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CHAPTER GRAVITATIONAL AND CENTRAL FORCES 6.1 m = ρ V = ρ π rs3 3m rs = 4πρ F= Gmm ( 2rs ) 2 Gm 4πρ G 4πρ 43 = = m 3m 4 F F Gm 4πρ 3 = = m W mg 4g F 6.672 × 10−11 N ⋅ m ⋅ kg −2 4π × 11.35 g ⋅ cm −3 kg 106 cm3 = × × × kg )3 ( W × 9.8 m ⋅ s −2 103 g m3 F = 2.23 × 10−9 W 6.2 (a) The derivation of the force is identical to that in Section 6.2 except here r < R This means that in the last integral equation, (6.2.7), the limits on u are R – r to R + r GmM R + r r − R F= Q 1 + ds s2 Rr ∫R − r R s GmM R2 − r R2 − r = + − − + − R r R r ( ) θ ψ Rr R+r R−r P r GmM 2r + R − r − ( R + r ) = F= Rr (b) Again the derivation of the gravitational potential energy is identical to that in Example 6.7.1, except that the limits of integration on s are ( R − r ) → ( R + r ) 2πρ R R + r ds rR ∫R − r 2πρ R = −G R + r − ( R − r ) rR 4π R ρ M = −G φ = −G R R For r < R , φ is independent of r It is constant inside the spherical shell φ = −G 6.3 F =− GMm eˆr r2 The gravitational force on the particle is due only to the mass of the earth that is inside the particle’s instantaneous displacement from the center of the earth, r The net effect of the mass of the earth outside r is zero (See Problem 6.2) M = π r3ρ F = − Gπρ mreˆr = − kreˆr r F The force is a linear restoring force and induces simple harmonic motion T= 2π ω = 2π m = 2π k 4Gπρ The period depends on the earth’s density but is independent of its size At the surface of the earth, GMm Gm = ⋅ π Re ρ Re2 Re 4Gπρ g = Re mg = T = 2π 6.4 Re 6.38 × 106 m hr = 2π × ≈ 1.4 hr g 9.8 m ⋅ s −2 3600 s Fg = − GMm eˆr , where M = π r ρ r The component of the gravitational force perpendicular to the tube is balanced by the normal force arising from the side of the tube The component of force along the tube is Fx = Fg cos θ The net force on the particle is … F = −iˆ Gπρ mr cos θ r cos θ = x ˆx F = −iˆ Gπρ mx = −ik As in problem 6.3, the motion is simple harmonic with a period of 1.4 hours 6.5 6.6 GMm mv GM v2 = = so r r r for a circular orbit r, v is constant 2π r T= v 4π r 4π T2 = r ∝ r3 = v GM 2π r v From Example 6.5.3, the speed of a satellite in circular orbit is … (a) T = gR v= e r T= 2π r g Re T gRe2 r = 4π 1 r T g 242 hr × 36002 s ⋅ hr −2 × 9.8 m ⋅ s −2 = = Re 4π Re 4π 6.38 × 106 m r = 6.62 ≈ Re (b) T= 2π r 2 g Re = 2π ( 60 Re ) 2 603 Re g = 2π g Re 2 603 × 6.38 × 106 m = 2π −2 −2 −2 2 2 9.8 m ⋅ s × 3600 s ⋅ hr × 24 hr ⋅ day T = 27.27 day ≈ 27 day 6.7 From Example 6.5.3, the speed of a satellite in a circular orbit just above the earth’s surface is … v = gRe T= 2π Re Re = 2π v g This is the same expression as derived in Problem 6.3 for a particle dropped into a hole drilled through the earth T ≈ 1.4 hours 6.8 The Earth’s orbit about the Sun is counter-clockwise as seen from, say, the north star It’s coordinates on approach at the latus rectum are ( x, y ) = ( ε a, −α ) The easiest way to solve this problem is to note that ε = orbit is almost circular! GM S m mv GM S ∴ = and v = r r r with r = α ≈ a ≈ b when ε ≈ GM S m v≈ = ⋅10 s α More exactly ml r × v = α v cos β = l , but α = k l α= , Since k = GM S m GM S is small The 60 (equation 6.5.19) hence l = α v cos β = (α GM S ) GM S Or v= α cos β The angle β can be calculated as follows: y x2 + =1 b2 a dy b2 x ∴ =− dx a y (see appendix C) and at ( x, y ) = ( ε a, −α ) so dy b ε a b ε b2 = = = ε since − ε = dx a α a (1 − ε ) a2 here dy = tan β = ε dx and GM S GM S ≈ v= as before α cos ε α 6.9 F ( r ) = Fs + Fd or β ≈ ε (small ε) GMm r2 GM d m Fd = − r2 The net effect of the dust outside the planet’s radius is zero (Problem 6.2) The mass of the dust inside the planet’s radius is: Fs = − 4 M d = π r3ρ GMm F ( r ) = − − πρ mGr r 6.10 1 − kθ = e r r du k = − e − kθ dθ r u= d 2u k − kθ = e = k 2u dθ r From equation 6.5.10 … d 2u + u = k 2u + u = − 2 f ( u −1 ) du ml u −1 2 f ( u ) = − ml ( k + 1) u f (r ) = − ml ( k + 1) r3 The force varies as the inverse cube of r From equation 6.5.4, r 2θ = l dθ l = e −2 kθ dt r l e kθ dθ = dt r kθ lt e = +C r 2k 2klt ln + C ′ 2k r θ varies logarithmically with t θ= k = ku 3 r From equation 6.5.10 … d 2u ku + u = − 2 ⋅ ku = − 2 dθ ml u ml d u k + 1 + u = dθ ml 6.11 f (r ) = k If 1 + < , ml k If 1 + = , ml du = C1 dθ u = c1 θ + c2 r= c1 θ + c2 k If 1 + > , ml u = A cos ( cθ + δ d 2u − cu = , dθ d 2u =0 dθ ) d 2u + cu = , dθ k r = A cos + θ + δ ml 6.12 c > , for which u = aebθ is a solution c>0 −1 1 = r r cos θ du sin θ = dθ r cos θ u= d 2u 2sin θ − cos θ = + = 1 + dθ r cosθ cos3 θ r cos θ cos θ d 2u = u ( 2r 2u − 1) = 2r 2u − u dθ Substituting into equation 6.5.10 … 2r 2u − u + u = − 2 f ( u −1 ) ml u −1 2 f ( u ) = −2r ml u − 1 = r cosθ cos θ 2r ml f (r ) = − r5 6.13 From Chapter 1, the transverse component of the acceleration is … aθ = rθ + 2rθ If this term is nonzero, then there must be a transverse force given by … f (θ ) = m(rθ + 2rθ ) For r = aθ , and θ = bt f (θ ) = 2mab ≠ Since f (θ ) ≠ , the force is not a central field For r = aθ , and the force to be central, try θ = bt n f (θ ) = m 2ab n 2t n − + ab n ( n − 1) t n − For a central field … f (θ ) = 2n + ( n − 1) = n= θ = bt (a) 6.14 Calculating the potential energy a2 dv − = f ( r ) = −k + dr r r a2 Thus, V = −k + 4r r The total energy is … 1 9k 9k E = T + V = v2 − k + = − = a a 2a 4a Its angular momentum is … 9k l = a 2v = = constant = r 4θ 2 Its KE is … dr 2 l 1 dr T = r + r 2θ = + r θ = +r 2 dθ dθ r The energy equation of the orbit is … a2 dr l + − T + V = = r k 2+ 4 dθ 4r r r ( ) dr 2 9k a2 = + − r k 2+ 4 4r r dθ 4r or dr 2 = (a − r ) dθ r = a cos φ Letting So dφ = dθ then dr dφ = −a sin φ dθ dθ ∴φ = θ r = a cos θ ( r = a @θ = ) 3π (b) at θ = r → the origin of the force To find how long it takes … av v l θ= 2= = 1 r a cos θ a cos θ 3 a dt = cos θ dθ v Thus T= 3π ∫ π a 3a 3π a cos θ dθ = cos φ dφ = ∫ v v 4v 9k Since v = 2a 1 π a2 T = πa = 4 k 9k (c) Since the particle falls into the center of the force v→∞ (since l = vr⊥ = const ) 6.15 From Example 6.5.4 … v 2r1 = vc r1 + r v we have V = Letting V = vc 1+ r r1 −2 