Example Problem with Complete Solution

1A-1 : Kinetic and Potential Energy of an Airplane in Flight 6 pts
Gravity is given as a function of altitude by g = 9.81 - 3.32 x 10-6 x  (m/s2), where x is the altitude above sea level.  An airplane is traveling at 900 km/h at an elevation of 10 km.  If its weight at sea level is 40 kN, determine:
               
a.) Its kinetic energy        
b.) Its potential energy relative to sea level
               
Read : Since gravitational acceleration is LESS at higher altitude, the gravitational potential energy of the airplane will not be quite as great as you might ordinarily expect.  We need the weight of the airplane at sea level in order to determine the mass of the airplane.  We need ot know the mass in order to calculate both the kinetic and gravitation potential energies of the plane.
Given : g = a - b * h v 900 km/h
a 9.81 m/s2 h        10,000 m
b 3.32E-06 s-2 Wsea level 40 kN
gc 1 kg-m/N-s2
Solution : Let's begin by determining the mass of the airplane from the weight at sea level.
Newton's 2nd Law of Motion: equation 1a
or equation 1b Eqn 1
At sea level, according to the eqn given in the problem statement, the acceleration of gravity is :
g 9.81 m/s2
Now, we can solve Eqn 1 for m and plug in values :
equation 2 Eqn 2
m 4077 kg
Now, we are ready to solve the rest of the problem.
Part a.) The definition of kinetic energy is :
equation 3 Eqn 3
Ek 1.274E+08 J
Ek 127 MJ
Part b.) The definition of gravitational potential energy, relative to sea level, is :
equation 4 Eqn 4
The problem is what value of g do we use ?  Do we simply use g at the altitude of the plane ?  Or do we use some sort of average value of g ?
Let's think about this part a bit more carefully. equation 5 Eqn 5
The differential increase in the potential energy of an object infinitessimally above sea level is:
equation 6 Eqn 6
So, the gravitational potential energy of an oject that is a distance h above sea level is :
equation 7 Eqn 7
Now, if m, g and gc are all constants, Eqn 6 simply reduces to Eqn 4 because the integral of dx from 0 to h is just h.  In our problem, however, g is NOT a constant.  Therefore :
equation 8
Eqn 8
Epot 3.993E+08 J Epot 399.3 MJ
So, what "average" value of g should we have used in Eqn 5 ?  Let's combine Eqns 5 and 8 and see what we get.
equation 9a or : equation 9b Eqn 9
So, the average effective value of the gravitational acceleration for determining the potential energy of the airplane in this problem is equal to the gravitational acceleration at HALF of the actual altitude of the airplane.  Would this be true if g = a - b x -c x2 ??  Nope.  What is special about the equation for g given in this problem that leads to the interesting result in Eqn 9 ?