Revision notes of Gravitation

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/ gcost

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