}^2 }{ 2G }$$ $$G ~=~ \frac{ R{\text p} \, {\class{purple}{v_2}}^2 }{ 2\class{brown}{M_{\text p}} }$$

Rearrange formula

What do the formula symbols mean?

Second cosmic velocity

$$ \class{purple}{v_2} $$ Unit $$ \frac{\mathrm m}{\mathrm s} $$

Second cosmic velocity is the velocity necessary to escape without propulsion from the gravitational field of a celestial body.

For a rocket to escape the Earth's gravitational field, it must have the following minimum velocity: \[ \class{purple}{v_2} ~=~ \sqrt{ 2 ~\cdot~ \frac{6.67 \cdot 10^{-11} \frac{\mathrm N \, \mathrm{m}^2}{\mathrm{kg}^2} ~\cdot~5.97 \cdot 10^{24}\,\mathrm{kg} }{6.38 \cdot 10^6 \,\mathrm{m}} } ~=~ 11.2 \, \frac{\mathrm{km}}{\mathrm s} \]

Radius of the celestial body

$$ R_{\text p} $$ Unit $$ \mathrm{m} $$

Radius of the celestial body you are trying to escape. For example, radius of the Earth.

Mass

$$ \class{brown}{M_{\text p}} $$ Unit $$ \mathrm{kg} $$

Mass of the celestial body. In the case of the Earth, the mass is: $ 5.972 \cdot 10^{24} \, \mathrm{kg} $.

Gravitational constant

$$ G $$ Unit $$ \frac{\mathrm{N} \, \mathrm{m}^2}{\mathrm{kg}^2} = \frac{\mathrm{m}^3}{\mathrm{kg} \, \mathrm{s}^2} $$

The gravitational constant is a physical constant that occurs in equations describing the interaction between masses. It has the following experimentally determined value: $$ G ~\approx~ 6.674 \, 30 ~\cdot~ 10^{-11} \, \frac{ \mathrm{m}^3 }{ \mathrm{kg} \, \mathrm{s}^2 } $$

Table of contents

Elon Musk's Starlink system consists of thousands of small satellites orbiting the Earth at a distance $ r $ (550 kilometers), serving people in areas without Internet access. Elon Musk must ensure that his satellite system, once placed in Earth orbit, does not crash and always maintains the specified distance $ r $ from Earth. Too large distance will cause link problems, while too small distance will be difficult to maintain. In addition, the Starlink system must avoid collisions with other satellites in space. To ensure a constant orbital radius $ r $, the Starlink system must have a certain orbital velocity $ \class{blue}{v} $. If the Starlink system orbits the Earth at this velocity, it will remain in a stable orbit.

What is this speed concretely?

To answer this question, we need only two ingredients:

The Newton's law of gravity - this law tells us what Earth's gravitational force $ F_{\text g} $ the Starlink system experiences at a distance $ r $.
The formula for the centripetal force $ F_{\text z} $ - this tells us what velocity is necessary at a certain distance to escape the gravitational force $ F_{\text g} $ (to be in free fall).

Newton's law of gravity links the mass $ \class{brown}{M} $ of the Earth (or other celestial body), the mass $ \class{brown}{m} $ of the Starlink system (or other orbiting object), their distance $ r $ from each other, and the gravitational constant $ G = 6.67 \cdot 10^{-11} \mathrm{m}^3/(\mathrm{kg}\cdot \mathrm{s}^2) $:

$$ \begin{align} F_{\text g} ~=~ G \, \frac{\class{brown}{M} }{r^2} \, \class{brown}{m} \end{align} $$

And the centripetal force $ F_{\text z} $ is a radial (to the center of the circle) directed force, which occurs, if a body (Starlink system) of the mass $ \class{brown}{m} $ orbits on a circular path with the radius $ r $ and with the orbital velocity $ \class{blue}{v} $:

Satellite orbits the Earth at a Fixed Distance without Propulsion

$$ \begin{align} F_{ \text z} ~=~ \frac{\class{brown}{m} \, \class{blue}{v}^2}{ r } \end{align} $$

The Starlink system experiences a gravitational force $ F_{\text g} $ toward the center of the Earth. It is therefore directed radially and thus corresponds to the centripetal force $ F_{\text z} $. Let us set the two forces 1 and 2 equal:

$$ \begin{align} G \, \frac{\class{brown}{M} }{r^2} \, \class{brown}{m} ~=~ \frac{\class{brown}{m} \, \class{blue}{v}^2 }{r} \end{align} $$

