Two-Body Orbital Mechanics
2.3 Constants of Motion
We begin to develop the analytical solution for two-body motion by determining constants associated with the two-body problem. The concepts presented in this section (momentum and energy) should be familiar to students with a background in basic mechanics. Many of the derivations that follow rely on vector operations such as the cross product and vector triple product; we summarize these operations in Appendix B.
2.3.1 Conservation of Angular Momentum
Linear momentum of a satellite is simply the product of its mass m and velocity vectorv.
Angular momentumH (or “moment of momentum”) is defined by the cross product of position vectorr and linear momentum mv:
H = r × mv (2.13)
The time derivative of angular momentum (for a satellite with constant mass) is
H = r × mv + r × mv (2.14)
Becauser = v, the first cross product in Eq. (2.14) is zero. The term mv is equal to the force F acting on the satellite. Hence, Eq. (2.14) becomes the familiar relationship
between the time-rate of angular momentum and the external torque produced by forceF
H = r × F (2.15)
For the two-body problem, the central-body gravitational force is the only force acting on the satellite. Furthermore, this attractive force is aligned with the position vectorr and hence the cross product in Eq. (2.15) is zero. Consequently, the satellite’s angular momentumH vector is constant for two-body motion.
We can arrive at the same result by performing vector operations on the governing two-body equation of motion (2.12). First, take the cross product of positionr with each side of Eq. (2.12):
r × r = r × −μ
r3 r (2.16)
Clearly, the right-hand side of Eq. (2.16) is zero because we are crossing two parallel vec-tors. Hence, Eq. (2.16) becomes r × r = 0. Next, we can carry out the following time derivative
d
dt r × v = r × v + r × v
=v × v + r × r
=r × r
(2.17)
Because Eq. (2.16) shows thatr × r = 0, the cross product r × v must be a constant vector.
Referring back to Eq. (2.13), we see thatr × v is angular momentum H divided by mass m.
The“specific angular momentum” or angular momentum per unit mass of a satellite in a two-body orbit is
h = r × v = constant vector (2.18)
Position and velocity vectors (r and v) will change as a satellite moves along its orbit but the angular momentum h remains a constant vector. Figure 2.5 shows an arc of a
Reference direction Local
horizon r
v
θ
θ θ γ
Satellite
Orbital path
Angular momentum (out of the page) r∙
vr =
r∙ v =
v r h= ×
Figure 2.5 Angular momentum and flight-path angle, γ.
satellite’s orbit where its current position and velocity vectors are denoted by r and v.
Because angular momentumh is the cross product r × v, it is perpendicular to the plane containing vectors r and v. Figure 2.5 shows counterclockwise satellite motion where vectors r and v are in the plane of the page and hence h is pointing out of the page.
Because vector h is constant, the plane containing the motion of the satellite (known as the orbital plane) is also fixed in space for two-body motion. The orbital plane passes through the center of the gravitational body because it contains position vectorr. The angleγ in Figure 2.5 is called the flight-path angle and it is measured from the local hori-zon (perpendicular tor) to the velocity vector v. Flight-path angle is positive when the satellite’s radial velocity component is positive, or r > 0 (as shown in Figure 2.5). Con-versely,γ < 0 if the length of the radius vector is decreasing, or r < 0. If the satellite moves in a circular orbit where the radius is constant (r = 0), then the flight-path angle is zero at all times and the velocity vectorv remains perpendicular to position vector r.
We can express the magnitude of the angular momentum vector in terms of radius r and the components of velocity vectorv. Let us define vectors r and v in terms of polar coordinates
r = rur (2.19)
v = vrur+ vθuθ (2.20)
where unit vectorurpoints in the radial direction and unit vectoruθpoints in the trans-verse direction (perpendicular tor or along the local horizon in the direction of motion;
see Section C.3 in Appendix C for additional details). The radial and transverse velocity components are vr= r = v sinγ and vθ= rθ = vcosγ, respectively (see Figure 2.5). Using Eqs. (2.19) and (2.20), the angular momentum is
h = r × v =
ur uθ uk
r 0 0
vr vθ 0
= rvθuk (2.21)
where the unit vectorukpoints normal to the orbital plane according to the right-hand rule. Equation (2.21) shows that the magnitude of the angular momentum vector is the product of the radius r and the transverse velocity component vθ.Therefore, a satellite with purely radial velocity will have zero angular momentum– it has no angular motion!
