# 7. Gravitation-Exercise

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SECTION (A) A 1. F = G m1m2 = 6.67 × 10–11 × 1.4  7.34 1022 6 2 = 4.8 × 10–5 N r 2 8 4 A 3. tan = 6 = 3  = 53° GmM (378  10 ) F = r 2 G 0.260  0.01 (0.1)2 2Fcos  a = m 0.260  3  = 2G (0.1)2  5  = 31.2 G m/s2   SECTION (B) v  B 1. Ex = – x v Ey = – x = – x  = – y (20x + 40y) = – 20 (20x + 40y) = – 40 → = E ˆi + E ˆj = – 20 ˆi – 40 ˆj Ans. E x y It is independent of co ordinates Force = → = m → = 0.25 {– 20 ˆi – 40 ˆj } = – 5 ˆi – 10 ˆj F E → magnitude of F = SECTION (C) = 5 N C 1. Potential energy at ground surface potential energy = –GMm R potential energy at a height of R is R potential energy = –GMm 2R When a body comes to ground Loss in potential energy = Gain in kinetic energy –GMm  – GMm  1 GMm 1  2R –   = mv2   2 2R = 2 mv2  GM   gR = v2  R2  g   v = SECTION (D) D 1. sin  = R R  h  R  or  = colatitude = sin–1  R  h  D 3. (a) F = GMm (2R)2 =   GM2 4R2 Mv 2 (b) R v = GM2 = 4R2 2R T = v = 2R = 4 (c) Angular speed 2 2  = T = = (d) Energy required to separate = – { total energy } = – { Kinetic energy + Potential energy }  1 2 1 2 GM2   2 GM2   GM GM2  = –  Mv   Mv 2 –  2R  = – Mv  –  2R  = – M 4R –  2R   GM2  GM2 = – –   = 4R  4R Ans. (e) Let its velocity = ‘v’ 1 Kinetic energy = mv2 2 Potential at centre of mass = – GM – R GM = – R 2GM R Potential energy at centre of mass = – 2GMm R For particle to reach infinity Kinetic energy + Potential energy = 0 1 mv2 × 2 2GMm R = 0 v = Ans. D 5. T2  r3 2 For the Saturn , r is more , So T will be more as  = T so  will be less, for the Saturn . GM and orbital speed Vo = r as r is more So VO will be less for the Saturn. SECTION (E)  2h  E 1. Field outside shell g1 = g 1– R    1– 2h  h 1 So 9 = g    R  1– h   R 20 Field inside shell E = g    R   1  E = 10 1– 20  E = 10 – 0.5 = 9.5 m/sec2   Solving we get x  9.5 m/s2 PART - II : SECTION (A) : A-1. Net force is towards centre Fnet = F2 + = F2 + F1 Gm2 F2  (2r)2  Gm2 4r 2     F Gm2 = + = mv2 F1  Gm2 Gm2   net 4r 2 2r 2 r  v = 2r 2  Ans. A-3. In horizontal direction Net force = G 3 mm 12 d2 cos30° – Gm2 4d2 cos60° Gm2 = 8d2 Gm2 – 8d2 = 0 in vertical direction G 3 m2 G 3 m2 Gm2 Net force = 12 d2 cos 60° + 3 d2 + 4 d2 cos30° 3 Gm2 3Gm2 3Gm2 1 8  3  3Gm2 = 24 d2 + 3 d2 + 8 d2 = d2  24  = 2d2 along SQ SECTION (B) B-1. dv = – Edr = Integrating both sides k dr r vv = k𝑙nrr r r v – vi = k 𝑙n di  v = v + k 𝑙n di Ans. B-3. 1m 2m 4m 8m v = v1 + v2 + v3 + v4 + ...... = – Gm – 1 Gm Gm – 4 2

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### 7. Gravitation-Exercise

