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GRAVITATION.Pdf GRAVITATION (For V SEMESTER ) (a) Newton’s law of gravitation: Every body in the universe attracts every other body with a force which is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centres. uur Gm1 m 2 $ Force on m2 due to m1 in vector form is F21 = − 2 r 12 . r12 Gm m In general F = 1 2 . r 2 (b) Gravitational Field: Gravitational field intensity at a point is defined as gravitational force experienced by a unit mass kept at that point. ur F ur E = [ E → intensity at P on m2 ] m2 Gm m Gm 1 2 $ ⇒ 1 $ F= − 2 r 12 E= − 2 r 12 r12 r12 (c) Gravitational potential: Gravitational potential at a point is the potential energy that a unit mass would possess at the point in the gravitational field. Potential difference (dv ) between points A and B dV= − Edr . dV Gravitational intensity E = − . dr Gm Also gravitational field is the negative potential gradient at a point is V = − . r Gm m Gravitational potential energy of a body at a point, PE = − 1 2 . r Expression for gravitational potential and field at a point due to a solid sphere: (a) Point P outside the sphere: Consider a solid sphere of mass M , radius R and density ρ. Consider it to be made of large number of thin spherical shells whose radii vary from O to R . Let x be radius of one such shell and dx its thickness. Mass of the shell =(4 πx2 dx ) ρ (mVwhereV=ρ =× Area thickness= 4 π× x2 dx ) The gravitational potential at point P due to shell is 2 GmG(4π x dx ) ρ G 4πρ dV=− =− =− xdx2 a a a The potential due to the entire sphere at P is GRAVITATION R G4πρR G 4 πρ x 2 V=−∫ xdx2 =− a0 a 2 ο R G4πρ x3 G 4 πρ V=− =− R 3 a3 ο 3 a G 4 3 GM 4 3 V=− π R ρ or V = − where M= π R ρ a 3 a 3 Gravitational field: dV dV d GM GM =− =− =− − =− E 2 dr da da a a GM E = − a2 (b) Point P on the surface of the sphere: GM GM 2 2 V= −[]Q a = R and E= − a = R R R2 (c) Point P inside the sphere: Let P be a point inside the sphere of radius R on a imaginary spherical surface of radius b . The gravitational potential at P due to all shells of radii between R and b (utside point P ) s R π2 ρ G(4 xdx ) 2 V1 = − ∫ mass of a shell =4 πx dx ρ ] b x R R 2 πρ x G4 2 2 VGxdxG1 =−πρ4∫ =−πρ 4 =− Rb − --------- (1) b 2 b 2 The potential at P due to shells within the sphere of radius b is G 4 3 4 3 V2 =− πρ b ----------- (2) [ πb ρ= m = mass of sphere of radius b ] b 3 3 The potential at P due to entire sphere is πρ G4 2 2 4 2 VVV=+=−1 2 Rb − − Gb πρ 2 3 Rbb222− Rbb 222 V=− G4 πρ + =− G 4 πρ − + 23 223 2 2 3R− b G4πρ 2 2 V=− G 4 πρ or V= −3 R − b ------------- (3) 6 6 G 4 =− πρ3 2 − 2 2 or V3 RRb ()3 [by multiplying and dividing equation (3) by R ] 2R 3 GM 4 = −2 − 2 Q = π3 ρ V3 3 R b M R 2R 3 VIJAYA COLLEGE Page 2 GRAVITATION Gravitational field: dV d GM =− =− −2 − 2 E3 ()3 R b db db2 R GMd GM GMb E=()3 Rb2 −=− 2 () 2 b or E = − 2Rdb3 2 R 3 R3 Kepler’s Laws of Planetary Motion: I law – Law of orbits : Every planet moves in an elliptical orbit with sun being at one of its foci. II Law – Law of areas: The radius vector drawn from the sun to the planet sweeps equal areas in equal intervals of time. i.e., the area velocity is constant. III Law – Law of periods: The square of the time period of revolution of a planet around the sun is proportional to the cube of the semi – major axis of the elliptical orbit. Derivation of Kepler’s Second Law: From Newton’s Law of gravitation ur GMm $ F= − r ------------- (1) r 2 M → mass of sun m → mass of planet r → distance between them r ur r dv The force experienced by the planet is F= ma = m -------------- (2) dt r r r dv GMm $ $ $ r From (1) and (2) m= − r as r= rr⇒ r = . dt r 2 r r dv GM r ∴ = − r dt r 3 r Taking left cross product on both sides of this equation with r . r rdv GM r r drr GM rr r× =−3 () r × r or ()rv× =−3 () rr × dt r dt r r r d r r r r As r× r = 0 ∴()r × v = 0 or r× v = constant dt r r r dr r dr Also r × = constant Q v = dt dt Multiplying and dividing LHS by 2, we get r 1 r dr 2 r × = constant --------- (1) 2 dt VIJAYA COLLEGE Page 3 GRAVITATION In a interval of time ∆t , the planet covers an area ∆A , then from the diagram, area SPP ′ is ur1 r r ∆A =() r ×∆ r . 