HCI 07B Gravitation Solutions (Discussion Questions)

Gravitation Tutorial 7B Discussion Question Suggested Solutions D1 (a) Total gravitational potential energy of the configuration, U = Work done by the external force in assembling the system                 13 2312 10 08 0 08 0 08 0 10 103 08 0 25 0 1 0 J Gm m Gm mGm mU .. . G. . . . (b) d cos30˚ = 0.80/2  d = 0.4619         10 -1103 4 332 10 J kg04 6 1 9 centre G. ..    10 10 10 43 3 2 1 0 0 43 3 1 0 J centre WD U m .. .             Gravitational force is attractive in nature and the gravitational potential is set to be zero at infinity. To move the 4th mass from infinity to the centre of the system, the force exerted on the mass by the external agent will be in opposite direction to the displacement of the mass. Thus negative work is done by the external force (agent). D2 2020 P1 Q12 A Increase in gravitational potential energy =  12 12 12 22 Gm m Gm m Gm m rrr     d  d  d  30˚

D3 (a) It represents the gravitational force on the body on Earth’s surface. p g dEF dr . The negative sign indicates that the gravitational force is in the direction of decreasing potential energy. (b) B. This amount of total energy allows the mass to just reach r = R. At r = R, TE = GPE and KE = 0. (c) C. The mass will go beyond r = R, comes to rest at r where the PE = TE and then turns back towards Earth – and that’s why it is “falling towards the Earth”. (d) D and E. For D, the mass will just reach infinity. For E, the mass will reach infinity where GPE = 0 and still have some KE.   D4 2021 P3 Q2 (a) Gravitational force is attractive. In displacing a mass from infinity towards M (at constant speed), an external force opposite in direction to the displacement needs to be applied. Work done by this external force is negative [OR positive work needs to be done by an external force to move a mass away from M, implying that the potential gets larger as one moves away from M.] Infinity is assigned a potential-value of zero. Hence the potential at any other point is negative. (b) (i)    11 23 7- 1 6 66 7 1 0 62 1 0 1 2 10 J kg 68 1 0 2 ..GM .r .          (ii) To travel to infinity, the total energy must be greater th an or equal to zero At surface, GPE + KE = (-1.22 × 10 7)(2.8) + ½ (2.8)(3800)2 = -1.4 × 107 J Hence, it does not escape, it returns to the planet. O R Initial KE of rock = ½ (2.8)(3800) 2 = 2.0 × 107 J To reach infinity, GPE to be gained =     772 8 0 1 22 10 3 4 10 Jfim. . .     Not enough KE to escape.

OR (calculate the escape speed) Loss in KE = Gain in GPE K E i – KEf = GPEf – GPEi    271 28 0 0 28 12 2 1 02 .v . .    3- 14 9 10 m sv. 3.8 × 10 3 m s-1 is not enough to escape. D5 Total energy of the space station in orbit = KE + GPE = ½ GPE of the space station at the orbital