• 1. SOLUTION OF GRAVITATION EXERCISE-1 : PART - I SECTION (A) A 1. F = G 2 2 1 r m m = 6.67 × 10–11 1 × 2 6 22 ) 10 378 ( 10 34 . 7 4 . 1    = 4.8 × 10–5 N A 3. tan = 6 8 = 3 4  = 53° F = 2 r GmM = 2 ) 1 . 0 ( 01 . 0 260 . 0 G  a = m cos F 2  = 2G 2 ) 1 . 0 ( 260 . 0       5 3 = 31.2 G m/s2 SECTION (B) B 1. Ex = – x v   = – x   (20x + 40y) = – 20 Ey = – x v   = – y   (20x + 40y) = – 40 E  = Ex î + Ey ĵ = – 20 î – 40 ĵ Ans. It is independent of co ordinates Force = F  = mE  = 0.25 {– 20 î – 40 ĵ } = – 5 î – 10 ĵ magnitude of F  = 2 2 5 10  = 5 5 N SECTION (C) C 1. Potential energy at ground surface potential energy = R GMm – potential energy at a height of R is R potential energy = R 2 GMm – When a body comes to ground Loss in potential energy = Gain in kinetic energy  R 2 GMm – –       R GMm – = 2 1 mv2  R 2 GMm = 2 1 mv2  gR = v2        g R GM 2   v = gR
• 2. SECTION (D) D 1. sin  = h R R  or  = colatitude = sin–1        h R R D 3. (a) F = 2 ) R 2 ( GMm = 2 2 R 4 GM (b) R Mv 2 = 2 2 R 4 GM v = R 4 GM T = v R 2 = R 4 GM R 2 = 4 GM R3 (c) Angular speed  = T 2 =           GM R 4 2 3 = 3 R 4 GM (d) Energy required to separate = – { total energy } = – { Kinetic energy + Potential energy } = –        R 2 GM – Mv 2 1 Mv 2 1 2 2 2 = –       R 2 GM – Mv 2 2 = –       R 2 GM – R 4 GM M 2 = –       R 4 GM – 2 = R 4 GM2 Ans. (e) Let its velocity = ‘v’ Kinetic energy = 2 1 mv2 Potential at centre of mass = – R GM – R GM = – R GM 2 Potential energy at centre of mass = – R GMm 2 For particle to reach infinity Kinetic energy + Potential energy = 0 2 1 mv2 × R GMm 2 = 0 v = R GM 4 Ans.
• 3. D 5. T2  r3 For the Saturn , r is more , So T will be more as  = T 2 so  will be less, for the Saturn . and orbital speed Vo = r GM as r is more So VO will be less for the Saturn. SECTION (E) E 1. Field outside shell g1 = g       R h 2 – 1 So] 9 = g       R h 2 – 1  20 1 R h  Field inside shell E = g       R h – 1 E = 10       20 1 – 1 E = 10 – 0.5 = 9.5 m/sec2 Solving we get x  9.5 m/s2 PART - II : SECTION (A) : A-1. Net force is towards centre Fnet = F2 + 2 F 2 1 = F2 + 2 F1            2 2 2 2 1 2 2 2 2 2 r 2 Gm ) r 2 ( Gm F r 4 Gm ) r 2 ( Gm F  Fnet = 2 2 r 4 Gm + 2 2 r 2 Gm 2 = r mv2  v = ) 2 2 1 ( r 4 Gm  Ans. A-3. In horizontal direction Net force = 2 d 12 mm 3 G cos30° – 2 2 d 4 m G cos60° = 2 2 d 8 m G – 2 2 d 8 m G = 0 in vertical direction Net force = 2 2 d 12 m 3 G cos60° + 2 2 d 3 m 3 G + 2 2 d 4 m G cos30° = 2 2 d 24 m G 3 + 2 2 d 3 m G 3 + 2 2 d 8 m G 3 = 2 2 d m G 3         24 3 8 1 = 2 2 d 2 m G 3 along SQ
• 4. SECTION (B) B-1. dv = – Edr = r k dr Integrating both sides  v vi v =  r i d r n k  v – vi = k n i d r  v = vi + k n i d r Ans. B-3. 1m 2m 4m 8m v = v1 + v2 + v3 + v4 + ...... = – 1 Gm – 2 Gm – 4 Gm – 8 Gm – 16 Gm – 32 Gm – ........ = – Gm           ....... 8 1 4 1 2 1 1  G.P. of infinite series = – m       2 / 1 – 1 1 v = – Gm (2) = – 2Gm Ans. B-5. If we take complete spherical shell than gravitational field intensity at P will be zero hence for the hemi spherical shell shown the intensity at P will be along c. SECTION (C) C 1. Initial total energy = Initial kinetic energy + initial potential energy = 2 1 m (0)2 +         0 R GMm – = 0 R GMm – Total energy, when it reaches the surface of earth = 2 1 mv2 +       R GMm – Applying energy conservation, 2 1 mv2 – R GMm = 0 R GMm – v =       0 R 1 – R 1 GM 2 Ans.
