Tiêu đề 07 Gravitation Tutorial Solutions.pdf ✅

St Andrew’s Junior College H2 Physics 7-3 13(a) Change in gravitational potential energy = m (f - I) = 50 [–60 – (–20)] = – 2000 MJ Since the change in gravitational potential energy is negative, it is a loss by the satellite. [1] [1] (b) Since a mass will move from a position of less negative (-20) or higher gravitational potential energy to one of more negative (-80) or lower gravitational potential energy when released, the satellite moves closer to the Earth. [1] [1] 17 Ans: D Since F = -dEp/dx, and R is already the rate of change of potential energy with distance, then R = -dEp/dx. Hence, F = R. Solutions to Mastery Questions 1(a)(i) Since g = − dr d , The field strength at A = 0 Nkg-1 [1] [1] (ii) The net force at A is zero. (Let RE be the centre-to-centre distance between the Earth and point A, and RM centre-to-centre distance between the Moon and point A.) 2 E E R GM = 2 M M R GM M E M M = 2 2 M E R R Since the mass of the Earth is larger than that of the moon, equilibrium is only possible when point A is closer to the moon. [1] [1] (b)  = [-3.9 × 106 – (- 62.3 × 106 )] = 58.4 × 106 J kg-1 U = 3000 × (58.4 × 106 ) = 1.752 × 1011 J [1] [1] (c) It is a gain in energy. [1] (d) Minimum KE needed is just to send B to A. After A the attractive force of the moon will draw the craft to the surface of the Earth.  = [-1.3 × 106 – (- 62.3 × 106 )] = 61.0 × 106 J kg-1 U = 3000 × (61.0 × 106 ) = 1.83 × 1011 J [1] [1] 2(a)(i) GM1M2/(R1 + R2) 2 [1] (a)(ii) M1R1ω2 or M2R2ω2 [1] (b) ω = 2π/(1.26 x 108 ) or 2π/T = 4.99 x 10-8 rad s-1 [1] [1] (c)(i) Reference to either taking moments (about C) OR same (centripetal) force M1R1 = M2R2 OR M1R1ω2 = M2R2ω2 hence M1/M2 = R2/R1 [1] [1]