7. We choose the coordinate system shown in the figure and find the force on the mass in the lower left corner. Because the masses are equal, for the magnitudes of the forces from the other corners we have F_{1}= F_{3}= Gmm/r_{1}^{2}= (6.67x10^{-11}N · m^{2}/kg^{2})(8.5 kg)(8.5 kg)/(0.70 m)^{2}= 9.83x10^{-9}N; F_{2}= Gmm/r_{2}^{2}= (6.67x10^{-11}N · m^{2}/kg^{2})(8.5 kg)(8.5 kg)/[(0.70 m)/cos q45°]^{2}= 4.92x10^{-9}N. From the symmetry of the forces we see that the resultant will be along the diagonal. The resultant force is F = 2F_{1}cos q=45° + F_{2}= 2(9.83x10^{-9}N) cos q=45° + 4.92x10^{-9}N = 1.9x10^{-8}N toward center of the square.

9. For the magnitudes of the forces on the mass m from the other masses we have F_{2}= Gm(2m)/x_{0}^{2}= 2Gm^{2}/x_{0}^{2}; F_{3}= Gm(3m)/(x_{0}^{2}+ y_{0}^{2}) = 3Gm^{2}/(x_{0}^{2}+ y_{0}^{2}); F_{4}= Gm(4m)/y_{0}^{2}= 4Gm^{2}/y_{0}^{2}. The force F_{3}is at an angleq above the x-axis, with sin q = y_{0}/(x_{0}^{2}+ y_{0}^{2})^{1/2}, and cos q = x_{0}/(x_{0}^{2}+ y_{0}^{2})^{1/2}. Thus the resultant force is F = (F_{2}+ F_{3}cos q)i + (F_{3}sin q + F_{4})j = Gm^{2}{(2/x_{0}^{2}) +[3/(x_{0}^{2}+ y_{0}^{2})][x_{0}/(x_{0}^{2}+ y_{0}^{2})^{1/2}]}i + Gm^{2}{[3/(x_{0}^{2}+ y_{0}^{2})][y_{0}/(x_{0}^{2}+ y_{0}^{2})^{1/2}] + (4/y_{0}^{2})}j = Gm^{2}{(2/x_{0}^{2}) +[3x_{0}/(x_{0}^{2}+ y_{0}^{2})^{3/2}]}i + Gm^{2}{[3y_{0}/(x_{0}^{2}+ y_{0}^{2})^{3/2}] + (4/y_{0}^{2})}j.

23. We take the positive direction upward. The spring scale reads the normal force expressed as an effective mass: F_{N}/g. We write Sum(F) = ma from the force diagram: F_{N}- mg = ma, or m_{effective}= F_{N}/g = m(1 + a/g). (a) For a constant speed, there is no acceleration, so we have m_{effective}= m(1 + a/g) = m = 56 kg. (b) For a constant speed, there is no acceleration, so we have m_{effective}= m(1 + a/g) = m = 56 kg. (c) For the upward (positive) acceleration, we have m_{effective}= m(1 + a/g) = m(1 + 0.33g/g) = 1.33(56 kg) = 75 kg. (d) For the downward (negative) acceleration, we have m_{effective}= m(1 + a/g) = m(1 - 0.33g/g) = 0.67(56 kg) = 38 kg. (e) In free fall the acceleration is - g, so we have m_{effective}= m(1 + a/g) = m(1 - g/g) = 0.

33. From Kepler's third law, T^{2}= 4p^{2}r^{3}/Gm_{E}, we can relate the periods of the satellite and the Moon: (T/T_{Moon})^{2}= (r/r_{Moon})^{3}; (T/27.4 d)^{2}= [(6.38x10^{6}m)/(3.84x10^{8}m)]^{3}, which gives: T = 0.0587 days (1.41 h).