dV r r −2 1 + − So: = dr1 2V r1 r12 dV r r 1 V = = Thus dr1 r r1 r r1 1 + 1 + r1 r1 r1 r 1+ r1 dV V ≈ r (b) dr1 = 2r1 dV = 60 1% = 120%! (a) ( ) dr1 r1 r1 r V r1 The approximation of a differential has broken down – a correct result can be obtained by calculating finite differences, but the implication is clear – a 1% error in boost causes rocket to miss the moon by a huge factor - ∼ 2! From section 6.5, ε = 0.967 and r = 55 ×106 mi 1+ ε From equations 6.5.21a&b, r1 = r 1− ε r 55 × 10 mi AU a = ( r + r1 ) = = × 1− ε − 0.967 93 × 10 mi a = 17.92 AU 6.16 From equation 6.6.5, τ = ca τ = 1yr ⋅ AU τ = 75.9 yr − 3 ×17.92 AU From equation 6.5.21a and 6.5.19 … α ml ε = − 1= −1 r0 kr0 ε= mr v − and k k = GMm GM 2 v = ( ε + 1) r From Example 6.5.3 we can translate the factor GM into the more convenient GM = ae ve2 … with ae the radius of a circular orbit and ve the orbital speed … a v 93 ×106 mi 2 v = 1.967 (ε + 1) = ( ) ve 55 × 10 mi r v0 = 1.824 ve Since l is constant … r1v1 = r v r 1− ε − 967 × 1.824 ve = 0.0306 ve v1 = v = v = r1 1+ ε 1.967 e e ve ≈ 2π ae τ = 2π × 93 ×106 mi = 66, 705 mph yr × 365 day ⋅ yr −1 × 24 hr ⋅ day −1 v = 1.22 × 105 mph and v1 = 2.04 × 103 mph 6.17 From Example 6.10.1 … 2 v r 2 ε = 1 + q − ( qd sin φ ) where q = and d = d ve ae are dimensionless ratios of the comet’s speed and distance from the Sun in terms of the Earth’s orbital speed and radius, respectively (q and d are the same as the factors V and R in Example 6.10.1) φ is the angle between the comet’s orbital velocity and direction vector towards the Sun (see Figure 6.10.1) The orbit is hyperbolic, parabolic, or elliptic as ε is > , = , or < … 2 i.e., as q − is > , = , or < R 2 q − is > , = , or < as q d is > , = , or < R Since l is constant, vmax occurs at r and vmin occurs at r1 , i.e vmax = v and vmin = v1 and form the constancy of l … v1r1 = v r r vmin vmax = v1v = v r1 k r v = ( ε + 1) (See Example 6.5.4) m k 2π a From equation 6.6.5 … = GM = a m τ 6.18 2π a a ( ε + 1) vmin vmax = ⋅ r1 τ 1+ ε With 2a = r + r1 : 1− ε a ( ε + 1) ( r + r1 )( ε + 1) r − ε = = + 1 ( ε + 1) = 2 r1 + ε r1 r1 From equation 6.5.21a&b … r1 = r vmin vmax 2π a = τ + 1 (ε + 1) = 6.19 As a result of the impulse, the speed of the planet instantaneously changes; its orbital radius does not, so there is no change in its potential energy V The instantaneous change in its total orbital energy E is due to the change in its kinetic energy, T, only, so δv δv 1 δ E = δ T = δ mv = mvδ v = mv = 2T v v 2 δE δv =2 T v 10 But the total orbital energy is k k E=− So δE = δa 2a 2a Since planetary orbits are nearly circular k k V ~− and T~ a 2a δa δE δa and = Thus, δ E ≅ T a T a δa δv =2 We obtain a v 6.20 (a) V = τ Vdt τ∫ k r From equation 6.5.4, l = r 2θ dθ l r dθ = or dt = dt r l τ 2π kr ∫0 Vdt = −∫0 l dθ From equation 6.5.18a … a (1 − ε ) r= + ε cos θ ka (1 − ε ) 2π τ dθ ∫0 Vdt = − l ∫0 + ε cosθ 2π a 1− ε From equation 6.6.4 … τ = l V (r ) = − k − ε 2π dθ ∫ + ε cos θ 2π a 2π dθ 2π k ∫0 + ε cosθ = − ε , ε < ∴V = − a V =− (b) This