Tiêu đề 08. Gravitation Medium Ans.pdf ✅

1. (c) Kepler’s second law linear momentum of the planet is related by the relations 2. (c) Kepler’s third law is also known as harmonic law. 3. (d) According to Kepler’s first law, every planet moves in an elliptical orbit with the sun situated at one of the foci of the ellipse. In options (a) and (b) sun is not at a forus while in (c) the planet is not in orbit around the sun. Only (d) represent the possible orbit for a planet. 4. (b) According to Kepler’s third law 2 3 T  R 3/ 2 2 1 2 1 R R T T          = Or 2 / 3 1 2 / 3 1 2 2 1 2 16 (R ) T T R (R )       =         = = 4R1 = 4R (given R1 = R ) ....(i) Orbital velocity, R GM v0 = 2 1 4R R R R v v 1 1 2 1 01 0 2  = = = (using (i)) Or 0 2 01 0 v 2 1 v 2 1 v = = 5. (d) Asteroids move in circular orbits like planets under the actions of central force. 6. (b) According to Kepler’s seconds law, equal areas are swept in equal intervals of time. As area SCD = 2 area SAB, hence 1 2 t = 2t 7. (a) Here, A E P RE r = 6R ,r = 2 The eccentricity of the orbit is A p A P r r r r e + − = 2 1 8 4 6R 2R 6R 2R e E E E E = = + − = 8. (b) According to law of conservations of angular momentum. Angular momentum at perigee = Angular momentum at apogee p p A A mv r = mv r Where m is the mass of the satellite 3 1 6R 2R r r v v E E A P P A  = = = 9. (d) Here, T 29.5T ,R 1.5 10 km,R ? s 8 s = E E =  = According to Kepler’s third law, 3 2 T  R 3 E 3 S 2 E 2 S R R T T  = 2 / 3 E s S E T T R R         = 1.4 10 km T 29.5T R 1.5 10 9 2 / 3 E 8 E s =          =  10. (d) As areal velocity of a planet around the sun is constant. Therefore, the desired time is timeperiod area of ellipse area ABS t AB         = If a = semi-major axis and b = semi-minor axis of ellipse, then, area of ellipse = ab Area ABS 4 1 = (area of ellipse) – Area of triangle ASO (ea) (b) 2 1 ab 4 1 =   −  ( )         = −        −   = 2 e 4 1 T T ab eab 2 1 4 ab t AB 11. (d) As a(1 e) a(1 e) v v a p − + = Where a is semi-major axis of the ellipse p a p p a a v (1− e) = v (1+ e) or v − ev = v + ev a p p a a p p a v v v v e(v v ) v v ore + − + = − = 12. (a) From Kepler’s seconds law of planetary motion, the linear speed of a planet is maximum when its distance from the sun is least. 13. (a) Gravitational force is a conservative force. Hence, work done in taking a mass rom one point to another in a gravitational field depends on the end points only.