37. We use Kepler's third law, T^{2}= 4p^{2}r^{3}/Gm_{E}, for the motion of the Sun: T^{2}= 4p^{2}r^{3}/Gm_{galaxy}; = 4p^{2}[(3x10^{4}ly)(3x10^{8}m/s)(3.16x10^{7}s/yr)]^{3}/(6.67x10^{-11}N · m^{2}/kg^{2})(4x10^{41}kg), which gives: T = 6x10^{15}s = 2x10^{8}yr.

57. (a) If we assume Eros is a cylinder, its mass is m_{eros}= pr^{2}h = (2.3x10^{3}kg/m3)p(3x10^{3}m)^{2}(40x10^{3}m) = 2.6x10^{15}kg. We assume NEAR orbits around the narrow waist of Eros and use Kepler's third law: T^{2}= 4p^{2}r^{3}/Gm_{eros}= 4p^{2}[(15x10^{3}m) + (3x10^{3}m)]^{3}/(6.67x10^{-11}N · m^{2}/kg^{2})(2.6x10^{15}kg), which gives T = 3.6x10^{4}s = 10 h. (b) With the same mass and density, the sphere must have the same volume: 4/3pr_{0}^{3}= (40 km)p(3 km)^{2}, which gives r_{0}= 6.5 km. (c) The acceleration due to gravity on the surface of a spherical Eros is g = Gm_{eros}/r_{0}^{2}= (6.67x10^{-11}N · m^{2}/kg^{2})(2.6x10^{15}kg)/(6.5x10^{3}m)^{2}= 4.2x10^{-3}m/s^{2}.

8. For the magnitude of each attractive gravitational force, we have: F_{V}= Gm_{E}m_{V}/r_{V}^{2}= Gf_{V}m_{E}^{2}/r_{V}^{2}= (6.67x10^{-11}N · m^{2}/kg^{2})(0.815)(5.98x10^{24}kg)^{2}/[(108 - 150)x10^{9}m]^{2}= 1.10x10^{18}N; F_{J}= Gm_{E}m_{J}/r_{J}^{2}= Gf_{J}m_{E}^{2}/r_{J}^{2}= (6.67x10^{-11}N · m^{2}/kg^{2})(318)(5.98x10^{24}kg)^{2}/[(778 - 150)x10^{9}m]^{2}= 1.92x10^{18}N; F_{Sa}= Gm_{E}m_{Sa}/r_{Sa}^{2}= Gf_{Sa}m_{E}^{2}/r_{Sa}^{2}= (6.67x10^{-11}N · m^{2}/kg^{2})(95.1)(5.98x10^{24}kg)^{2}/[(1430 - 150)x10^{9}m]^{2}= 1.38x10^{17}N. The force from Venus is toward the Sun; the forces from Jupiter and Saturn are away from the Sun. For the net force we have: F_{Net}= F_{J}+ F_{Sa}- F_{V}= 1.92x10^{18}N + 1.38x10^{17}N - 1.10x10^{18}N = 9.6x10^{17}N away from the Sun.

11. The weight of objects is determined by g, which depends on the mass and radius of the Earth: g = Gm_{E}/r_{E}^{2}. If the density remains constant, the mass will be proportional to the radius cubed: m_{E}'/m_{E}= (r_{E}'/r_{E})^{3}. If we form the ration of g's, we have g'/g = (m_{E}'/m_{E})/(r_{E}'/r_{E})^{2}= (m_{E}'/m_{E})/(m_{E}'/m_{E})^{2/3}= (m_{E}'/m_{E})^{1/3}= 2^{1/3}= 1.26.

