Escape  velocity

In the universe massive bodies have bounded all small bodies with bound energy, Now to escape small bodies from the surface of massive bodies gravitational field influence, You must required a minimum velocity to free small object from massive bodies surface, this minimum velocity is know as escape velocity.

What is escape velocity 

Escape velocity of a planet is that minimum velocity given to an object on that planet due to which it will just escape the planet and will never come back again on that planet.


When you through an object from the surface of the Earth, it goes up some height and then start fall back towards  the Earth surface, This happen due to gravitational field attraction by Earth around it.

Earth gravity acts vertically downward continuously and due to this object speed decrease continuously at per second  9.8 m/s and finally reaches at a height at which final speed becomes zero, but gravity acceleration acting downward and then object start coming back towards surface of the Earth similarly every second speed increase 9.8 m/s and finally reaches the Earth Surface. 


But when you will through the object with so much high velocity that it will reach very high where gravity pulling effect vanishes and then the object velocity becomes zero that that point object will stop in space and will never come back on the Earth surface that minimum through velocity is escape velocity.


See the below picture.

                                                                 

escape velocity,what is escape velocity
Escape Velocity

                                                                       

Path of escape velocity

When you through an object horizontally with some velocity that object fall on the Earth surface, but as you increase the greater velocity that object does not fall on the Earth surface but it rotates around the Earth surface in an elliptical path.
But this is not the escape velocity because object is in the gravitational field influence not crosses the gravitational field. 

Again you through the object with more velocity then object does not fall on the Earth but rotate in circular path, but this is not escape velocity, object is in the gravitational field influence, it has not crosses the gravitational field.

Again you increase more velocity and thrown object horizontally it will rotates in elliptical path around the Earth but does not fall on the Earth due to Earth rotation as the distance fall by object Earth displace the other location.
Hence object fall always the distance but never fall on the Earth surface.


Now you through the object with very high velocity such that the object crosses the gravitational field, So in this case the path of the object will be parabola and it will not rotates around the Earth surface it will reach at infinity. It has escaped or crosses the gravitational field now Earth gravitational field will not attract the object and this velocity is called escape velocity.
See the picture below.
                                                                     

escape velocity,escape velocity curve
Escape velocity curve

Where does Earth gravitational field becomes zero ? Well Earth gravitational field only and only zero at infinity, So to escape the gravitational field thrown object must reach infinity, now at infinity the object velocity will be zero, but at this point Earth will not attract the object.

Imagine before infinity if object velocity will be zero then again Earth will attract that object and will continue to fall and rotate around the Earth.

Now this is assumption that in space Earth is available only otherwise other planet will also attract that object. So to escape any object that means that object must reach at infinity.       

How escape velocity is calculated ? 

Escape velocity is calculated using energy conservation, Total initial mechanical   energy equal to total final mechanical energy. You can refer previous post for potential energy concept, I have discussed at infinity potential energy is zero because there is no gravitational field effect.   

KEi + PEi   = KEf  + PEf 
1/2mve² – GmMe/Re  =  0 + 0    (gravitational field at infinity is zero so potential energy is zero and at that point object velocity will also be zero, so its kinetic energy will be also zero)

So   1/2mve² – GmMe/Re = 0 
1/2mve²  = GmMe/Re  or  ve²  = 2GMe/Re   Vₑ  =√2GMe/Re here G is universal constant Me is mass of the Earth and Re is radius of the Earth so escape velocity depend upon only planet so for different planets its value will be different, it does not depend upon any other things even projected mass.

You know g = GMe/Re² so GMe  = gRe² put this value in the above equation Vₑ  =√2GMe/Re  =  Vₑ   = √2gRe   now put the value g 9.8m/s² and Re = 6400 km  you will get  Vₑ = 11.2 km/s .


Vₑ   = 11.2 km/s this is escape velocity . If any mass given this velocity at surface of the Earth it will not come back again on the surface of the Earth.

Escape velocity question

                                                               

escape velocity,escape velocity question
See the above picture an object of mass m thrown from height h = Re, then find the escape velocity .

You know that to calculate escape velocity for any planet, you must use mechanical energy conservation concept, and for escape velocity just remember that object must reach infinity.
So use these two concept, you can find velocity.

Hence mechanical energy conservation  Ui  = Uf  or  (KEi + PEi )  =  ( KEf + PEf )  
1/2mV²e  + (-GMem)/2Re  = 0 + 0 
 1/2mV²e  =     GMem)/2Re  or  V²e = GMe/Re  V’e  = √GMe/Re , you know that Ve = √2GMe/Re hence V’e  = Ve/√2 = 11.2/1.414 = 7.92 km/s
Hence escape velocity v’e  = 7.92 km/s.

Q What will be Escape velocity find in the figure given ?
                                                                     

escape velocity,escape velocity questions

  
As from figure mass m is to escape from the surface of the Earth, and away from the Sun along the line joining Earth and Sun.

Where Mass of Sun is given Ms = 3×10⁵Me and r is distance between Earth centre and Sun centre and given r = 2.5×10⁴Re , Where Me is mass of Earth and Re is radius of Earth.
it is also given that normal escape velocity from the surface of Earth is Ve = 11.2 km/s .

Now suppose that escape velocity is Ve’, now you know that only in case of Earth escape velocity of any mass is 11.2km/s, but here Sun is also available so you have to consider Earth and Sun together as system and apply conservation of mechanical energy.

Hence Ui  = Uf 
(PEi + KEi )  = ( PEf + KEf )
-GMem/Re -GMsm/(Re+r) +1/2mV’²e = 0 + 0  ( final kinetic and potential energy is zero at ∞)

-GMe/Re – GMs/r + 1/2V’²e  = 0  ( here Re is very very small than r so (Re+r)→r )
 1/2V’²e  =  GMe/Re + Gx 3×10⁵Me/2.5×10⁴Re  ( Put value of Ms and r from above )
  1/2V’²e = GMe/Re +   12GMe/Re  = 13GMe/Re 
  V’²e  = 13x2GMe/Re 
  V’e  = √13x√2GMe/Re  = √13×11.2  (  √2GMe/Re = 11.2 km/s )
  V’e  = 3.6×11.2  =  40.2 km/s
Hence escape velocity will be V’e  = 40.2 km/s.


I will continue this post with more important concept, I hope you have enjoyed learning escape velocity. I want your feedback comments, share and likes. Learn more and grow thanks for sharing.

Dated 20th Jan 2019