Answers to homework problems:

Chapter 5 
                                    
 8)  Centripetal acceleration  a = v^2/r
     Velocity is equal to circumference of Earth at a given latitude divided by 24 h
     a) at the equator:
         v = 2*pi*r(Earth)/24h = 2*pi*6.38 10^6m/(24*3600)s = 464 m/s
         a = v^2/r = (464 m/s)^2/6.38 10^6 m = 0.0337 m/s^2
     b) at latitude 30 degrees: the radius of a circle describing person's trajectory 
        is r = r(Earth)*cos(30 degrees) =5.525 10^6 m
           v = 2*pi*r/(24*3600s) = 401.8 m/s
           a = v^2/r = 0.0292 m/s^2

22)  The angle is given by the formula (5.4)
        tan(theta) = v^2/(r*g) = (25 m/s)^2 /(150m * 9.8 m/s^2) = 0.4252
        theta = atan(0.4252) = 23 degrees

35)  The period of the satelite is given by formula (5.6)
        T = (2*pi*r^(3/2))/sqrt(G*M(Earth))
        The satelite A is at a radius r_A 
        The satelite B is at a radius r_B
        The ratio of the periods  T_A/T_B = (r_A/r_B)^(3/2)
        We can relate the radius of the satelite's orbit to its velocity through
        the equation (5.5) 
         v = sqrt((G*M(Earth))/r)
        The ratio of the velocities is
         v_A/v_B = sqrt(r_B/r_A)  (ratio of square roots is equal to square root of the ratio)
          ===> r_A/r_B = (v_B/v_A)^2
        After insertion into the expression for ratio of periods we get
             T_A/T_B = (v_B/v_A)^3
        Answer: If the orbital spped of satelite A is twice that of satelite B: v_A = 2*v_B
                then the period of satelite A will be 1/8 that of the satelite B.

38)  The normal force exerted on a pilot by his seat at a bottom of a vertical circle 
     is equal to centripetal force + his weight.
           F= m*v^2/r+mg
     The requirement that the overall force does not exceed 3*mg means that the centripetal 
     acceleration cannot exceed 2g. The maximum is given by v^2/r = 2*9.8 m/s^2
      ====> r = v^2/(2*9.8 m/s) = (230 m/s)^2/(2*9.8 m/s^2) = 2699 m