# The Escape Velocity of Earth

PhysicsCelestial bodies

#### Class 11th Physics - Elasticity

6 Lectures 1 hours

#### Class 11th Physics - Oscillations

12 Lectures 2 hours

#### Class 11th Physics - Waves

14 Lectures 3 hours

## Introduction

It is often that we wonder how it is possible for any mass to deny the gravitational pull of the earth. The mass of the earth is $\mathrm{5.972\:\times\:10^24\:kg}$ is quite heavy and based on its mass, its pull will be much stronger in nature.

This leads to the question of its possibilities and that provides the answer that the rockets induce the principle of huge velocities that are better known to be escape velocities. With this force exerted, the rockets are able to suppress the high gravitational pull of the earth and travel to outer space.

### What are escape speed and its unit?

Escape speed is referred to as the minimum amount of speed that is required by a mass of the body that will propel the body from the surface of the earth. Escape speed is also known to be the escape velocity that leads to free respective body masses from the gravitational force (Haug, 2021).

However, the unit of escape velocity is stated as, meters per second or denoted by $\mathrm{m^{*}s^{-1}}$, which is similar to the SI unit of escape speed. The dimensional formula to the escape velocity is denoted as, [$\mathrm{M^{o}L^{1}T^{-1}}$].

Figure 1: Escape velocity

### What speed is known as Escape Velocity?

In order to get an understanding of the speed associated with the escape velocity, we need to consider the earth to be a huge mass. More to this it needs to be understood that the object that wants to escape the gravitational takes into consideration the mass of the earth and its respective distance from the object that wants to escape (Müller, 2018).

Therefore, the speed will be derived based on these two aspects. This in turn states that with the increase in the mass and decrease in the distance of the object that wants to escape, the escape velocity needs to be invariably greater.

### Derivation of escape speed

In order to derive the escape speed, the object and its velocity are to be considered. When the object tends to move, then the kinetic energy and its gravitational potential energy is considered to be zero. This in turn defines, that the according to the principle of conservation of energy it can be stated as

(K+Ug)i = (K+Ug)f

, where, K is the kinetic energy and is represented by the formula, $\mathrm{1/2\:m^{*}v^{2}}$. The letter U denotes energy of gravitational potential and that represents, $\mathrm{(GMm)/r\:(Qrg.northwestern,\:2022)}$. Therefore, the minimum escape velocity that is required is represented by $\mathrm{Va_{e}}$ which equals $\mathrm{\sqrt{2\:g*}}$ and that in turn equals ($\mathrm{\sqrt{2GM/r}}$). In this equation, g is stated as the universal gravitational constant, where, g is represented by $\mathrm{GM/r^2}$ respectively.

Figure 2: Derivation of Escape velocity

### Value for escape velocity of the earth

The value for escape velocity can be determined by the respective values that include acceleration due to gravity for the earth, that is, $\mathrm{g\:=\:9.8m/s^2}$, the radius of the earth is, $\mathrm{R\:=\:6.4\:*\:106m}$ (Letstalkscience, 2022). Therefore, the value of escape velocity is determined by $\mathrm{ve\:=\:11.2\:km/s}$ that can be considered as an approximation.

Figure 3: Value of Escape velocity

### Escape velocity for several objects

It needs to be noted that as escape velocity is primarily dependent on the mass of the planet and also the distance that exists between the planet and the object that tends to escape, the escape velocity for objects changes depending upon the mass of the planets within the solar system (Vanderbilt, 2022).

For example, the escape velocity of Jupiter is the highest and on the other hand, the escape velocity of Mars is the lowest. However, for the sun the escape velocity is $\mathrm{618 km /s}$, for the moon the value of escape velocity is $\mathrm{2.38 km/s}$ (Scienceabc, 2022). In addition, Ceres the value for escape velocity is $\mathrm{0.64km/s}$, respectively.

## Conclusion

In this tutorial, it is noticed that an in-depth study has been conducted that has dealt with the notion of escape velocity. It is often to our wonder that we do not understand why the rockets of such negligible mass than the earth can easily escape the surface of the earth. However, it needs to be understood that there lays the simple physics that denotes the escape velocity, which is dependent on the aspects of mass and distance. Therefore, it is understood that the escape velocity is different for different celestial bodies.

### FAQs

Q1. What is the difference noticed between orbital velocity and escape velocity?

There lies a striking difference between orbital velocity and escape velocity. The orbital velocity refers to the speed that is necessary for reaching out the orbit of any planets or any stars within the solar system. On the other hand, escape velocity is defined as the speed that is necessary to leave the respective orbits.

Q2.What is the relationship between the two notions of escape velocity and orbital velocity?

For any object, escape velocity is referred to as the function that is associated with the velocity within the orbits. The products noted from the orbital velocities and their square root of 2 can achieve Escape velocity. In addition, the gravitational force that tends to control the object can also be obtained. For example within our solar system, the planet of Jupiter seems to have the highest escape velocity and on the other, the planet Mercury seems to have the lowest escape velocity

Q3. What are the factors on which escape velocity depends?

The factors, on which the escape velocity for a particular object depends on earth, include the mass of the planet earth and also the mean density which is denoted with the symbol ρ. However, it needs to be noted that the escape velocity does not depend on the direction of the projected body and its respective mass.

Updated on 13-Oct-2022 11:19:47