14. (a) Time period of a revolution of a planet. s 3/ 2 s Gm 2 r r GM 2 r v 2 r T  =  =  = Where Ms is the mass of the sun Squaring both sides ,we get s 2 2 2 GM 4 r T  = The graph between 2 T and 3 r is a straight line whose slope is S 2 GM 4 . 15. (d) The period of moon’s rotation around the earth does not depend upon the mass of the moon. 16. (a) Draw a perpendicular AD to the side BC. l 2 3 AD = ABsin 60 = Distance AO of the centroid O from A is AD. 3 2 3 1 l 2 3 3 2 AO =          = By Symmetry, AO = BO = CO 3 1 = Force on mass 2m at O due to mass m at A is alongOA l 6Gm (l / 3) Gm(2m) F 2 2 OA 2 = = Force on mass 2m at O due to mass m at B is alongOB l 6Gm (l / 3) Gm(2m) F 2 2 OB 2 = = Force on mass 2m at O due to mass m at C is alongOC l 6Gm (l / 3) Gm(2m) F 2 2 OC 2 = = Draw a line PQ parallel to BC passing through O. Then BOP = 30 = COQ Resolving FOB and FOC into two components. Resolving acting along OP and OQ are equal in magnitude and opposite in directions. So, they will cancel out while the components acting along OD will add up.  The resultant force on te mass 2m at O is F F (F sin30 F sin30 ) R OA OR OC = − +  0 2 1 l 6Gm 2 1 l 6Gm l 6Gm 2 2 2 2 2 2 =        = −  +  17. (c) The gravitations force is independent of the intervening medium. In other words, the force between two mass remains the same whether they are in air, vacuum, water or separated by a brick wall. Hence, (a) is an incorrect statement. The gravitational force is a conservative force. Hence, (b) is an incorrect statement. The gravitational force obeys newton’s third law of motion. Thus, it form action – reactions pair. Hence, (c) is a correct statement. The gravitational force is a central force. Hence, (d) is an incorrect statement. 18. (a) The gravitational forces are mutually equal and opposite. 19. (a) From figure, 3l 2 3 AC = AM + MC = 2AM = 2l cos30 = 2l = Similarly, AE = 3l, AD = AO + ON + ND = l sin 30 + l + l sin 30 2l 2 1 l l 2 1 = l + +  = AB = AF = l Force on mass m at A due to mass m at B is alongAB l Gmm (AB) Gmm FAB 2 2 = = Force on mass m at A due to mass m at C is along AC 3l Gm ( 3l) Gmm (AC) Gmm F 2 2 AC 2 2 = = = Force on mass m at A due to mass m at D is alongAD 4l Gm (2l) Gmm (AD) Gmm F 2 2 AD 2 2 = = = Force on mass m at A due to mass m at E is along AE 3l Gm ( 3l) Gmm (AE) Gmm F 2 2 AE 2 2 = = = Force on mass m at A due to mass m at F is along AF l Gm (AF) Gmm F 2 2 AF 2 = =
Resultant force due to FAB and FAF is F = F + F + 2F F cos120 AB AF 2 AF 2 R1 AB       −                 +         +         = 2 1 l Gm l Gm 2 l Gm l Gm 2 2 2 2 2 2 2 2 2 2 along AD l Gm 2 2 = Resultant force due to FAC and FAF is FR = F + F + 2F F cos60 AC AE 2 AE 2 2 AC                       +         +         = 2 1 3l Gm 3l Gm 2 3l Gm 3l Gm 2 2 2 2 2 2 2 2 2 2 along AD 3l Gm 3l 3Gm 2 2 2 2 = = Net force on mass m along AD is 2 2 2 2 2 2 R R R AD 4l Gm 3l Gm l Gm F F F F 1 2 = + + = + +       = +       = + + 3 1 4 5 l Gm 4 1 3 1 1 l Gm 2 2 2 2 20. (c) The situations is as shown in the figure. Gravitational force on an object of mass m at point p distance d from centre O lying on the axis of the circular ring of radius R and mass M is given by 2 2 3/ 2 (R d ) GMmd F + = When d = r, then 2 2 3/ 2 2 3/ 2 3/ 2 3 2 2 2R GMm 2 R GMmR (2R ) GMmR (R R ) GMmR F = = = + = 21. (b) It will remains the same as the gravitational force is independent of the medium separating the masses. 