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Revision Notes on Gravitation and Projectile

Gravitation:-


	

	
Kepler’s first law (law of elliptical orbit):- A planet moves round the sun in an elliptical orbit with sun situated at one of its foci.
	
	

	
Kepler’s second law (law of areal velocities):- A planet moves round the sun in such a way that its areal velocity is constant.
	
	

	
Kepler’s third law (law of time period):- A planet moves round the sun in such a way that the square of its period is proportional to the cube of semi major axis of its elliptical orbit.
	


T2 ∝ R3

Here R is the radius of orbit.

T2 = (4π2/GM)R 3


	

	
Newton’s law of gravitation:-
	


Every particle of matter in this universe attracts every other particle with a forcer which varies directly as the product of masses of two particles and inversely as the square of the distance between them.

F= GMm/r2

Here, G is universal gravitational constant. G = 6.67 ´10 -11 Nm2 / kg2


	

	
Dimensional formula of G: G = Fr2/Mm =[MLT-2][L2]/[M2] = [M-1L3T-2]
	



	

	
Acceleration due to gravity (g):- g = GM/R2
	



	

	
Variation of g with altitude:- g' = g(1- 2h/R),  if h<<R. Here R is the radius of earth and h is the height of the body above the surface of earth.
	



	

	
Variation of g with depth:- g' = g(1- d/R). Here g' be the value of acceleration due to gravity at the depth d.
	



	

	
Variation with latitude:-
	


At poles:- θ = 90°, g' = g

At equator:- θ = 0°, g' = g (1-ω2R/g)

Here ω is the angular velocity.


	

	
As g = GMe/Re2 , therefore gpole > gequator
	
	

	
Gravitational Mass:- m = FR2/GM
	
	

	
Gravitational field intensity:-
	


E = F/m

= GM/r2


	

	
Weight:- W= mg
	
	

	
Gravitational intensity on the surface of earth (Es):- 
	


Es = 4/3 (πRρG)

Here R is the radius of earth, ρ is the density of earth and G is the gravitational constant.


	

	
Gravitational potential energy (U):- U = -GMm/r
	


(a) Two particles: U = -Gm1m2/r

(b) hree particles: U = -Gm1m2/r12 – Gm1m3/r13 – Gm2m3/r23


	

	
Gravitational potential (V):- V(r) =  -GM/r
	


At surface of earth,

Vs=  -GM/R

Here R is the radius of earth.


	

	
Escape velocity (ve):- 
	


It is defined as the least velocity with which a body must be projected vertically upward   in order that it may just escape the gravitational pull of earth.

ve = √2GM/R

or, ve = √2gR = √gD

Here R is the radius of earth and D is the diameter of the earth.


	

	
Escape velocity (ve) in terms of earth’s density:- ve = R√8πGρ/3
	



	

	
Orbital velocity (v0):-
	


v0 = √GM/r

If a satellite  of mass m revolves in a circular orbit around the earth of radius R and h be the height of the satellite above the surface of the earth, then,

r = R+h

So, v0 = √MG/R+h = R√g/R+h

In the case of satellite, orbiting very close to the surface of earth, then orbital velocity will be,

v0 = √gR


	

	
Relation between escape velocity ve and orbital velocity v0 :- v0= ve/√2  (if h<<R)
	



	

	
Time period of Satellite:- Time period of a satellite is the time taken by the satellite to complete one revolution around the earth.
	


T = 2π√(R+h)3/GM = (2π/R)√(R+h)3/g

If h<<R, T = 2π√R/g


	

	
Height of satellite:- h = [gR2T2/4π2]1/3 – R
	



	

	
Energy of satellite:-
	


Kinetic energy, K = ½ mv02 = ½ (GMm/r)

Potential energy, U = - GMm/r

Total energy, E = K+U

= ½ (GMm/r) + (- GMm/r)

= -½ (GMm/r)


	

	
Gravitational force in terms of potential energy:- F = – (dU/dR)
	



	

	
Acceleration on moon:-
	


gm = GMm/Rm2 = 1/6 gearth 

Here Mm is the mass of moon and Rm is the radius of moon.