• 5. C 2. (a) 120° F m F F m m 120° 120° Due to geometry net force is zero. (b) By geometry, x2 + 4 a2 = a2 and F1 = F2 F F2 F1 x a x2 = 4 a 3 2 x = 2 a 3 Fnet = F = 2 2 x Gm = 3 4 2 2 a Gm (c) Initial potential energy = –             a Gm a Gm a Gm 2 2 2 = – 2 3 a Gm2 Work done on system = Final potential energy – intial potential energy = – 2 3 a Gm2 –           a Gm 3 – 2 = 2 3 a Gm2 Ans. (d) Initial kinetic energy = 0 Initial potential energy = – a Gm2 – a Gm2 = – a Gm 2 2 Total initial energy = – a Gm 2 2 Now, kinetic energy = 2 1 mv2 Potential energy = – 2 / a Gm 2 2 – 2 / a Gm2 = – a Gm 4 2 Total energy = 2 1 mv2 – a Gm 4 2  a Gm 2 2 = 2 1 mv2  a Gm 4 = v v = a Gm 2 Ans.
• 6. SECTION : (D) D-1. 2 r GMm = r mv2 v = 2 r GM M r V m T = v r 2 = GM r 2 2 3  = 3 2 3 r 3 4 G r 2     T   1 Ans. D-3*. Escape velocity = R GM 2 = Ve Orbital velocity = R GM = V0 Escape velocity = 2 × orbital velocity  (A) 2 1 mVe 2 = 2 × 2 1 mV0 2  (C) D-5*. PE = –G m1 m2 /r, ME = – G m1 m2 / 2r On decreasing the radius of orbit PE and ME decreases SECTION (E) E 1. we = 50 × 10 = 500 N  wp = 50 × 5 = 250 N Hence option A is correct PART - III 1. (A) Till the particle reaches the centre of planet, force on both bodies are in direction of their respective velocities, hence kinetic energies of both keep on increasing . After the particle crosses the centre of planet, forces on both are retarding in nature. Hence as the particle passes through the centre of the planet, sum of kinetic energiesof both the bodiesismaximum. Therefore statement-1 isTrue, Statement-2 isTrue;Statement- 2 is a correct explanation for Statement-1. EXERCISE-2 PART - I 1. E  = î 3 – ĵ 4 N / kg Then F  = mE  = (1 kg) ( î 3 – ĵ 4 ) N / kg = 3 î – 4 ĵ N Work done = W = F  . S  S  = Displacement vector displacement is along y = 4 3 x + 4 9 Any point on this line can be written as k1 î +        4 9 k 4 3 1 ĵ = k2 î +        4 9 k 4 3 2 ĵ S  = Displacement vector = (k2 – k1 ) î + 4 3 (k2 – k1 ) ĵ Now, W = (3 î – 4 ĵ ) .        ĵ ) k – k ( 4 3 î ) k – k ( 1 2 1 2 = 3 (k2 – k1 ) – 3 (k2 – k1 ) = 0 Ans.
• 7. 3. (a) Force will be due to the mass of the sphere upto the radius r In case (i) 0 < r < b ; Mass M = 0, therefore F(r) = 0 In case (ii) b < r < a ; Mass M =  3 4 (r3 – b3 ), therefore F(r) =           2 3 r b r Gpm 3 4 (iii) a < r <  ; Mass M =  3 4 (a3 – b3 ), therefore F(r) =           2 3 3 r b a Gpm 3 4 (b) Uf – Ui = –  2 1 r r c dr . F  (i) 0 < r < b ; u(r) = - 2 Gm(a2 — b2 ) (ii) b < r < a ; u(r) = r 3 m G 2    (3ra2 - 2b3 - r3 ) (iii) a < r <  ; u (r) = ) b a ( r 3 m G 4 3 3     5. (a) The gravitation field isuniform inside the cavity and isdirected along ´ OO . Hence the particle will strike atA. (b) The gravitational field at any point P inside cavity. |E  | = 3 4 G – 3 4 G = 3 4 Gy ´ OO = 3 2 GR Total workdone = m |E  | . S  = m . 3 2 GR . 2 R Applying work - energy theorem Workdone by all force = Change in kinetic energy m . 3 2 GR . 2 R = 2 1 mv2 v = 3 R G 2 2   Ans.
