Tiêu đề 8.GRAVITATION - Explanations.pdf ✅

2 (c) g = GM R2 = GM0 (D0/2) 2 = 4GM0 D0 2 3 (a) Time period, T = 2πR √ GMm R = 2πR 3/2 √GMm Where the symbols have their meaning as given in the question Squaring both sides, we get T 2 = 4π 2R 3 GMm 4 (c) (i) Tst = 2π√ (R+h) 3 GM = 2π√ R g [As h << R and GM = gR 2 ] (ii) Tma = 2π√ R g (iii) Tsp = 2π√ 1 g( 1 l + 1 R ) = 2π√ R 2g [As l = R] (iv) Tis = 2π√ R g [As l = ∞] 5 (b) The relation between mass and density of earth is given form Newton’s law of gravitation, according to which Me = gRe 2 G where Me is mass of earth, G the gravitational constant, Re the radius of earth and g the acceleration due to gravity. Also, mass = volume × density g = G × volume × density R2 Assuming spherical shape of earth volume = 4 3 πR3 g = G × 4 3 πR3 R2 ρ ⟹ g = G ∙ 4 3 πRρ Hence, increases in radius would dominate. 6 (b) Weight of body on the surface of earth mg = 12.6 N At height h, the value of g′ is given by g ′ = g R 2 (R + h) 2 Now, h = R 2 ∴ g ′ = g ( R R + (R/2) ) 2 = g 4 9 Weight at height h = mg 4 9 = 12.6 × 4 9 = 5.6 N 7 (c) Angular momentum of the earth around the sun is L = MEv0r ⇒ L = ME√ GMs r r (∵ v0 = √ GMs r ) ⇒ L = [ME 2GMsr] 1/2 Where, ME = Mass of the earth Ms = Mass of the sun r = Distance between the sun and the earth ∴ L ∝ √r 8 (c) g = GM r 2 ∴ log g = log G + log M − 2 log r Differentiating both sides w.r.t. t 1 g = dg dt = 0 − 2 × 1 r dr dt ( dr dt × 100 = −1) ⇒ 1 g ( dg dt × 100) = −2 × 1 r ( dr dt × 100) ⇒ dg dt × 100 = −2 × (−1) = 2 ∴ g increasing by 2% 9 (c) Applying law of conservation of energy for asteroid at a distance 10 Re and at earth’s surface. Ki + Ui = Kf + Uf ... . (i) Now, Kf = 1 2 mvi 2 and Ui = − GMem 10Re Kf = 1 2 mvf 2 and Uf = − GMem Re Substituting these values in Eq. (i), we get 1 2 mvi 2 − GMem 10Re = 1 2 mvf 2 − GMem Re ⟹ 1 2 mvf 2 = 1 2 mvf 2 + GMem Re − GMem 10Re ⟹ vf 2 = vi 2 + 2GMe Re − 2GMe 10Re ∴ vf 2 = vi 2 + GMem Re (1 − 1 10) 12 (b) g ∝ ρR
13 (c) Landsats 1 through 3 operated in a near polar orbit at an altitude of 920 km with an 18 day repeat coverage cycle. These satellites circled the earth every 103 min completing 14 orbits a day. 14 (c) When gravitational force becomes zero, then centripetal force on satellite becomes zero and therefore, the satellite will become stationary in its orbit. 15 (c) The earth moves around the sun is elliptical path, so by using the properties of ellipse r1 = (1 + e)a and r2 = (1 − e)a ⇒ a = r1+r2 2 and r1r2 = (1 − e 2)a 2 Where a = semi major axis b = semi minor axis e = eccentricity Now required distance = semi latusrectum = b 2 a = a 2 (1 − e 2 ) a = (r1r2) (r1 + r2)/2 = 2r1r2 r1 + r2 16 (c) The value of acceleration due to gravity changes with height (ie, altitude). If g′ is the acceleration due to gravity at a point, at height h above the surface of earth, then g ′ = GM (R + h) 2 but, g = GM R2 ∴ g ′ g = GM (R + h) 2 × R 2 GM = R 2 (R + h) 2 Here, g ′ = GM (R + h) 2 = GM (R + 3R) 2 = GM (4R) 2 = GM 16R2 = ge 16 17 (c) gR 2 (R + h) 2 = g (1 − h R ) or (1 − h R ) (1 + h 2 R2 + 2h R ) = 1 or h 3 R3 + h 2 R2 − h R = 0 or h R ( h 2 R2 + h R − 1) = 0 or h R = −1±√1+4 2 = √5−1 2 or h = √5R−R 2 18 (a) Time period of satellite T = 2π√ (R + h) 3 GMe where R + h = orbital radius of satellite, Me = mass of earth. Thus, time period does not depend on mass of satellite. 