25. The acceleration due to gravity is g = Fgrav/M = Gm/r^{2}= (6.67x10^{-11}N · m^{2}/kg^{2})(7.4x10^{22}kg)/(4.1x10^{6}m)^{2}= 0.29 m/s^{2}. We take the positive direction toward the Moon. The apparent weight is measured by the normal force. We write Sum(F) = Ma from the force diagram: - F_{N}+ Mg = Ma, (a) For a constant velocity, there is no acceleration, so we have - F_{N}+ Mg = 0, or F_{N}= Mg = (75 kg)(0.29 m/s^{2}) = 22 N (toward the Moon). (b) For an acceleration toward the Moon, we have - F_{N}+ Mg = Ma, or F_{N}= M(g - a) = (75 kg)(0.29 m/s^{2}- 2.6 m/s^{2}) = - 1.7x10^{2}N (away from the Moon).

45. (a) The gravitational field is g = Gm_{S}/D_{SE}2 = (6.67x10^{-11}N · m^{2}/kg^{2})(2.0x10^{3}0 kg)/(1.50x10^{11}m)^{2}= 5.9x10^{-3}N/kg. (b) Because the field due to the Sun is about 1000 times smaller than the field from the Earth, the effect is not significant.

60. We relate the speed to the period of revolution from v = 2pr/T, where r is the distance to the center of the Milky Way. The gravitational attraction provides the centripetal acceleration: Gm_{galaxy}m_{S}/r^{2}= m_{S}v^{2}/r = m_{S}(2pr/T)^{2}/r = m_{S}4p^{2}r/T^{2}, so we have: m_{galaxy}= 4p^{2}r^{3}/GT^{2}= 4p^{2}[(30,000 ly)(9.5x10^{15}m/ly)]^{3}/ (6.67x10^{-11}N · m^{2}/kg^{2})[(200x10^{6}yr)(3.16x10^{7}s/yr)]^{2}= 3.4x10^{41}kg. The number of stars ("Suns") is (3.4x10^{41}kg)/(2.0x10^{30}kg) = 1.7x10^{11}.

62. The gravitational attraction from the core must provide the centripetal acceleration for the orbiting stars: Gm_{star}m_{core}/R^{2}= m_{star}v^{2}/R, so we have: m_{core}= v^{2}R/G = (780x10^{3}m/s)^{2}(5.7x10^{17}m)/(6.67x10^{-11}N · m^{2}/kg^{2}) = 5.2x10^{3}9 kg. If we compare this to our Sun, we get m_{core}/m_{Sun}= (5.2x10^{3}9 kg)/(2.0x10^{3}0 kg) = 2.6x10^{9}X.

17. For a stationary object at the equator, we know that the effective weight is w_{0}= mg - mv_{E}^{2}/r_{E}, where v_{E}= 2pr_{E}f is the speed of a point on the equator. When the ship moves with respect to the surface at speed v, the effective speed will be v_{E}± v, depending on the direction. The positive sign is used for a ship traveling eastward; the negative sign for a ship traveling westward. Thus the apparent weight is: w = mg - m(v_{E}± v)^{2}/r_{E}= mg - (mv_{E}^{2}/r_{E})[1 ± (v/v_{E})]^{2}= mg - (mv_{E}^{2}/r_{E})[1 ± 2(v/v_{E}) + (v/v_{E})^{2}]. Because v << v_{E}, we can drop the last term to get w = mg - (mv_{E}^{2}/r_{E}) - 2(mv_{E}v/r_{E}). If we divide this by the effective weight for a stationary object, we have w/w_{0}= [mg - (mv_{E}^{2}/r_{E}) - 2(mv_{E}v/r_{E})]/(mg - mv_{E}^{2}/r_{E}) = 1 - 2(mv_{E}v/r_{E})/(mg - mv_{E}^{2}/r_{E}) = 1 - 2(v_{E}v/gr_{E})/(1 - v_{E}^{2}/gr_{E}). Because v_{E}^{2}/gr_{E}<< 1, we get w/w_{0}~ 1 - 2(v_{E}v/gr_{E}) ~ 1 ± 2(v_{E}v/gr_{E}) = 1 ± 4pfv/g, if we reverse the meaning of the sign for the direction of the ship: positive sign is used for a ship traveling westward; the negative sign for a ship traveling eastward.