Two-Body Orbital Mechanics
2.5 Elliptical Orbit
2.5.2 Flight-Path Angle and Velocity Components
Figure 2.9 also shows the satellite’s flight-path angle, γ. Recall that we measure flight-path angle from the local horizon (perpendicular to the radius vectorr) to the velocity vector v. Flight-path angle γ is always between –90 and +90 and it is zero when the satellite is at perigee or apogee. When a satellite is traveling from periapsis to apoapsis (as shown in Figure 2.9), its true anomaly is between zero and 180 and its flight-path angle is positive.
Apogee
Earth Satellite
ra
r θ
Perigee v
rp
γ
Local horizon
Figure 2.9 Elliptical orbit about the Earth.
Conversely, flight-path angle is negative when a satellite approaches periapsis (i.e., 180 <θ < 360 ).
The reader should note that Eq. (2.22) is the only formula we have developed thus far for computing flight-path angle. Using Eq. (2.22) to findγ requires taking the inverse cosine of the positive quantity h/(rv). A calculator or computer inverse-cosine operation of a positive argument will always place the angle in the first quadrant (i.e., 0 <γ < 90 ).
From the previous discussion, we know thatγ < 0 for half of an elliptical orbit when the satellite approaches periapsis. We can resolve this quadrant ambiguity by expressing the tangent of the flight-path angle as the ratio of the radial velocity component vr
and the transverse velocity component vθ tanγ =vr
vθ (2.66)
Figure 2.5 shows this simple geometric relationship between the velocity components.
We obtain the radial velocity by taking the time derivative of the trajectory equation (2.45) via the chain rule:
vr= r =dr dθ
dθ
dt = pesinθ 1 + e cosθ 2
dθ
dt (2.67)
Equation (2.22) shows that the time rate of true anomaly is dθ
dt = h
r2=h 1 + e cosθ 2
p2 (2.68)
Note that the trajectory equation (2.45) has been squared and substituted for r2 in Eq. (2.68). Finally, substituting Eq. (2.68) for dθ/dt in Eq. (2.67) and using p = h2/μ we obtain a simplified expression for radial velocity
vr=μ
hesinθ (2.69)
Equation (2.69) shows that r = 0 at periapsis (θ = 0) and apoapsis (θ = 180 ) as expected.
Transverse velocity can be computed directly from the ratio of h and r vθ= rθ =h
r=h 1 + e cosθ
p (2.70)
Again, the trajectory equation (2.45) has been substituted for r. Using p = h2/μ in Eq. (2.70) yields
vθ=μ
h 1 + e cosθ (2.71)
Dividing Eq. (2.69) by Eq. (2.71) yields the tangent of the flight-path angle tanγ = esinθ
1 + e cosθ (2.72)
Applying a calculator’s or computer’s inverse-tangent function to Eq. (2.72) will place the flight-path angle in the correct range (i.e.,–90 < γ < 90 ).
Finally, let us determine the locations in the orbit where velocity is minimum and max-imum. Equation (2.22) shows that angular momentum h is the product of radius r and
the transverse velocity vθ (i.e., the velocity component perpendicular to r). Hence, the angular momentum at periapsis and apoapsis is
h= rpvp= rava (2.73)
where vpand vaare the velocities at periapsis and apoapsis, respectively. Because h is constant and periapsis rp is the minimum radius, the satellite’s maximum velocity is at periapsis. Conversely, the satellite’s slowest velocity occurs when it is at apoapsis or the farthest position in its orbit.
Table 2.2 summarizes the values (or range of values) for true anomaly and flight-path angle for positions within an elliptical orbit. True anomalyθ is a key element because it is needed to determine whether flight-path angle is positive or negative [remember that using Eq. (2.22) to compute flight-path angle does not determine its sign]. The fourth column in Table 2.2 is the dot productr v = rr which can be used to determine the range for true anomaly. When r v > 0, radial velocity must be positive and the satellite is approaching apoapsis. Ifr v < 0, the satellite is approaching periapsis (r < 0).
It is useful to summarize the orbital relationships that we have developed at this stage of the chapter (most of these relationships are valid for all conic sections).
1) Semimajor axis (a) determines total energy (this is true for all conic sections).
2) Parameter (p) determines the angular momentum magnitude h (this is true for all conic sections).
3) The three geometric characteristics (a, e, and p) are not independent. Knowledge of two characteristics can be used to determine the missing element. From a, e, and p we can determine total energy, angular momentum, and periapsis radius (this is true for all conic sections). For elliptical orbits we can determine the apoapsis radius.