19 (d) F ∝ 1 r 2 . If r becomes double then F reduces to F 4 20 (c) Force acting on a body of mass M at a point at depth d. Inside the earth is F = mg ′ = mg (1 − d R ) = mGM R2 ( R − d R ) = GM m R3 r (∵ R − d = r) So, F ∝ r; Given F ∝ r n n = 1 21 (b) Weight on surface of earth, mg = 500N and weight below the surface of earth at d = R 2 mg ′ = mg (1 − d R ) = mg (1 − 1 2 ) = mg 2 = 250N 22 (a) GMM L 2 = MV 2 L ⇒ V = √ GM L 23 (b) Using g = GM R2 we get gm = g/5 24 (a) Acceleration due to gravity at a height above the earth surface g ′′ = g ( R R + h ) 2 g g′′ = ( R + h R ) 2 g g′′ = ( R + nR R ) 2 g g′′ = (1 + n) 2
25 (b) The period of revolution of geostationary satellite is the same as that of the earth. Orbital velocity vo = √gRe Escape velocity ve = √2gRe where Re is radius of earth. % increase = ve − vo vo × 100 % increase = √2gRe − √gRe √gRe × 100 = (√2 − 1) × 100 = (1.141 − 1) × 100 = 41.4% 26 (d) Weight on mars = mg ′ = mG(m/10) (R/2)2 = m × 4 10 mg = 4 10 × 200 = 80 N 27 (a) Let R be the radius of earth and ρ its density, then since shape of earth is assumed spherical we have Mass of earth = volume × density M = 4 3 πR 3 × ρ ... . (i) The acceleration due to gravity which arises in the body due to gravitational force of attraction is given by g = GM R2 ... . (ii) Putting the value of M from Eq.(i), we get g = G 4 3 πR 3ρ R2 = G 4 3 πRρ ... . (iii) Given, ρp = ρ, Rp = 0.2Re ∴ gp = G 4 3 πRpρp = G × 4 3 π × 0.2Rρ = 0.2g 28 (d) Acceleration due to gravity at a height h form the surface of the earth g ′ = g 1 (1 + h R ) 2 Given, h = 2R ∴ g ′ = g 1 (1 + 2) 2 or g ′ = g 9 29 (b) VA =( Potential at A due to A) + (Potential at A due to B) ⇒ VA = − Gm1 R − Gm2 √2R Similarly, VB =(Potential at B due to A) + (Potential at B due to B) ⇒ VB = − Gm2 R − Gm1 √2R Since, WA→B = m(VB − VA) ⇒ WA→B = Gm(m1 − m2 )(√2 − 1) √2R 30 (c) The velocity of the spoon will be equal to the orbital velocity when dropped out of the space- ship 31 (d) F = mR ω2 = 6 × 1024 × (1.5 × 1011)(2 × 10−7) 2 = 36 × 1021N 32 (c) g = GM r 2 . Since M and r are constant, so g = 9.8m/s 2 33 (a) At a height h, (Taking h < < R) from the surface of earth gh = g (1 − 2h R ) or gh g = 1 − 2h R = 90 100 or 2h R = 1 − 99 100 = 1 100 or g = R 100 = 6400 200 = 32 km 34 (a) v = √2gR ⇒ vp ve = √ gp ge × Re Rp = √2 × 1 4 = 1 √2 ∴ vp = ve √2 36 (a) At an altitude h the acceleration due to gravity is g ′ = g (1 − 2h Re ) or mg ′ = mg (1 − 2h Re ) ie, w ′ = w (1 − 2h Re ) 99 100 w = w (1 − 2h Re ) ie, h = 0.005Re At point below the surface of earth at depth h. The weight of body given by w ′ = w (1 − 2h Re ) w ′ w = 0.995 %∆w = (1 − 0.995)w w × 100 %∆w = 0.5%(decreases) 37 (c) The variation of g with angular velocity (ω) is given by