### Escape velocity – Definition, Derivation of equation and Values

##### November 3, 2017

# Escape velocity

**Escape velocity** is defined as the minimum velocity with which if an object is thrown from the surface of the earth or any other planet or satellite then the object defeats the gravitational attraction working on it and can go beyond this attraction.

From the term Escape Velocity itself we get an idea of its definition. It’s a velocity that helps to escape – isn’t it? But Escape from what and how? Should we use the term velocity or we should use the term speed? This is what we are going to discuss today in our post on escape velocity. In the next paragraphs we will derive the escape velocity equation or formula and calculate the escape velocity of the earth.

If we throw something vertically upwards or in any other direction, after sometime it it comes down towards the surface of the earth.We know this happens due to the gravitational attraction of the earth on the object thrown.

Now if this initial speed (the speed with which it is being thrown) is gradually raised, that object seems to traverse more and more before coming back to the earth’s surface.

## what is escape velocity?

At some point, this initial speed can reach such a value that the object doesn’t come back; it defeats the gravitational force without any need of any additional propulsion and goes beyond it to the infinite distance.This speed is called the escape velocity. To note that esc. velocity actually is a speed because it doesn’t mention any direction, provided its path is not intersecting with the earth’s surface.

This is to be noted that this concept of escape velocity is not only applicable for the earth, but also it is equally applicable for all other planets and satellites. But the value of this esc. velocity would be different for them. As we derive the equation of this escape velocity, this point will get clearer.

## Escape Velocity Derivation – Deriving the formula

We will derive the equations using the following * condition*:

**The initial kinetic energy of the object would at least equalize the amount of work done to send the same object from the surface of the earth to infinite distance.**

Let’s find the amount of work done first.

### Work done to send the object from the surface of the earth to infinite distance

Say mass of the earth is M and its radius is R.

If an object of mass m is positioned at a distance x from the centre of the earth, then the Force of **Gravitational Attraction** on the object is expressed as

F = (G M m)/x²

Now if this object is moved a small distance dx against this gravitational force, along the line of action of the same gravitational force then the **work** done is

**W = F dx= ((G M m)/x²) dx**

So to get the expression of the work done to send the object from the surface of the earth to infinite distance, we need to integrate the above expression with 2 limits, R and infinity. [ R representing the lower limit as it’s the distance of the surface of the earth from its centre]

**By integrating with lower limit R and upper limit infinity:**

** W= ∫((G M m)/x²)dx = (GMm)/R ………………….. (1)**

### Initial Kinetic Energy

Now if the escape velocity of earth is V, then the initial kinetic energy**(KE) of the object would be = ½ .m.V^2** …………. (2)

### Getting the expressions now

If this initial KE (equation 2) can at least equalize the work done (in eqn 1), then the object can really escape the earth’s gravitation.

So equalizing equation 1 and 2,

**½ .m.V^2 = (GMm)/R**

V^2 = (2GM)/R

Escape Velocity**V = √( 2GM/R) ……… (3) [formula 1]**

As the expression of the acceleration due to gravity on the earth’s surface is **g = (GM)/R^2**

[for this equation read this: acceleration due to gravity]

i.e., **g.R^2 = GM**,

therefore we can rewrite equation number 3 as follows:

**V = [ (2 g R^2) / R ]^(1/2) = (2gR)^(1/2) = √(2gR)**

**Escape Velocity**

V**=√(2gR)**…………** (4) [formula 2]**

**It’s to be noted that this velocity doesn’t depend on the mass of the object which is being thrown.**

But this esc. velocity certainly depends on the mass and the radius of the planet or satellite or star from where the object is being thrown.(see equation 3 above)

### 2 Equations derived

V = √( 2GM/R)

V= √(2gR)

### Esc. Velocity is **√2** times of Orbital velocity for nearby orbits

For an orbit which is pretty close to the earth, we can ignore the height above the surface and can consider only the radius of the earth as the distance between the satellite and the earth’s centre. For this scenario we get an equation for orbital speed

**V _{orbital} = √(gR) [ to find this derivation please refer to our post: Orbital Velocity]**

Whereas as found, ** V _{escape} = √(2gR)**

Very clearly its visible that **the Escape velocity is √2 times of Orbital velocity for nearby orbits.**

## Value of the earth’s Escape Velocity

We know the value of g on the surface of the earth. It’s 9.8 m/s^2.

The radius of the earth (R) is 6400 Km = 6.4 X 10^6 meter.

So from these data, we get the **esc. velocity of the earth** from equation 4 above:

Esc Velocity of earth =V = √(2 X 9.8 X 6.4 X 10^6) m/s =11200 m/s =**11.2 km/s = 7 mile/second**

So if an object is thrown upwards with a velocity of 11.2 Km/Second from the earth’s surface, it will be able to escape i.e. go beyond the gravitational field of the earth.

### Escape Velocity for moon and..

Escape Value of the moon: 2.38 km/sec

and that of Jupiter: 59.5 km/sec

#### Suggested Reading:

**Orbital Velocity**

**Universal Law of Gravitation**