4) Given any two of the three geometric characteristics (a, e, and p) and true anomalyθ (i.e., angular position in the orbit), we can determine radius r using the trajectory equation. Velocity magnitude v can be determined from the total energy. Flight-path angleγ can be determined from angular momentum or by using Eq. (2.72) (this is true for all conic sections).
5) Position and velocity vectors (r,v) determine every orbital constant and the satellite’s position in the orbit. Note that we calculate true anomaly using the trajectory equa-tion where the proper quadrant forθ is determined by checking the sign of the dot productr v (this is true for all conic sections).
The following examples illustrate many of these relationships for a satellite in an elliptical orbit.
Table 2.2 True anomaly and flight-path angle values on an elliptical orbit.
Orbital position True anomaly Flight-path angle Dot product
Periapsis θ = 0 γ = 0 r v = 0
Approaching apoapsis 0ο<θ < 180ο γ > 0 r v > 0
Apoapsis θ = 180ο γ = 0 r v = 0
Approaching periapsis 180ο<θ < 360ο γ < 0 r v < 0
Example 2.2 An Earth-orbiting satellite has semimajor axis a = 7,758 km and param-eter p = 7,634 km. Dparam-etermine (a) orbital energy, (b) angular momentum, and (c) whether or not the satellite will pass through the Earth’s appreciable atmosphere (i.e., altitude less than 122 km).
a) We compute total energy using Eq. (2.63) withμ = 3.986(105) km3/s2 ξ = − μ
2a= – 25 6896 km2/s2
Energy is negative because the satellite is following an elliptical orbit (a > 0).
b) We determine angular momentum using parameter p and Eq. (2.46) h= μp = 55,162 6 km2/s
c) Computing the perigee radius rpwill determine the satellite’s closest approach. We may use either Eq. (2.50) or Eq. (2.51). In either case, we need to determine the orbital eccentricity from a and p
e= 1−p
a= 0 1264
This calculation also verifies that the orbit is elliptical. Perigee radius is rp= p
1 + e= 6, 777 2 km
Perigee altitude is rp– RE= 6,777.2– 6,378 = 399.2 km which is greater than 122 km.
Therefore, this satellite will not pass through the Earth’s appreciable atmosphere.
Figure 2.10 shows the elliptical orbit about the Earth with the proper scale. The Earth’s atmosphere is the very thin shaded region surrounding the Earth in Figure 2.10. The
Orbit Orbit
Apogee
Earth’s atmosphere
Earth Zoomed -in view
Figure 2.10 Elliptical orbit and Earth’s atmosphere with zoomed-in view near perigee (Example 2.2).
zoomed-in view near perigee (shown on the right-hand side of Figure 2.10) illustrates the proximity between perigee and the upper atmosphere: the perigee pass is less than 280 km above the significant atmosphere. Computing the apogee radius ra yields an apogee altitude of 2,361 km, which is a substantial distance above the Earth’s atmosphere as shown on the left-hand side of Figure 2.10.
Example 2.3 The Chandra X-ray Observatory (CXO) is an Earth-orbiting satellite with perigee and apogee altitudes of 14,308 km and 134,528 km, respectively (Figure 2.11).
Calculate (a) angular momentum of the orbit, (b) total energy of the orbit, and (c) radius, velocity, and flight-path angle at true anomalyθ = 120 .
a) First, let us compute the perigee and apogee radii from the given altitude information:
Perigee radius rp= 14,308 km + RE= 20,686 km Apogee radius ra= 134,528 km + RE= 140,906 km
where RE= 6,378 km is the Earth’s radius. It is extremely important for the reader to remember that radius is the value used in the energy, momentum, and trajectory equations and not the altitude above a planet’s surface!
We can determine semimajor axis (a) and eccentricity (e) using perigee and apogee radii and Eqs. (2.64) and (2.65):
a=rp+ ra
2 = 80,796 km e=ra−rp
rp+ ra
= 0 7440
We can compute angular momentum directly from parameter p. Using Eq. (2.49), the parameter is determined to be
p= a 1−e2 = 36, 075 81 km Angular momentum is
h= μp = 119,915 89 km2/s
θ
Apogee altitude
134,528 km Perigee altitude
14,308 km
γ
r
v Satellite
Figure 2.11 Chandra X-ray Observatory orbit (Example 2.3).
