P45    
W = PE = mgh = 1.3 x 10**7 J,     m = 80 kg
h = PE/mg = 1.3 x 10**7 / 80 x 9.8 = 16.6 km

P51   
(a) PE = mgh = 10 N x 10 m = 100 J = 0.1 kJ
(b) No rope on the floor
(c) PE increase by 0.1 kJ due to the work done by the monkey.

P73   
A satellite is in a circular orbit at a speed of V = 1.5 km/s.

The Law of Universal Gravitation: FG = GmM/r**2 = mg(r) so that

   g(r) = GM/r**2 = GMRe**2 / R**2
          = GM / (Re**2 x R**2)
    g(R) = 9.8/R**2 (see section 6.3.2 of lecture notes)

To stay in orbit, the centripetal force acceleration must balalnce the gravitational force, leading to

   Vo**2/r = g(r) ------------------------------------------------------------------ (1)  

   Vo**2/R = 9.8 Re/R**2 --------------------------------------------------------- (2)  

   where R is the radius of the orbit in units of earth radius.

   R = 9.8 x Re / Vo**2

The escape speed from the orbit is determined by

    KE + PE(R) = 0 (shown in lecture notes) ---------------------------------------- (3)

    Substituting PEG = - m 9.8 Re/R for PE(R) in (3), we obtain

    (1/2) m Vesc**2 = m g(R) = m 9.8 Re/R   or

    Vesc = (2 x 9.8 x Re/9.8 x Re / Vo**2)**(1/2) = (2)**0.5 Vo = 2.1 km/s

Show that the orbital speed at R = 1 is Vo = 7.9 km/s.

P79
P = F dot V = mg V = 1.0 N x 50 m/10 s = 5.0 Nm/s = 5.0 J/s 5.0 w