How to Calculate Escape Velocity of a Satellite

What is Escape Velocity?

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. In this article, you will how to calculate escape velocity. I will derive the formula, and then solve a few problems for you.

The symbol for escape velocity is v_e while v_s is the speed of the satellite in orbit. The unit of escape velocity is meters per second (ms^-1). The escape velocity formula is v_e = √2gR

How to Calculate Escape Velocity of a Satellite

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 mv²/r [ where m = mass of the object, v = velocity of the object, and r = radius]

Also, the formula for gravitational force is Gm₁m₂/r² [ where G = gravitational constant, m₁and m₂ are the masses of the two bodies apart, and r = radius]

Therefore, we can now equate centripetal force (mv²/r) and the gravitational force (Gm₁m₂/r²)

mv²/r = Gm₁m₂/r²

by making m₁ = m, and m₂ = m_e [ where m_e is the mass of the earth], the above equation will become

mv²/r = Gmm_e/r²

one r on the right-hand side (RHS) will cancel one r on the left-hand side (LHS), leaving us with

mv² = Gmm_e/r

We are making v the subject of the formula. Therefore, we can divide both sides by m

v² = Gm_e/r

We can now take the square root of both sides to make v subject of the formula

√v² = √(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_s = which is the velocity by which a satellite circles the earth. And we can now write it as:

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/r²

Therefore w is now equal to

w = (Gmm_e/r²) 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 (mv²)

we can now write

1/2 (mv²) = Gmm_e/r

But v = v_e = escape velocity

This shows that we now have

1/2 (mv_e²) = Gmm_e/r

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

1/2 (v_e²) = Gm_e/r

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

(v_e²)r = 2Gm_e

Now, we divide each side by r to obtain

v_e² = 2Gm_e/r

We can now take the square root of both sides to make v_s subject of the formula, which is

√v_e² = √(2Gm_e/r)

Therefore, the above expression becomes

v_e = √(2Gm_e/r)

However, m_e = gr²/G

By inserting the above expression into v_e = √(2Gm_e/r)

we will have v_e = √(2G/R x gR²/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

Solved Problems: How to Calculate Escape Velocity

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

Problem 1

Calculate the escape velocity of a satellite launched from the earth’s surface if the radius of the earth is 6.4×10⁶m.

Solution

Data

The radius of the earth, r = 6.4×10⁶m

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 6.4×10⁶ = 11.3km/s or 11314m/s

Thus, the escape velocity is 11314m/s

Problem 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 = 6.4×10⁶m. Calculate the speed of the satellite.

Solution

Data

The radius of the earth, R = r = 6.4×10⁶m

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_e²/G

we have v_s = √gr [where r = R =r_e]

Thus,

v_s = √9.8 x 6.4×10⁶ = 8×10³ms^-1

Therefore, the speed of the satellite is 8×10³ms^-1

Problem 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^-11Nm²kg^-2, R = 6.4×10⁶m, and the mass of the earth is 6.0×10²⁴kg.

Solution

Data:

m_e = 150kg

Gravitational constant, G = 6.67×10^-11Nm²kg^-2

Radius of the earth, R = 6.4×10⁶m

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] /6.4×10⁶) = √(2.001 x 10^-8 / 6.4×10⁶) = √(2.001 x 10^-8 / 6.4×10⁶) =√(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.

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