	

	
Gravitational field:-
	


(a)    Inside:-

 

(b)   Outside:-

 


	

	
GRAVITATIONAL POTENTIAL & FIELD DUE TO VARIOUS OBJECTS
	



	

		

			

			
Causing Shape
			
			

			
Gravitational Potential (V)
			
			

			
Gravitational Field (I or E)
			
			

			
Graph

			
V vs R
			
		
		

			

			
POINT MASS
			
			

			
 
			
			

			
 
			
			

			
 
			
		
		

			

			

			
			

			


			
 
			
			

			


			
 
			
			

			

			
		
		

			

			
AT A POINT ON THE AXIS OF RING
			
			

			
 
			
			

			
 
			
			

			
 
			
		
		

			

			
 
			
			

			


			
 

			
  
			
			

			


			
 
			
			

			
 

			

			
		
		

			

			
ROD
			
		
		

			

			
1. AT AN AXIAL POINT
			
			

			
 
			
			

			
 
			
			

			
 
			
		
		

			

			
 

			
 
			
			

			
 

			
 

			
V = –
			
			

			
 

			

			
			

			
 

			

			
		
		

			

			
2. AT AN EQUATORIAL POINT
			
			

			
 
			
			

			
 
			
			

			
 
			
		
		

			

			
 

			

			
			

			
 

			
 

			
 

			


			
 
			
			

			
 

			
 

			
 

			

			
			

			


			

			

			 
			
		
		

			

			
CIRCULAR ARC
			
			

			
 
			
			

			
 
			
			

			
 
			
		
		

			

			
 

			
 
			
			

			
 

			
 

			


			
 

			
 
			
			

			
 

			
 

			


			
 
			
			

			
 

			
 

			

			
		
		

			

			
HOLLOW SPHERE
			
			

			
 
			
			

			
 
			
			

			
 
			
		
		

			

			


			
 
			
			

			


			

			

			

			
 

			
 
			
			

			


			

			

			
 
			
			

			
 
			
		
		

			

			
SOLID SPHERE
			
			

			
 
			
			

			
 
			
			

			
 
			
		
		

			

			


			
 
			
			

			


			

			

			


			
 

			
 

			
 
			
			

			


			

			


			
 

			
 

			
 
			
			

			

			
		
		

			

			
LONG THREAD
			
			

			
 
			
			

			
 
			
			

			
 
			
		
		

			

			

			
			

			
V = ∞
			
			

			
 

			
 

			
 

			

			
			

			
 
			
		
	


Projectile:- 


	

	
Projectile fired at angle α  with the horizontal:- If a particle having initial speed u is projected at an angle α (angle of projection) with x-axis, then,
	


Time of Ascent, t = (u sinα)/g

Total time of Flight, T = (2u sinα)/g

Horizontal Range, R = u2sin2α/g

Maximum Height, H = u2sin2α/2g

Equation of trajectory, y = xtanα-(gx2/2u2cos2α)

Instantaneous velocity, V=√(u2+g2t2-2ugt sinα)

and         

β = tan-1(usinα-gt/ucosα)


	

	
Projectile fired horizontally from a certain height:-
	


Equation of trajectory: x2 = (2u2/g)y

Time of descent (timer taken by the projectile to come down to the surface of earth), T = √2h/g

Horizontal Range, H = u√2h/g. Here u is the initial velocity of the body in horizontal direction.

Instantaneous velocity:-

V=√u2+g2t2

If β be the angle which V makes with the horizontal, then,

β = tan-1(-gt/u)


	

	
Projectile fired at angle α with the vertical:-
	


Time of Ascent, t = (u cosα)/g

Total time of Flight, T = (2u cosα)/g

Horizontal Range, R = u2sin2α/g

Maximum Height, H = u2cos2α/2g

Equation of trajectory, y = x cotα-(gx2/2u2sin2α)

Instantaneous velocity, V=√(u2+g2t2-2ugt cosα) and β = tan-1(ucosα-gt/usinα)


	

	
Projectile fired from the base of an inclined plane:-
	


Horizontal Range, R = 2u2 cos(α+β) sinβ/gcos2α

Time of flight, T = 2u sinβ/ gcosα

Here, α+β=θ

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