2 ur r r r ∆A1 ∆ r r1 r dr The area velocity of the planet h===lim lim r × or h= r × ---------- (2) or ∆→t0∆t ∆→ t 0 2 dt 2 dt r r r 1 h= r × v 2 1 Comparing (1) and (2) we get h = × constant or h = constant. 2 Thus area velocity of the planet is a constant. Hence IInd law. Derivation of Kepler’s First Law: From second law, areal velocity = constant r1 rr rrr or h=()() rv × ⇒ 2 hrv= × 2 r r r r $ $d $ dr d $ As r= r r ∴=+2h rr( rr ⋅ ) Q v= =() r ⋅ r dt dt dt r $ $dr dr $ 2hrrr= × + r dt dt r $ 2 $dr dr $ $ 2hrr=×+ r() rr × dt dt r $ 2 $ 2 $ dr Q $ $ r$ dr 2h= r r × r× r = 0 or h= r × ---------- (1) dt 2 dt r Taking right cross product on both sides of (2) with h r r r r dv GM$ dv GM $ =−r −−−−− (2) ⇒ × hrh =− × −−−−− (3) dt r2 dt r 2 r Substituting for h from (1) in RHS of equation (3) r 2 $ $ dvr GM r dr GM dr ×=−$$ × × =− $$ ×× h2 rr rr --------- (4) dt r2 dt 2 dt ur ur ur ur ur ur ur ur From vector identity A××=⋅( BC) ( ACB) −⋅( ABC) r r r r $ $ dr dr $ $ $ dr ∴××rr =⋅ r rrr −⋅() dt dt dt r r r $ dr dr $ $ As r is perpendicular to (velocity) ∴⋅r =0, r . r = 1 dt dt r $ r r dr dr ∴×r r × =− dt dt VIJAYA COLLEGE Page 4 GRAVITATION r $ dvr GM dr Thus equation (4) becomes ×=−h − dt2 dt r $ $ dvr GM dr dr r GM dr or ×h = or ()v× h = ------------ (5) dt2 dt dt2 dt Integrating equation (5) r r r GM $ vh× = rc + 2 r Taking dot – product on both sides of the above equation with r , we get, rrr r rr GM $ rvh⋅×=() rrrc ⋅+⋅ 2 GM $$ $ =rr ⋅+ r rc cos θ 2 r r r GM rvh⋅×=() rrc +cos θ ----------- (6) 2 ur ur ur ur ur ur ur ur ur using vector identity ABC⋅×( ) =⋅× BCA( ) =⋅ CAB( × ) rrr rrr rvh⋅( ×) =⋅ hrv( × ) r r r =⋅hh()2Q 2 hrv =× = 2h2 Thus equation (6) can be written as GM GM 1GM c cos θ 2 = + θ= + θ = + 2h rrc cos r c cos or 2 2 2 2 r4 h 2 h 4h2 Multiplying throughout by GM 4h2 2 C 4h2 2 c =1 + cos θ let l=, e = GMr M GM GM l Then =1 +e cos θ ----------- (7) r This is the equation of a conic section with l → semi latus rectum and e → eccentricity. If θ < 1 , then the above equation is the equation of ellipse with sun at one of its foci. This proves Kepler’s first law. Derivation of Kepler’s third law: The path of the planet with sun at one of the foci is as shown a → semi major axis, b → semi minor axis, l → semi latus rectum of the ellipse. For an ellipse b2= a 2(1 − e 2 ) …….(1) and l= a(1 − e 2 ) ---------- (2) l 1 b2 Dividing (2) / (1) = or l = b2 a a VIJAYA COLLEGE Page 5 GRAVITATION 4h2 4 h 2 b 2 4h2 a As l= ⇒ l = = or b2 = ----------- (3) GM GM a GM πab The period of revolution of the planet round the sun is T is given by T = . h area of ellipse h = period of revolution [Q areal velocity ] πab = T π2a 2 b 2 or T 2 = ------------- (4) h2 substituting for b2 from (3) in (4) π22a4 ha 2 4 π 23 a T 2 = = h2 GM GM T 24π 2 or = = constant or T2∝ a 3 a3 GM Hence the Kepler’s third law. Inertial mass and gravitational mass: ur r If a force F acts on aur body r of mass m , and the acceleration of the body is a , then from Newton’s second law F= mi a , where mi is called the inertial mass. This body is also acted upon by gravitational force. Then the mass of the body is called gravitational mass mg . The gravitational force on the body by the earth is ur GMe m g F = 2 --------- (1) R e From Newton’s second law F= mi g ---------- (2) GMe m g mi GM e Comparing (1) and (2) mi g = 2 or = 2 ----------- (3) Re mg gR e At a place, RHS of above equation is a constant.
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