## What is Escape Velocity

In this post, I will help you to master how to calculate escape velocity. I will define escape velocity, derive the formula, and then solve a few examples for you.

**Definition of Escape Velocity:** Escape velocity is the minimum amount of kinetic energy that an object needs in order to permanently escape the influence of gravitation on earth or an astronomical body.

The symbol for escape velocity is v_{e} while v_{s} is the speed of the satellite in the orbit. The unit of escape velocity is ms^{-1}. The Formula for calculating escape velocity is v_{e} = **√**2gR

### Explanation of Escape Velocity

Of course, we currently have thousands of man-made satellites moving around the earth in order to send and receive information.

These satellites circle the earth continuously. Additionally, these satellites receive gravitational attraction, and there is a centripetal force that keeps them moving around the orbit non-stop.

The engineers behind the invention of those satellites applied the knowledge of escape velocity in order to keep them circling our favorite planet.

Therefore, for the body to continue moving, we need a specific velocity that will help it to escape the gravitational influence and it is the escape velocity of the body.

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## How to Derive the Formula for Escape Velocity

Here is a derivation of a formula that will help you to understand how to calculate the escape velocity of the earth, moon, mars, and the sun:

### First Method of Deriving the Formula for Speed of Satellites

Here is how to derive the formula for the speed of light. It is important to know that in this case, **centripetal force is equal to the gravitational force**.

And **the formula for centripetal force is mv ^{2}/r** [ where m = mass of the object, v = velocity of the object, and r = radius]

Also, the formula for **gravitational force is Gm _{1}m_{2}/r^{2}** [ where G = gravitational constant, m

_{1}and m

_{2}are the masses of the two bodies apart, and r = radius]

Therefore, we can now equate **centripetal force** (**mv ^{2}/r**) and

**the gravitational force**(

**Gm**)

_{1}m_{2}/r^{2}**mv ^{2}/r** =

**Gm**

_{1}m_{2}/r^{2}by making **m _{1}** =

**m, and**[ where m

**m**=_{2}**m**_{e}_{e}is the mass of the earth], the above equation will become

**mv ^{2}/r** =

**Gmm**

_{e}/r^{2}one r on the right-hand side (RHS) will cancel one r on the left-hand side (LHS), leaving us with

**mv ^{2}** =

**Gmm**

_{e}/rWe are making v the subject of the formula. Therefore, we can divide both sides by m

**v ^{2}** =

**Gm**

_{e}/rWe can now take the square root of both sides to make v subject of the formula

√** v^{2} = √(Gm_{e}/r**)

and the square root will cancel the square on the side of v to give us

** v = √(Gm_{e}/r**)

and if r = r**_{e}** = radius of the earth, and v = v

**= which is the velocity by which a satellite circles the earth. And we can now write it as:**

_{s}v_{s} = √(Gm_{e}/r**_{e}**) [v

_{s}= speed of satellite]

### Second Method of Deriving the Formula for Escape Velocity

We can also derive the formula for escape velocity by equating the work done in moving the mass of an object from the earth through a distance r.

Thus, since work (w) = force (F) x distance (r)

we now have w = F x r

But F = Gmm_{e}/r^{2}

Therefore w is now equal to

w = (Gmm_{e}/r^{2}) x r

r will cancel the square leaving only one r to give us the result of work done (w)

w = Gmm_{e}/r

By equating the above work done (w) with kinetic energy (k.e) of the body, we will have

k.e = Gmm_{e}/r

since we know k.e = 1/2 (mv^{2})

we can now write

1/2 (mv^{2}) = Gmm_{e}/r

But v = v_{e} = escape velocity

This shows that we now have

1/2 (mv_{e}^{2}) = Gmm_{e}/r

m will cancel each other on both sides to leave us with

1/2 (v_{e}^{2}) = Gm_{e}/r

By cross-multiplying the above equation, we will now have

(v_{e}^{2})r = 2Gm_{e}

Now, we divide each side by r to obtain

v_{e}^{2} = 2Gm_{e}/r

We can now take the square root of both sides to make v_{s} subject of the formula, which is

**√**v_{e}^{2} = **√**(2Gm_{e}/r)

Therefore, the above expression becomes

v_{e} = **√**(2Gm_{e}/r)

However, m_{e} = gr^{2}/G

By inserting the above expression into v_{e} = **√**(2Gm_{e}/r)

we will have v_{e} = **√**(2G/R x gR^{2}/G) [ where R = radius of the earth = r]

The above equation will become

v_{e} = **√**2gR

Therefore, the formula for escape velocity is v_{e} = **√**2gR

## Examples

Here are a few examples to help you understand the calculation of escape velocity.

### Example 1

Calculate the escape velocity of a satellite launched from the earth’s surface if the radius of the earth is 6.4×10^{6}m.

**Solution**

**Data**

The radius of the earth, r = 6.4×10^{6}m

Gravitational acceleration, g = 10ms^{-2}

Escape velocity is v_{e} = ?

And the formula for calculating escape velocity, v_{e} = **√**2gR

We now substitute our data into the above formula to have

v_{e} = **√**2 x 10 x 6.4×10^{6} = 11.3km/s or 11314m/s

Thus, the escape velocity is 11314m/s

### Example 2

For a satellite that is close to the earth’s surface, the radius of the orbit may be assumed to be equal to the earth’s radius, r = 6.4×10^{6}m. Calculate the speed of the satellite.

**Solution**

**Data**

The radius of the earth, R = r = 6.4×10^{6}m

Gravitational acceleration, g = 9.8ms^{-2}

v_{s} = ?

We need to apply the formula below

v_{s} = √(Gm_{e}/r**_{e}**) and since m

_{e}= gr

_{e}

^{2}/G

we have v_{s} = √gr [where r = R =^{ }r_{e}]

Thus,

v_{s} = √9.8 x 6.4×10^{6} = 8×10^{3}ms^{-1}

Therefore, the speed of the satellite is 8×10^{3}ms^{-1}

### Example 3

A rocket of mass 150-kilogram is fired from the earth’s surface so that it just escapes from the gravitational influence of the earth. Calculate the velocity with which it escapes where G = 6.67×10^{-11}Nm^{2}kg^{-2}, R = 6.4×10^{6}m, and the mass of the earth is 6.0×10^{24}kg.

**Solution**

**Data:**

m_{e} = 150kg

Gravitational constant, G = 6.67×10^{-11}Nm^{2}kg^{-2}

Radius of the earth, R = 6.4×10^{6}m

We apply the formula v_{e} = **√**(2Gm_{e}/r)

By substituting our data into the above equation, we will have

v_{e} = **√**([2 x 6.67×10^{-11} x 150] /6.4×10^{6}) = **√**(2.001 x 10^{-8} / 6.4×10^{6}) = **√**(2.001 x 10^{-8} / 6.4×10^{6}) =**√**(3.127 x 10^{-15})

v_{e} = **√**3.127 x 10^{-15} = 5.592ms^{-1}

Therefore, the escape velocity of the rocket is 5.592 meters per second.

Drop a comment on what you don’t understand so that we can provide you with more resources.

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