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 ve while vs is the speed of the satellite in orbit. The unit of escape velocity is meters per second (ms-1). The escape velocity formula is ve = √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.
You may also like to read:
Solved Problem: How to Calculate the Wavelength of Radiation
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 = me [ where me is the mass of the earth], the above equation will become
mv2/r = Gmme/r2
one r on the right-hand side (RHS) will cancel one r on the left-hand side (LHS), leaving us with
mv2 = Gmme/r
We are making v the subject of the formula. Therefore, we can divide both sides by m
v2 = Gme/r
We can now take the square root of both sides to make v subject of the formula
√v2 = √(Gme/r)
and the square root will cancel the square on the side of v to give us
v = √(Gme/r)
and if r = re = radius of the earth, and v = vs = which is the velocity by which a satellite circles the earth. And we can now write it as:
vs = √(Gme/re) [vs = 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 = Gmme/r2
Therefore w is now equal to
w = (Gmme/r2) x r
r will cancel the square leaving only one r to give us the result of work done (w)
w = Gmme/r
By equating the above work done (w) with kinetic energy (k.e) of the body, we will have
k.e = Gmme/r
since we know k.e = 1/2 (mv2)
we can now write
1/2 (mv2) = Gmme/r
But v = ve = escape velocity
This shows that we now have
1/2 (mve2) = Gmme/r
m will cancel each other on both sides to leave us with
1/2 (ve2) = Gme/r
By cross-multiplying the above equation, we will now have
(ve2)r = 2Gme
Now, we divide each side by r to obtain
ve2 = 2Gme/r
We can now take the square root of both sides to make vs subject of the formula, which is
√ve2 = √(2Gme/r)
Therefore, the above expression becomes
ve = √(2Gme/r)
However, me = gr2/G
By inserting the above expression into ve = √(2Gme/r)
we will have ve = √(2G/R x gR2/G) [ where R = radius of the earth = r]
The above equation will become
ve = √2gR
Therefore, the formula for escape velocity is ve = √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 ve = ?
And the formula for calculating escape velocity, ve = √2gR
We now substitute our data into the above formula to have
ve = √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
vs = ?
We need to apply the formula below
vs = √(Gme/re) and since me = gre2/G
we have vs = √gr [where r = R = re]
Thus,
vs = √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:
me = 150kg
Gravitational constant, G = 6.67×10-11Nm2kg-2
Radius of the earth, R = 6.4×106m
We apply the formula ve = √(2Gme/r)
By substituting our data into the above equation, we will have
ve = √([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)
ve = √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.
You may also like to read:
How to Calculate Capacitive Reactance
How to Calculate the Work Function of a Metal