b) We can calculate total energy using semimajor axis a = 80,796 km ξ = − μ
2a= – 2 4667 km2/s2
c) Using the trajectory equation (2.45) and the computed values of p and e, the radius at true anomalyθ = 120 is
r= p
1 + e cosθ= 57, 444 31 km
We can manipulate the energy equation (2.29) and obtain the velocity:
ξ =v2 2−μ
r v= 2 ξ +μ
r = 2 9907 km/s
We can compute flight-path angle using either the angular momentum equa-tion (2.22) or Eq. (2.72). First, let us determine γ using Eq. (2.22) and h= 119,915.89 km2/s, r = 57,444.31 km, and v = 2.9907 km/s:
h= rv cosγ γ = cos−1 h
rv = 45 73
Flight-path angle is positive becauseθ = 120 (the CXO is approaching apogee; see Figure 2.11). Let us re-compute flight-path angle using Eq. (2.72) with e = 0.7440 and θ = 120 :
tanγ = esinθ
1 + e cosθ= 1 025931 γ = 45 73 same result
Example 2.4 The Apollo command module (CM) is being tracked by an Earth-based radar station during its return from the moon. The station determines that the CM orbit has a semimajor axis a = 424,587 km with eccentricity e = 0.9849.
a) Determine the flight-path angle and radial velocity of the CM when true anomaly is 330 .
b) Determine the vehicle’s velocity and flight-path angle at the so-called “entry inter-face” (EI) or start of the atmospheric entry phase (EI is defined as 122 km above the Earth’s surface).
a) We can compute flight-path angle using Eq. (2.72) with e = 0.9849 andθ = 330 : tanγ = esinθ
1 + e cosθ=– 0 265766
Taking the inverse tangent, we find flight-path angle γ = –14 8832 One way to compute radial velocity is to use Eq. (2.69):
r=μ hesinθ
where angular momentum h is computed from parameter p
p= a 1−e2 = 12, 725 72 km h= μp = 71,221 28 km2/s Using these values, the radial velocity atθ = 330 is
r=μ
hesinθ = – 2 7561 km/s
Radial velocity is negative because the CM is approaching Earth (or, approaching per-igee).
Let us show an alternate solution method using the total energy. Because radial velocity is vr= v sinγ, we must determine the vehicle’s velocity at true anomaly θ = 330 . First, compute the energy from semimajor axis:
ξ = − μ
2a=– 0 469397 km2/s2
Total energy is the sum of kinetic and potential energy. Potential energy depends on radius r which in turn can be determined using the trajectory equation (2.45) with p= 12,725.72 km, e = 0.9849, andθ = 330 :
r= p
1 + e cosθ= 6, 867 82 km Next, solve the energy equation (2.29) for velocity:
v= 2 ξ +μ
r = 10 7303 km/s
Therefore, the radial velocity is vr= v sinγ = – 2 7561 km/s (same result).
b) We can write an energy equation by combining Eqs. (2.29) and (2.63)
ξ = v2EI 2 − μ
rEI
=− μ
2a=– 0 469397 km2/s2
where rEIand vEIare the radius and velocity at entry interface, respectively. It should be clear that energy is constant along the return trajectory. Using rEI= 122 km + RE= 6,500 km, the velocity at EI is
vEI= 2 ξ + μ
rEI = 11 0321 km/s
The flight-path angle at EI can be determined from angular momentum, Eq. (2.22):
h= rEIvEIcosγEI γEI= cos−1 h rEIvEI
Using the previously computed values for h, rEI, and vEI, we obtain γEI=– 6 6841
Note that flight-path angle is negative at entry interface because the CM is approaching Earth (i.e., r < 0).
Let us show an alternate solution for EI flight-path angle using Eq. (2.72):
tanγEI= esinθEI
1 + e cosθEI
We can obtain the true anomaly at EI (θEI) using the trajectory equation:
rEI= p 1 + e cosθEI
cosθEI=1 e
p
rEI−1 = 0 972487
The true anomaly at EI isθEI=–13.47 (or, 346.53 ). True anomaly at EI is in the fourth quadrant because the CM is approaching perigee. Finally, the tangent of the flight-path angle is
tanγEI= esinθEI
1 + e cosθEI
=– 0 117192
The inverse-tangent operation yieldsγEI=–6.6841 as previously computed.
One advantage of using Eq. (2.72) to determine the flight-path angle is that a quadrant check is unnecessary. However, we must correctly determine the satellite’s true anomaly θ by using the trajectory equation (2.45) and additional information regarding the satellite’s position in the orbit. In this example, we are told that the Apollo command module is returning to Earth and therefore it is approaching perigee (180ο<θ
< 360οandγ < 0). The reader should make a special note regarding this example!