P9   
      V = 2 x (2pr) m/s = 2 x (2 x 3.1416 x 6 x 0.3) m/s = 22.6 m/s
      F = mV**2/r          m = 16 x 4.45 / 9.8 = 7.26 kg

      F = 7.26 x 22.6 x 22.6 / (6 x 0.3) = 2060 N

This HW problem can be rephrased as follows:
A hammer thrower whirls the ball at K revolutions per second, revolving a m kg ball at the end of a cable of R meters effective radius. Show that the inward force is F = mR(2pK)**2.

Solution in equation form: The inward force is the centripetal force given by

       F = mV**2/R.

The speed of the revolving ball can be calculated from V = K2pR

       F = m(K 2pR)**2/R = mR(2pK)**2

P23
       Re = 6370 km, WS = 100 N at south pole where centripetal force is zero.

       V = 465 m/s due to earth rotation at the equator

       WE = WS - mV**2/Re
            = 100 - (100/9.8) x (465)**2/(6.37 x 10**6)
            = 100 N - 0.34 N = 99.66 N

P25  

        W = mg = GmM/r**2
        If m' = 2m and r' = 2r,
        then W' = Gm'M/r'**2 = G2mM/(2r)**2 = W/2

P33  

        V = Vi - gt ------------------------------------------------------------------- (1) 
        z = zi + Vi t - (1/2) g t**2 --------------------------------------------------- (2)
        When z = h (maximum height), V = 0 so that Vi = gt ------------------------(3)
        Substituting (3) into (2),
        z = h = 0 + gt**2 - (1/2) g t**2 = (1/2) g t**2 so that
        t = (2h/g)**(1/2) --------------------------------------------------------------(4)

substituting (4) into (3) gives Vi = (2gh)**(1/2)
The same person is jumping on Earth and on Venus, the initial velocity must be the same.
so that (2gh)**(1/2) = (2g'h')**(1/2) where prime denotes quantities on Venus.

The above equation can be rewritten to show that

             gh = g'h' or h' = (g/g') h = (1/0.88) h = 1.14 h

This HW problem can be rephrased as "Show that the height of a jumper is inversely proportional to the gravity."