JEE Physics/ Lakshmikanta Satapathy/ Gravitation QA part 3/ JEE question on finding the escape velocity from the surface of a planet from velocity of projection and height gained by a bullet

1. Physics Helpline
L K Satapathy Gravitation QA 3
Escape Velocity
2
e
GM
v
R


2. Physics Helpline
L K Satapathy Gravitation QA 3
Question : A bullet is fired vertically upwards with velocity v from the surface of a
spherical planet . When it reaches its maximum height , its acceleration due to the
planet’s gravity is (1/4)th of its value at the surface of the planet . Then the escape
velocity from the surface of the planet is
( ) 2 ( ) 2 ( ) ( ) ( 2 1)
2
v
a v b v c d v
Answer :
2
' . . . (2)
( )
GM
g
R h


2 2
4 ( ) 4
g GM GM
g
R h R
   

Mass of earth = M , Radius of earth = R , Let mass of bullet = m
Acceleration due to gravity on the surface of earth is 2
. . . (1)
GM
g
R

At a height h , acceleration due to gravity becomes
It is given that
2 2
( ) 4R h R  
2R h R h R    

3. Physics Helpline
L K Satapathy Gravitation QA 3
Correct option = (b)
21
2 2
GMm GMm
mv
R R
    
21
2 2
GMm
mv
R
 
GM
v
R
 
]2 [
2
e
GM
v v
R
Ans 
Velocity of projection from surface = v 21
2
K mv 
Potential energy on the surface is
GMm
U
R
 
 Total energy on the surface is 21
2
GMm
E mv
R
  
At height h = R , it stops.  KE = 0 and PE
2
GMm
R
 
 Total energy
2
GMm
R
 
 Conservation of energy gives
 Escape velocity from the surface is

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L K Satapathy
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