22. (c) if point mass is placed inside a uniform spherical shell, the gravitational force on the point mass is zero. Hence, the gravitational force exerted by the shell on the point mass is zero. 23. (a) The situation is as shown in the figure. According to Newton’s law of gravitations, gravitational force between two bodies of masses m1 and m2 is 2 1 2 r Gm m F = Where r is the distance between the two masses. Hence, 1 2 r = r + r 2 1 2 1 2 (r r ) Gm m F +  = 24. (b) G has different value in different system of units. In SI system the value of G is 11 2 2 6.67 10 Nm Kg − −  whereas in CGS its values is 8 6.67 10−  dyne 2 2 cm g − . The value of G is same throughout the universe. The value of G was first experimentally determined by English scientist Henry Cavendish. G is a scalar quantity. 25. (c) G gR or M R GM g 2 E 2 E E E = = Cavendish determined the value of G. 26. (b) Accelerations due to gravity on earth is 2 E E R GM g = ..... (i) As 3 E E 3 E E R 3 4 M R 3 4 M  =     = Substituting this value in Eq. (i), we get E 2 E E 3 E 4 GR 3g GR or 3 4 R R 3 4 G g  =   =         = 27. (d) The tidal wave in sea is due to gravitational pull of moon on the earth. 28. (b) as 2 E E 2 E g G R g R GM g  =  = 29. (c) Accelerations due to gravity at a altitude h above the earth’s surface is 2 E 2 E h (R h) gR g + = .....(i)
Where g is the acceleration due to gravity on the earths’ surface and RE is the radius of the earth. Eq. (i) Shows that accelerations due to gravity decreases with increasing altitude. Acceleration due to gravity at a depth d below the earth’s surface is       = − d E R d g g 1 ...... (ii) Eq. (ii) shows that acceleration due to gravity decreases with increasing depth. Accelerations due to gravity at latitude   = −   2 2 g g Rg cos ..... (iii) Where  is the angular speed of rotation of the earth. Eq. (iii) shows that acceleration due to gravity increase with increasing latitude. Accelerations due to gravity of body of mass m is placed on the earth’s surface is 2 E E R GM g = ..... (iv) Eq. (iv) shows that accelerations due to gravity is independent of the mass of the body but it depends upon the mass of the earth. 30. (a) Value of g is maximum at poles. 31. (d) The acceleration due to gravity at a depth d below the surface of earth is R r g R R d g R d g` g 1  =      −  =      = − ..... (i) Where R – d = r = distance of locations form the centre of the earth. When r = 0, g` = 0 From (i) g  r till R = r, for which g` = g For r > R, 2 2 2 2 2 r 1 org` r gR (R h) gR g` =  + = Here, R + h = r Therefore, the variation of g with distance r from centre of earth will be as shown in figure (4).Thus, options (d) is correct. 32. (c) Gravitational potential energy at any point at a distance r form the centre of the earth is r GM m U E = − Where ME and m be masses of earth and body respectively. At the surface of the earth, RE r = E E 1 R GM m U = − At a height h from the surface, E E E RE r = R + h = R + nR = (1+ n)  E E 2 (n 1)R GM m U + = − Changes in potential energy is         − − +  = − = − E E E E 2 1 R GM m (n 1)R GM m U U U E E E E (n 1)R GM mn (n 1) 1 1 R GM m + =        + = −         = + = 2 E E E R Gm g (n 1) n mgR  33. (c) Gravitations potential on the surface of the shell is V = Gravitational potential due to particle (V1 ) + Gravitational potential due to shell itself ( V2 ) R 4Gm R G3m R Gm  = −      = − + − 34. (d) Let P is the midpoint of the line joining the centres of the spheres. The gravitational potential at point P is r 4GM r 2GM r 2GM r / 2 GM r / 2 Gm VP = − − = − − = − 35. (b) The escape velocity is independent of mass of the body and the direction of projection. It depends upon the gravitational potential at the point from where the body is launched. Since this potential depends slightly on the latitude and height of the point, the escape velocity depends slightly on these factors. 36. (d) Moon has no atmosphere because the rms speed of gas molecules is greater than the escape speed (i.e., v v ). rms  e 37. (a) The escape velocity from the surface of the earth is         = = = 2 E E E E E e R GM 2gR g R 2GM v  38. (d) Kinetic energy = 2r GMEm A –s Potential energy r GMEm = − B – r