problem is an example of the virial theorem which, for a bounded, periodic τ system, relates the time average of the quantity ∫∑ p⋅r i derive it for planetary motion as follows: τ τ τ 1 p ⋅ rdt = mr ⋅ rdt = ∫ ∫ ∫ F ⋅ rdt τ τ τ 11 i to its kinetic energy T We will Integrate LHS by parts τ τ 1 mr ⋅ r τ0 − ∫ mr dt = ∫ F ⋅ rdt τ τ τ 0 The first term is zero – since the quantity has the same value at and τ denote time average of the quantity within brackets Thus T = − F ⋅ r where but − r ⋅ F = r ⋅∇V = r dV k = =− V dr r hence T = − V but hence and Thus: V V + V = 2 k E=− = constant V =2 E but 2a τ k k E = ∫ Edt = E = − so 2E = − a τ0 2a k k V = − as before and therefore T = − V = a 2a E = T + V =− The energy of the initial orbit is k k mv − = E = − r 2a k 2 1 v2 = − (1) m r a Since = a (1 + ε ) at apogee, the speed v1 , at apogee is 6.21 v12 = k k (1 − ε ) − = m a (1 + ε ) a ma (1 + ε ) To place satellite in circular orbit, we need to boost its speed to vc such that k k mvc − = − since the radius of the orbit is ra 2ra k k = vc2 = mra ma (1 + ε ) Thus, the boost in speed ∆v1 = vc − v1 (2) 2 k ∆v1 = 1 − (1 − ε ) ma (1 + ε ) Now we solve for the semi-major axis a and the eccentricity ε of the first orbit From (1) above, at launch v = v at r = RE , so 12 k 1 − m RE a and solving for a RE a= noting that … RE − mv k k GM E = = gRE (3) mRE RE RE R a= = E = 4.49 ⋅103 km v 1.426 2− gRE v2 = The eccentricity ε can be found from the angular momentum per unit mass, l, equation 6.5.19, and the data on ellipses defined in figure 6.5.1 … ka (1 − ε ) kα l = r θ = v ( RE sin θ ) = = m m where v , θ are the launch velocity, angle Solving for ε (using (3) above) v2 v2 ε = 1− 2− sin θ = 0.795 gRE gRE ∴ ε = 0.892 Inserting these values for a, ε into (2) and using (3) gives (a) (b) 6.22 R a ∆v1 = gRE E 1 − (1 − ε ) = 4.61 ⋅103 km ⋅ s −1 + ε h = a (1 + ε ) − RE = 2.09 ⋅10 km {altitude above the Earth … at perigee} −be − br e −br e − br − = f ′ ( r ) = −k k r3 r2 r f ′(a) 2 = −b + f (a) a f ′(a) From equation 6.14.3, ψ = π 3 + a f (a) ψ= − 2 b + r = π 3 − ( ab + ) π − ab 13 − 6.23 From Problem 6.9, f ( r ) = − GMm − πρ mGr r2 2GMm − πρ mG r3 4πρ a −3 GMma mG πρ − − + f ′(a) 3M = = f ( a ) −GMma −2 − πρ mGa 4πρ a a 1 + 3M f ′(r ) = f ′(a) From equation 6.14.3, ψ = π 3 + a f (a) 4πρ a −2 + 3M ψ = π 3 + 4πρ a 1+ 3M − 1+ c 2 ψ =π , + 4c 6.24 c= − 4πρ a + 3M =π 4πρ a + 3M − 4πρ a 3M We differentiate equation 6.11.1b to obtain mr = − For a circular orbit at r = a , r = so dU r =a = dr For small displacements x from r = a , r=x r = x+a and From Appendix D … x2 f ( x + a ) = f ( a ) + xf ′ ( a ) + f ′′ ( a ) + … dU d 2U , f ′(r ) = Taking f ( r ) to be dr dr Near r = a … dU dU d 2U = + x r =a r =a + … dr dr dr d 2U mx = − x r = a dr dU ( r ) dr This represents a “restoring force,” i.e., stable motion, so long as 14 d 2U > at r = a dr 6.25 f ′(r ) = 2k 4ε + r3 r5 From equation 6.13.7, the condition for stability is f ( a ) + k ε a 2k 4ε − + +
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