The mass $ \class{brown}{m} $ of the Starlink system is not relevant here because it cancels out:

$$ \begin{align} G \, \frac{\class{brown}{M} }{r^2} ~=~ \frac{\class{blue}{v}^2 }{r} \end{align} $$

The radius can also be canceled once:

$$ \begin{align} G \, \frac{\class{brown}{M} }{r} ~=~ \class{blue}{v}^2 \end{align} $$

Let's just take the root on both sides of the equation to transform the equation with respect to the orbital velocity $ \class{blue}{v} $:

$$ \begin{align} \class{blue}{v} ~=~ \sqrt{ G \, \frac{\class{brown}{M} }{r} } \end{align} $$

We have derived a formula with which we can calculate the orbital velocity $ \class{blue}{v} $ of the Starlink system when it orbits the Earth of mass $ \class{brown}{M} $ on the circular path with distance $ r $ to Earth.

So we only have to insert the desired distance $r $ to the earth center and the earth mass $ \class{brown}{M} = 5.97 \cdot 10^{24} \, \mathrm{kg} $ into the formula to calculate concretely the orbital velocity $ \class{blue}{v} $ with which the Starlink system will neither crash into the earth nor change its fixed distance to the earth. The distance to the center of the earth here is composed of the earth radius $ R = 6370000 \, \mathrm{m} $ and the distance $ 550000 \, \mathrm{m} $ from the earth surface to the Starlink system: $ r = 6920000 \, \mathrm{m} $.

So the orbital velocity of the Starlink system must be $ \class{blue}{v} = 7586 \, \mathrm{m}/\mathrm{s} $. This velocity corresponds approximately to $ 7.6 \, \mathrm{km}/\mathrm{s} $.

First cosmic velocity: A stable orbit

We have considered the Starlink system at quite a large distance from the Earth. Of course, it is theoretically possible to find out what orbital velocity is necessary to orbit the Earth directly at the Earth's surface without crashing into the Earth. This orbital velocity, which is necessary to orbit a celestial body directly at the surface in a stable orbit, is called first cosmic velocity $ \class{blue}{v_1} $.

For this we only have to insert for the circular orbit radius $ r $ in equation 6 the radius $ R_{\text p} $ of the celestial body for which the first cosmic velocity is to be calculated.

$$ \begin{align} \class{blue}{v_1} ~=~ \sqrt{\frac{G\, \class{brown}{M_{\text p}} }{R_{\text p}}} \end{align} $$

Stable Circular Orbit of a Satellite at the Earth's Surface

The following table lists the first cosmic velocities for various celestial bodies:

Table : First cosmic velocity for different celestial bodies.
Celestial body	First cosmic velocity $ \class{blue}{v_1} $
Earth	7.9 km/s
Moon	1.7 km/s
Jupiter	42 km/s
Sun	236 km/s

Second cosmic velocity: How to escape the Earth's gravitational field?

Elon Musk wants to colonize Mars. To do so, his SpaceX rocket must first reach space without falling back to earth beforehand. A certain minimum velocity is required for this. The escape velocity, which is required to escape the gravitational attraction of a celestial body, is referred to as the escape velocity (or second cosmic velocity) $ \class{purple}{v_2} $.

We can find out this escape velocity with the help of one of the most powerful tools of physics, namely with the conservation of energy.

The gravitational potential energy $ W_{\text{pot}} $ of the SpaceX rocket at the ground can be derived from Newton's law of gravity and is given by:

$$ \begin{align} W_{\text{pot}} ~=~ G \, \frac{\class{brown}{M_{\text p}}}{R_{\text p}} \, \class{brown}{m} \end{align} $$

Since the rocket is on the ground before the escape attempt, we use the earth radius $ R_{\text p} $ in formula 8 for gravitational energy of the rocket. Here $ \class{brown}{M_{\text p}} $ is the mass of the Earth and $ \class{brown}{m} $ is the mass of the SpaceX rocket.

This gravitational energy must be compensated by the rocket. By definition, the rocket must have the escape velocity $ \class{purple}{v_2} $ for this. The kinetic energy $ W_{\text{kin}} $ of the rocket having this velocity is given by the following formula:

$$ \begin{align} W_{\text{kin}} ~=~ \frac{1}{2} \, \class{brown}{m} \, (\class{purple}{v_2})^2 \end{align} $$

This kinetic energy must be at least equal to the gravitational potential energy 8 of the rocket. If we equate the kinetic energy 9 with the gravitational potential energy 8, then the velocity $ \class{red}{v_2} $ corresponds exactly to the minimum velocity required to escape the gravitational field of the Earth.

