Page 46 - Intro to Space Sciences Spacecraft Applications

Solution: Orbital Principles 33

T = (27.32 days) x (86,164 sedday) = 2,354,000 sec

Solving equation 2-6 for the semi-major axis:

a3 = (4) ~2 (2 - 7)
4n:

we find that for the period computed above:

a = 1.26 x lo9 ft = 237,658.8 mi (using 5,280 ft/mi)
= 382,469.6 km (using the metric values)

The semi-major axis represents the mean distance of an orbiting body
from the center of the earth. If the moon’s orbit were circular, this distance
from the earth would remain constant (as we have already shown). In fact,
the moon’s orbit is slightly elliptical, so its distance from the earth is con-
stantly changing as the following problem shows.
Note: the apoapsis and periapsis points of an orbit described around the
earth are called apogee and perigee respectively.

Example Problem:

The moon is in a slightly elliptical orbit around the earth (e =
0.055). Determine the moon’s apogee and perigee distances in both
miles and kilometers.

Solution:

Recalling the values obtained earlier for the moon’s semi-major
axis, and using the relationships obtained from the conic equation at
apoapsis and periapsis (eq. 2-5):

r, = a(l + e) = 250,730.0 mi = 403,505.4 km

rp = a( 1 - e) = 224,587.6 mi = 361,433.7 km