• 8. 7. Applying angular momentum conservation : mv0  = mvd v0 = vd .......... (i) Intial energy = 2 1 mv0 2 + 0 Final energy = 2 1 mv2 – d GMs Applying energy conservation , 2 1 mv0 2 = 2 1 mv2 – d m GMs v0 2 = v2 – d GM 2 s .......... (ii) From equation (i) and (ii) : v0 2 = 2 2 2 0 d v  – d GM 2 s d2 + 2 0 s v GM 2 d – 2 = 0 Solving this quadratic d = – 2 0 s v GM + 2 2 2 0 s v GM           = 2 0 s v GM                    1 – GM v 1 2 2 0  Ans. PART - II 1. Gravitational field at ‘m’ due to hollowed - out lead sphere = { Field due to solid spere } – { Field due to mass that was removed } Field due to solid sphere = 2 d GM = E1 = 2 R 4 GM Field due to removed mass = 2 x ' GM = E2 M’ = 3 R 3 4 M  × 3 2 R 3 4        = 8 M And x = d – 2 R So, E2 = 2 2 R – d 8 GM       = 2 2 R 3 8 GM       = 2 R 18 GM Enet = E1 – E2 = 2 R GM       18 1 – 4 1 = 2 R 36 GM 7 Fnet = mEnet = 2 R 36 GMm 7 Ans.
• 9. 3. F =  cos ) R 2 ( GMm 2 = 2 3 ) R 2 ( GMm 2  = 2 R 8 GMm 3 5. Ve = R GM 2 V = kVe = R GM 2 K Initial total energy = 2 1 mv2 – GMm R = 2 1 m.k2 R GM 2 – GMm R Final total energy = 2 1 m02 – GMm x Applying energy conservation 2 1 mk2 . R GM 2 – GMm R = 0 – GMm x x 1 = R 1 – 2 k R  x = 2 k – 1 R Ans. 7. inside a uniform spherical shell Ein = 0 Vin = constant = R Gm 9. Fg = 3 R GMmr pressing force = Fg cos  = 3 R cos GMmr  = 2 R 2 GMm = constant a = m sin Fg  = 3 R sin GMr  a = 3 R GMy 11.* S = 5 . 1 2 , E = 24 2 west to east = 2       24 1 – 5 . 1 1 Twest to east = east to west 2   = 1.6 hours Similarly east to west = 2        24 1 5 . 1 1  Teast to west = 17 24 hours
• 10. PART - III 1. P.E. = – r GMm  K.E. = 2 1 mV2 Total energy = r GMm  + 2 1 mV2 T.E. = 0 if 0 mV 2 1 r GMm 2     v = r GM 2 For v < r GM 2 , T.E. is – ve for v > r GM 2 , T.E. is + ve If V is r GM i.e. equal to orbital velocity, path is circular. . If T.E. is negative, path is elliptical. If T.E. is zero, path is parabolic. If T.E. is positive, path is hyperbolic. PART - IV 4. Let M and R be the mass and radius of the earth respectively. If m be the mass of satellite, then escape velocity from earth e = ) Rg 2 ( Velocity of satellite s = 2 e  = 2 / ) g R 2 ( ......... (1) Further ] s =       r GM =          h R g R2  2 s  = h R g R2  h = R = 6400 km 5. T2 = 3 2 x Gm 4 Hence time period of revolution T is T = Gm x 2 3  (Put x = 2R )  T = g R 8 2 6. Now total energy at height h = total energy at earth's surface (from principle of conservation of energy)  0 – G M h R m  = 2 1 m2 – GM R m or 2 1 m2 = R m GM – R 2 GMm ( h = R)  v = gR
• 11. 7 to 9. Let theangular speed of revolution of both starsbeaboutthecommoncentre , that is, centreof massof system. The centripetal force on star of mass m is m2 3 d 2 = 2 d ) m 2 ( Gm . Solving we get T= 3 2 d Gm 3 4 The ratio of angular momentum is simply the ratio of moment of inertia about center of mass of system. 2 3 d m 2 3 d 2 m L L 2 2 M m M m                  I I Similarly, The ratio of kinetic energy is simply the ratio of moment of inertia about center of mass of system. 2 3 d m 2 3 d 2 m K K 2 2 2 M 2 m M m                  I 2 1 I 2 1 EXERCISE- 3 PART - I 1. (A) Gravitational field isa conservative force field. In a conservative force field work done ispath independent.  WI = WII = WIII 2. speed of particle at A VA = escape velocity on the surface of moon = R GM 2 At highest point B, VB = 0 From energy conservation. 2 A mV 2 1 = VB – VA = m       m U – m U A B or        m U – m U 2 V A B 2 A , also 3 A R 2 GM – m U  [3R2 – r2 ]                          2 2 3 100 R – R 5 . 0 – R 5 . 1 R GM – – h R GM – R GM or 2 1 –1 3 1 99 1 – R R h 2R 2 100 R          h = R ) 99 ( ) 100 ( ) 99 ( ) 100 ( 2 2 2 2    h = R 199 19801 or h = 99.5 R  99R Ans 3. A B mA mB com rA rB C  2 B A B A ) r r ( m Gm  = mA rA 2 A 2 T 4 = mB rB 2 B 2 T 4  2 B B B 2 A A A T r m T r m  As C is com  mA rA = mB rB hence TA =TB