The following expression may be used to determine the angular momentum magnitude:
h= rvθ= r2θ = rvcosγ (2.22)
From the equivalence of the terms above, it is easy to reconcile that the satellite’s trans-verse velocity component is vθ= rθ = vcosγ (see Figure 2.5).
2.3.2 Conservation of Energy
We demonstrated the conservation of angular momentum by taking the vector (or cross) product of the governing two-body equation and positionr. Next, we will obtain a scalar result by taking the scalar (or dot) product of the velocity vectorr and both sides of the governing two-body equation of motion (2.12):
r r = r −μ
r3 r (2.23)
The left-hand side of Eq. (2.23) is the dot productv v while the right-hand side involves the dot productr r. Therefore both sides contain a dot product between a vector and its time derivative. Figure 2.5 shows that the dot productr v = r r involves the projection of velocity vectorv in the direction of position r, that is
r r = rvr= rr (2.24)
Using this result, we obtainv v = vv, or the product of the velocity magnitude and the rate of change of the length of vectorv. Equation (2.23) becomes
vv= −μ r3 rr=−μ
r2 r (2.25)
Note that each side of Eq. (2.25) can be written as a time derivative:
d dt
v2
2 = vv and d dt
μ r =−μ
r2 r (2.26)
Therefore, Eq. (2.25) becomes d dt
v2 2 = d
dt μ
r (2.27)
or
d dt
v2 2−μ
r = 0 (2.28)
The bracketed term in Eq. (2.28) must be a constant. Integrating Eq. (2.28), we obtain ξ =v2
2−μ
r= constant (2.29)
whereξ is the specific energy (total energy per unit mass) of the satellite in its orbit. The reader should be able to identify the first term on the right-hand side (v2/2) as kinetic energy per unit mass. The second term (−μ/r) is the potential energy of the satellite per unit mass. A satellite’s potential energy increases as its distance from the attracting body increases (similar to the“mgh” potential energy discussed in a university physics course). However, a satellite’s minimum potential energy (occurring when r is equal to the radius of the attracting body) is negative and its maximum potential energy approaches zero as r ∞. We shall soon see that adopting this convention means that a satellite in a closed (or repeating) orbit has negative total energy while a satellite fol-lowing an unbounded open-ended trajectory has positive energy. In either case, Eq. (2.29) tells us that the satellite’s total energy ξ remains constant along its orbital path. The sat-ellite may speed up during its orbit and gain kinetic energy but in doing so it loses poten-tial energy so that total energyξ remains constant.
We can also demonstrate the conservation of energy by using a gravitational potential function. From the gradient of a scalar potential function U, we can determine the grav-itational force (or gravgrav-itational acceleration)
r = ∇U (2.30)
where the “del” operator is a vector differential operation of partial derivatives with respect to position coordinates (such as XYZ Cartesian coordinates). The two-body potential function is
U=μ
r (2.31)
Note that the potential function U is the negative of the potential energy. Computing the gradient of potential function U =μ/r leads to the right-hand side of Eq. (2.12), the governing equation of motion for the two-body problem (we will present the details of the gradient operation in Chapter 5). Next, we may use the chain rule to write the time derivative of the scalar potential function:
dU dt =∂U
∂r dr dt
The first term on the right-hand side is∇U = r. Therefore, the time derivative is dU
dt =r r = v v = vv (2.32)
Equation (2.26) shows that vv is the time derivative of kinetic energy. Defining specific kinetic energy as T = v2/2, we can write Eq. (2.32) as
dU dt =dT
dt or
d
dt T−U = 0 (2.33)
Equation (2.33) shows that T– U is constant. Because potential energy V is the negative of the potential function (i.e., V =−μ/r = −U), Eq. (2.33) shows that the sum of kinetic energy and potential energy (ξ = T + V ) is constant.