$$ \begin{align} G \, \frac{\class{brown}{M_{\text p}}}{R_{\text p}} \, \class{brown}{m} ~=~ \frac{1}{2} \, \class{brown}{m} \, (\class{purple}{v_2})^2 \end{align} $$

The mass $ \class{brown}{m} $ of the SpaceX rocket is not important here because it cancels out in 10:

$$ \begin{align} G \, \frac{\class{brown}{M_{\text p}}}{R_{\text p}} ~=~ \frac{1}{2} \, (\class{purple}{v_2})^2 \end{align} $$

Let's just bring the factor 1/2 to the other side and take the root on both sides of the equation. Then we get a formula for the escape velocity:

$$ \begin{align} \class{purple}{v_2} ~=~ \sqrt{\frac{2G\, \class{brown}{M_{\text p}} }{R_{\text p}}} \end{align} $$

Escape Velocity of a Rocket to Leave the Gravitational Field of the Earth

We have considered here the earth as an example. Of course, any other round planet or celestial body can be considered in the same way. Only its radius and mass must be used. The following table shows some escape velocities for different celestial bodies:

Table : Zweite kosmische Geschwindigkeit (Fluchtgeschwindigkeit) für verschiedene Himmelskörper.
Celestial body	Second cosmic velocity $ \class{purple}{v_2} $
Earth	11.2 km/s
Moon	2.3 km/s
Jupiter	60 km/s
Sun	617 km/s

As you can see from the table, the SpaceX rocket on Earth must have at least an escape velocity of 11.2 km/s to enter weightlessness. In reality, it will be higher because air friction slows down the rocket.

Third cosmic velocity: How to escape the gravitational field of the solar system?

The third cosmic velocity $ \class{red}{v_3} $ is an escape velocity that is required to escape the gravitational field of the solar system from a planet within the solar system.

So if the SpaceX rocket wants to not only escape the Earth's gravitational field, but leave the entire solar system, it must also overcome the gravitational field of the sun. More precisely, it must also overcome the gravitational field of individual planets in the solar system. However, since these have a low mass compared to the sun, they can be neglected for the sake of simplicity.

To calculate the escape velocity $ \class{red}{v_3} $ of the SpaceX rocket, which launches from Earth and aims to leave the solar system, we utilize one of the most fundamental principles of physics known as the superposition principle. According to this principle, the gravitational fields of individual celestial bodies combine to form a total gravitational field of the solar system. And it is this overall gravitational field that the SpaceX rocket must overcome to break free from the solar system. Additionally, it should be taken into account that the rocket already has a component of the escape velocity due to the Earth's rotation around the Sun.

The escape velocity $ \class{red}{v_3} $ of the rocket to leave the solar system from a planet P can be approximated by the following formula:

$$ \begin{align} \class{red}{v_3} ~\approx~ \sqrt{ G \, \left( (\sqrt{2}-1)^2 \, \frac{ \class{brown}{M_{\text{s}}} }{ R } ~+~ 2\frac{ \class{brown}{M_{\text p}} }{ R_{\text p} } \right) } \end{align} $$

Escape Velocity of a Rocket to Leave the Solar System

Here, $ \class{brown}{M_{\text s}} $ represents the mass of the Sun, $ \class{brown}{M_{\text p}} $ represents the mass of the planet from which the rocket attempts to escape, $ R_{\text p} $ denotes the radius of the planet, and $ R $ represents the mean distance between the planet and the Sun. The mean distance is considered because planets orbit the Sun on an elliptical path rather than a circular one. For Earth, the mean distance is given by $ R = 1.496 \cdot 10^{11} \, \mathrm{m} $. The factor $ (\sqrt{2}-1)^2 $ takes into account the pre-existing velocity of the rocket due to the planet's rotation around the Sun.

The following table lists the third cosmic velocity for various planets in the solar system:

Table : Third cosmic velocity (escape velocity) for different planets of the solar system
Planet	Third cosmic velocity $ \class{red}{v_3} $
Earth	16.6 km/s
Mercury	20.3 km/s
Jupiter	60.4 km/s
Uranus	21.6 km/s

As you can see from the table, Mercury is much lighter than Earth, but because of its proximity to the Sun, it is more difficult to leave the solar system from Mercury.

Fourth cosmic velocity: How to escape the Milky Way's gravitational field?

The fourth cosmic velocity $ v_4 $ is an escape velocity that is required to escape the gravitational field of the Milky Way from Earth. This velocity can only be calculated numerically. It is approximately equal to $ v_4 \approx 550 \, \mathrm{km}/\mathrm{s} $.