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
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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
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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
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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
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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. m ∴ratio i =K = constant mg If K = 1 , then G=6.67 × 10 −11 Nm 2 kg − 2 . For any other value of K , G is different. Thus
mi= m g . Thus inertial mass and gravitational mass are equivalent. Satellite motion: Objects that move around a planet is called a satellite. Moon is a natural satellite of earth. Man made satellite that revolve around the planets are called artificial satellites.
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Elements of satellite motion: 1. Orbital velocity: The velocity with which a satellite revolves round a planet is called orbital velocity. Consider a satellite of mass m , moving around the earth in an orbit of
radius r . Let M be the mass of the earth and vο be the orbital velocity of the satellite. The necessary centripetal force acting on the satellite is provided by the gravitational force of attraction on the satellite. 2 mv ο GMm i.e., Fc= F g Fc= F g = r r 2
mv 2 GMm GM GM ο = ⇒ v2 = or v = ------(1) r rο r ο r If g → acceleration due to gravity on the satellite near the surface of earth, then GMm m g = . R2 or gR2 = GM ------(2)
gR 2 Substituting for GM from (2) in (1), v = ο r If the satellite is close to earth, then r= R (radius of earth)
vο = gR
2. Time period of the satellite orbit: GM The orbital velocity of the satellite is v = ------(1) ο r
If ω is the angular velocity of the satellite, then vο = r ω ------(2) GM Comparing (1) and (2) rω = r GM 2π or ω = by definition ω = where T is the time period of the satellite. r 3 T 2π GM r 3 Thus = or T =2 π T r 3 GM If h is the height of the satellite from the surface of earth and R → radius of earth, then
()R+ h 3 r= R + h Thus T =2 π GM
()R+ h 3 As GM= gR 2 T =2 π gR 2 If the satellite is very near surface of earth h<< R
R ∴T =2 π g
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Escape velocity: It is defined as the minimum velocity with which an object has to be projected from the surface of the planet so that it escapes the planet’s gravitational force of attraction. If m is mass of an object projected from surface of earth, and M → mass of earth, R → radius of earth and v is the velocity of projection, then total energy of the object at the surface of the earth (just after projection) is 1 GMm TE=+() KE PE = mv 2 − ------(1) r= R 2 R 1 −GMm where KE= mv 2 and PE = 2 R
If ve is the velocity of the object just enough to take the object to infinity (v= v e ) , as both kinetic and potential energies are zero at infinity, then
TE=( KE + PE )r=∞ = 0 ------(2) From the law of conservation of energy (1) = (2) 1 GMm 1 GMm mv 2 − = 0 or mv 2 = 2 e R 2 e R 2GM 2GM 2GM or v2 = or v = In general v = e R e R e r
The condition for launching artificial satellites: GM (1) When v= v = ⇒ Satellite goes in a circular orbit. ο r GM (2) When v > but less than escape velocity r 2GM v = , then satellite goes in a elliptical path. e r 2GM (3) When v= v = , the satellite takes a parabolic path and escapes to infinity. e r 2GM (4) When v > , the satellite takes a hyperbolic path and escapes to infinity. r
Geostationary satellite: A satellite that appears to be fixed at a position above a certain distance from earth having same period of rotation of earth (24 hr) is called geo stationary satellite. The satellite rotates in the plane of the equator with the same rotation as that of earth and thus appears to be stationary to an observer on the earth. 2π The angular velocity of the satellite is ω = . T where T → time period of geo stationary satellite. 2π As T = 24 hrs = 86,400 sec, ω= =7.3 × 10−5 rad s − 1 . 86,400
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GM The orbital velocity v = . ο r GM GM GM As v= r ω = ⇒ ω = or ω2 = ο r r 3 r 3
1 GM GM 3 or r 3 = or r = ω2 ω2 G=×6.6710−11 Nm 2 kg − 2 , M =× 5.9810 24 kg and ω =7.3 × 10−5 rad s − 1
1 3 6.67× 10−11 × 5.98 × 10 24 r = −5 2 ()7.3× 10 r=42 × 10 6 m The height of the geo stationary satellite above the surface of the earth is given by h= r − R [Q r= R + h ] R → radius of earth R=6.37 × 10 6 m
6 6 h =42 × 10 − 6.37 × 10 h=35.63 × 10 6 m or h= 35,630 km
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