• 12. 4. (A) It is similar equation as v = 2 2 x a  ω in SHM. (B) Particle on positive x-axis move towards origin with speed decreasing as x decreasing. (C) It is spring mass system performing SHM. (D) Object moves away from Earth so its speed will decrease, since its speed is greater than escape velocity so it will never return back. 5.  = 0 r < R = 0 r > R Case Ir < R FC = r mV2 mg r mV R r 2  (g = acceleration due to gravity at surface of sphere) V = R g r for r < R Case II r > R 2 r GMm = r mV2 V = R r g r GM  So Ans. (C) 6. If only gravitational force acts on astronaut (that is in state of free fall), he shall feel weightless. Thus statement-2 is correct explanation of statement-1. 7. Wext = U – UP Wext = 0 –        1 . x Gdm – – Wext = G 2 2 2 r R 16 rdr 2 R 7 M      = 2 R 7 GM 2   2 2 r R 16 rdr = 2 R 7 GM 2  z zdz = 2 R 7 GM 2 [Z] r 3R 4R dr P x Wext = 2 R 7 GM 2 R 4 R 3 2 2 r R 16        Wext = 2 R 7 GM 2   R 5 – R 2 4 Wext = 2 R 7 GM 2   5 – 2 4 . 8. A B c.m. 5d 6 d 6 11 Ms 2.2 Ms c.m. about B of momentum Angular c.m. about momentum angular Total =                                         6 d 6 d ) M 11 ( 6 d 6 d ) M 11 ( 6 d 5 6 d 5 ) M 2 . 2 ( s s s = 6.
• 13. 9. g = 2 R GM = 2 3 R R 3 4 ) G (         ; g R g ' g =                 R ' R ' =       3 2       R ' R = 11 6 Given R ' R = 22 6 3 Ve = R GM = R R 3 4 )) ( ) G ( 3                  Ve  R  ; Ve = 3 km/hr. 10. Ve = 0 v 2 KE = 2 e mv 2 1 =  2 0 v 2 m 2 1 = mv0 2 11. Ves = R GM 2 = R R 3 4 . G . 2 3   = 3 G 4  R Ves  R Sarface area of P = A = 4RP 2 Surface area of Q = 4A = 4 RQ 2  RQ = 2Rp mass R is MR = MP + MQ 3 R R 3 4   = 3 P R 3 4   + 3 Q R 3 4    RR 3 = RP 3 + RQ 3 = 9RP 3 RR = 91/3 RP  RR > RQ > RP Therefore VR > VQ > VP P R V V = 91/3 and Q P V V = 2 1 12. 0 0 mv 2 1 L m 2 . GM 2      L GM 4 v  M M m L L PART - II 1. 2 1 T T = 2 / 3 2 1 r r         or 5 T = (4)3/2 or T = 40 hr. 2. Escape velocity is independent of direction of projection. 3. X R mv2  = 2 2 2 R R ) X R ( GM   or V = 2 / 1 2 X R R g          4. T  R3/2 Time period is independent of mass of satellite 5. PE = R GMm – R 2 GMm = 2 mgR 6. g       R d – 1 = g       R h 2 – 1 or d = 2h 7. W = r GMm = 6.67 × 10–11 × 100 × 10 / 0.1 = 6.67 × 10–7 J
• 14. 8. Electric charge on the moon = electric charge on the earth 9. V = R GM 2 = 10 R 10 GM 2 e e  = 10 e e R GM 2 = 110 k m/s 10. Acceleration due to gravity at leight h from earth surface. g' = 2 R h 1 g         2 R h 1 g 9 g          h = 2R 11. 2 x Gm = 2 ) x r ( ) m 4 ( G  x 1 = x r 2  r – x = 2x 3x = 3 r x = 3 r 3 / r 2 ) m 4 ( G 3 / r Gm   r Gm 6 r Gm 3   = r Gm 9 Ans. 12. 2 2 ) R 2 ( Gm = m2 R 3 2 R 4 Gm = 2  = 3 R 4 Gm v = R v = 3 R 4 Gm × R = R 4 Gm 13. W = 0 – R GMm R GMm         = gR2 × R m = mgR = 1000 × 10 × 6400 × 103 = 64 × 109 J = 6.4 × 1010 11. R 6 GMm 1 2 1 R 3 GMm R 3 GMm R 3 GM m 2 1 R 3 GMm mv 2 1 E 2 0 f               R GMm Ei   + K Ei = Ef R 6 GMm 5 K 
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