CHAPTER 12

  1. (a) The gravitational force due to 5.0-kg mass on the 20.0-kg mass is
          F1 = (6.67 x 10-11 N·m2/kg2)(5.0 kg)(20.0 kg)/(0.1 m)2i = (6.67 x 10-7 N)i
         The gravitational force due to 10.0-kg mass on the 20.0-kg mass is
          F2 = (6.67 x 10-11 N·m2/kg2)(10.0 kg)(20.0 kg)/(0.2 m)2j = (3.335 x 10-7 N)j
          The net force is
          F = F1 + F2 = [(6.67 x 10-7)i + (3.335 x 10-7)j] N.
          F = [(6.67 x 10-7)2 + ((3.335 x 10-7))2]1/2 N = 7.46 x 10-7 N
          θ = tan-1[(3.335 x 10-7) N/(6.67 x 10-7) N] = 26.6°, or 90° - 26.6° = 63.4° cw to y-axis.
    (b) The gravitational force due to 20.0-kg mass on the 5.0-kg mass is 
          F1 = -(6.67 x 10-11 N·m2/kg2)(5.0 kg)(20.0 kg)/(0.1 m)2i = -(6.67 x 10-7 N)i
          The magnitude of the gravitational force due to 10.0kg mass on the 5.0-kg mass is 
         F2 = (6.67 x 10-11 N·m2/kg2)(5.0 kg)(10.0 kg)/[(0.1 m)2 + (0.2 m)2]= 6.67 x 10-8 N
          F2 = F2 (-cosθ i + sinθj), where cosθ = (10 cm)/[(10 cm)2 + 20 cm)2]1/2 = 0.4472, θ = 63.4°, and sinθ = 0.8944
          F2 = -(2.98 x 10-8 N)i + (5.97 x 10-8 N)j
          F = F1 + F2 = -(6.67 x 10-7 N + 2.98 x 10-8 N)i  + (5.97 x 10-8 N)j = [-(9.65 x 10-8)i + (5.97 x 10-8)j] N.
          F = [(-9.65 x 10-8)2 + (5.97 x 10-8)2]1/2 N = 1.13 x 10-7 N
          φ = tan-1[(5.97 x 10-8 N)/(-9.65 x 10-8 N) = 58.2°, or 90° -58.2° = 31.8° ccw to y-axis
  2. a. DU = -(6.67 x 10-11 N·m2/kg2)(1.0 kg)(5.98 x 1024 kg)/x|6870 km6370 km
              = -4.557 x 106 J.
            DU = -DK = - 1/2mv2
                v = [2(-DU)/m]1/2 = [2(-4.557 x 109 J)/(1.0 kg)]1/2 = 3.02 x 103 m/s.
    b. 1/2mv2 = mgy;
              v = Ö2gy = Ö2(9.8 m/s2)(500000 m) = 3.1 x 103 m/s.
    c. error = (3.1 x 103 m/s -3.02 x 103 m/s)/(3.02 x 103 m/s) x100% = 2.7%
    d. No, because the hammer is moving around the circular orbit.
  3. Conservation of energy:
    1/2Mv12 + 1/2(2M)v22 - GM(2M)/(2R)= 0 - GM(2M)/(10R)
    Conservation of momentum:
       Mv1 + (2M)v2 = 0
    Sole the two equations 
     v1 = 2[(2/15)(GM/R)]1/2
     v2 = -[(2/15)(GM/R)]1/2
  4. Use Kepler's third law.
    T2 = (4π2/GM)r3
    r = (GMT2/4π2)1/3 = {(6.67 x 10-11 N·m2/kg2)(5.98 x 1024 kg)[(2)(24 h)(3600 s/h)]2/4π2}1/3
      = 3.12 x 107 m
  5. The two satellites have the same speed before the collision but in opposite direction.
    From conservation of momentum we find the speed of the debris after the collision
    m1v1 + m2v2 = (m1 + m2)V
    (400 kg)v + (100 kg)(-v) = (400 kg + 100 kg)V;
    V = (3/5)v, As V < v, Fc = mV2/r < GMm/r2
    the debris will not be held on the orbit and crashes to the earth.