Bate, Mueller, and White - Fundamentals of Astrodynamics ... - UL FGG

18 TWO·BODY ORBITAL MECHANICS, Ch. 1
from the center of the earth and "down" means toward the center of
the earth. So the local vertical at the location of the satellite coincides
with the direction of the vector r. The local horizontal plane must then
be perpendicular to the local vertical. We can now define the direction
of the velocity vector, v, by specifying the angle it makes with the local
vertical as 'Y (gamma), the zenith angle. The angle between the velocity
vector and the local horizontal plane is called ¢ (phi), the flight·path
elevation angle or simply "flight·path angle." From the definition of
the cross product the magnitude of h is
h = N sin 'Y,
We will find it more convenient, however, to express h in terms of the
flight·path angle, ¢. Since 'Y and ¢ are obviously complementary angles
I h = rv cos ¢. (1.4.4)
The sign of ¢ will be the same as the sign of r • v.
EXAMPLE PROBLEM. In an inertial coordinate system, the
position and velocity vectors of a satellite are, respectively, (4. 1852 1+
6.2778 J + 10.463 K) 10 7 ft and (2.5936 1+ 5.1872 1) 10 4 ft/sec
where I, J and K are unit vectors. Determine the specific mechanical
energy, &, and the specific angular momentum, h. Also find the
flight·path angle, ¢.
r = 12.899 X 107 ft, V = 5.7995 X 104 ft/sec
&=-L= 1.573x 109 ft2/sec2
2 r
h = r X v= (-5.4274I+2.7 1 37J+O.54273K)l012ft2 Isec
h = 6.0922 X 1012 ft2/sec
h = rv cos ¢, cos ¢= Jl. = 0.8143
rv
r • v > 0, therefore:
¢ = arccos 0.8143 = 35.420