Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 12 | Note to obtain the numerical results given in this chapter the following numerical values of certain physical quantities have been used G x 10 N m2 kg2 g m s2 and mE x 1024 kg. Use of other tabulated values for these quantities may result in an answer that differs in the third significant figure. The ratio will be the product of the ratio of the mass of the sun to the mass of the earth and the square of the ratio of the earth-moon radius to the sun-moon radius. Using the earth-sun radius as an average for the sun-moon radius the ratio of the forces is 2 z a . X x 1030 kg _ I---------. 1 x 1024 kgJ Use of Eq. gives F G 10-11 N m7kg 5 9 x kgX2150 104 R g r2 x 105m x 106 m 2 The ratio of this force to the satellite s weight at the surface of the earth is x 104 N 2150 kg m s2 79 . This numerical result requires keeping one extra significant figure in the intermediate calculation. The ratio which is independent of the satellite mass can be obtained directly as Gmmlr2 Gm R Y E E E _ 2 _ 1 1 mg r g r J yielding the same result. G nm1 nm2 G m1m2 nr12 r12 F 1 12. The separation of the centers of the spheres is 2R so the magnitude of the gravitational attraction is GM2 2R 2 GM2 4R2. a Denoting the earth-sun separation as R and the distance from the earth as x the distance for which the forces balance is obtained from GMm Mm S _ E R-x 2 x2 which is solved for R x MS 1 - W E x 108 m. b The ship could not be at equilibrium for long in that the point where the forces balance is moving in a circle and to move in that circle requires some force. The spaceship could continue toward the sun with a good navigator on board. a Taking force components to be positive to the right use of Eq. twice gives F 6 673 x 10-11 N- m2 kg2 0 100 kg - Fg x 10 Nm kg _ 2 2 x 10-11 N with the minus sign indicating a net force to the left. b No
