Module 1 ─ Orbit Dynamics I Study Notes Introduction Orbital mechanics extends celestial mechanics to artificial satellites. Ideal (Keplerian) vs. real (perturbed) orbits:Perturbations (non-spherical gravity, atmospheric drag, third-body effects, solar radiation pressure, etc.) can hinder or aid orbit control. Historical context:Johannes Kepler (1571–1630) supplied empirical laws. Isaac Newton (1643–1727) provided physical derivations via differential equations and universal gravitation. Developing the Model: Kepler & Newton Johannes KeplerAssistant to Tycho Brahe to obtain precise planetary data (esp. Mars). Derived three planetary laws geometrically, no dynamics. Hypothesized a central force toward the Sun. Corrected tidal theory, yet ignored by contemporaries (Galileo, Descartes). Waited ~60 years for Newtonian explanation. Isaac NewtonInvented differential calculus; modeled nature with equations. Derived Kepler’s laws from F = G m < e m > 1 m < / e m > 2 / r 2 F=Gm<em>1m</em>2/r^2 F = G m < e m > 1 m < / e m > 2/ r 2 and F = m a F=ma F = ma . Kepler’s Three Laws (Empirical) Law of OrbitsPlanets travel in ellipses with the Sun at one focus. Ellipse parameters:
• a a a – semi-major axis
• b b b – semi-minor axis
• e e e – eccentricity. Law of Areas (Constant Areal Velocity)Equal areas swept in equal times ⇒ A ˙ = const \dot A = \text{const} A ˙ = const for a given body. Quantitative form (derivable later): A ˙ = μ ( 1 − e 2 ) 2 h = const \dot A = \frac{\mu(1-e^2)}{2h} = \text{const} A ˙ = 2 h μ ( 1 − e 2 ) = const . Law of PeriodsT 2 / a 3 = 4 π 2 / μ T^2/a^3 = 4\pi^2/\mu T 2 / a 3 = 4 π 2 / μ ((\mu\equiv G M_{central})).Indicates sweep rate differs between orbits (larger (a) ⇒ longer (T)). Newton’s Four Fundamental Laws (Applied to Orbital Motion) Inertia: constant velocity (or rest) unless external force acts. Force–Momentum: F = d p d t , p = m v \mathbf F = \dfrac{d\mathbf p}{dt},\; \mathbf p=m\mathbf v F = d t d p , p = m v ; for constant (m): F = m a \mathbf F = m\mathbf a F = m a . Action–Reaction: F < e m > 12 = − F < / e m > 21 \mathbf F<em>{12} = -\mathbf F</em>{21} F < e m > 12 = − F < / e m > 21 . Universal Gravitation: F = G m < e m > 1 m < / e m > 2 r 2 F = G\dfrac{m<em>1 m</em>2}{r^2} F = G r 2 m < e m > 1 m < / e m > 2 , with G = 6.67 × 10 − 11 m 3 kg − 1 s − 2 G = 6.67\times10^{-11}\,\text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2} G = 6.67 × 1 0 − 11 m 3 kg − 1 s − 2 . Work & Energy in Gravitational Fields Differential work: d W = F ⋅ d r dW = \mathbf F\cdot d\mathbf r d W = F ⋅ d r (only component along motion contributes). Total work over path (c): W < e m > 12 = ∫ < / e m > r < e m > 1 r < / e m > 2 ! F ⋅ d r W<em>{12}=\int</em>{r<em>1}^{r</em>2}!\mathbf F\cdot d\mathbf r W < e m > 12 = ∫ < / e m > r < e m > 1 r < / e m > 2 ! F ⋅ d r . Work–Kinetic-Energy theorem: W < e m > 12 = 1 2 m v < / e m > 2 2 − 1 2 m v 1 2 = Δ K E . W<em>{12}=\tfrac12 m v</em>2^2 - \tfrac12 m v_1^2 = \Delta KE. W < e m > 12 = 2 1 m v < / e m > 2 2 − 2 1 m v 1 2 = Δ K E . Conservative field ⇒ potential energy (U): F = − ∇ U \mathbf F = -\nabla U F = − ∇ U .Gravitational potential between two masses (zero at (\infty)):U ( r ) = − G m < e m > 1 m < / e m > 2 r . U(r) = -\dfrac{G m<em>1 m</em>2}{r}. U ( r ) = − r G m < e m > 1 m < / e m > 2 . Conservation of mechanical energy: Δ K E + Δ P E = 0 \Delta KE + \Delta PE = 0 Δ K E + Δ P E = 0 ⇒ K E + P E = constant . KE+PE = \text{constant}. K E + P E = constant . Two-Body Problem (Idealized) Only two masses interact by inverse-square gravity; third-body forces neglected. Equations in vector form for masses 1 & 2: F < e m > 1 = − G m < / e m > 1 m < e m > 2 r 3 r , F < / e m > 2 = + G m < e m > 1 m < / e m > 2 r 3 r . \mathbf F<em>1 = -G\dfrac{m</em>1 m<em>2}{r^3}\mathbf r,\quad \mathbf F</em>2 = +G\dfrac{m<em>1 m</em>2}{r^3}\mathbf r. F < e m > 1 = − G r 3 m < / e m > 1 m < e m > 2 r , F < / e m > 2 = + G r 3 m < e m > 1 m < / e m > 2 r . Angular (Moment of) Momentum For a particle of mass (m) in orbit:h = r × m v \mathbf h = \mathbf r\times m\mathbf v h = r × m v . Magnitude (scalar) for planar motion: h = m r v < e m > t = m r 2 θ ˙ , h = m r v<em>t = m r^2 \dot\theta, h = m r v < e m > t = m r 2 θ ˙ , where (vt) is transverse component. In a central force field, h \mathbf h h is conserved (direction & magnitude). General Orbit Equation (Conic Section) From Newtonian dynamics: r = h 2 / μ 1 + e cos ( θ − θ < e m > 0 ) r = \frac{h^2/\mu}{1+e\cos(\theta-\theta<em>0)} r = 1 + e c o s ( θ − θ < e m > 0 ) h 2 / μ
or equivalently r = p 1 + e cos ν , r = \frac{p}{1+e\cos\nu}, r = 1 + e c o s ν p , with semilatus rectum p = h 2 / μ p = h^2/\mu p = h 2 / μ and true anomaly (\nu\equiv\theta-\theta0). The parameter e e e determines conic type:e = 0 e=0 e = 0 ⇒ circle.0<e<1 ⇒ ellipse. e = 1 e=1 e = 1 ⇒ parabola.e>1 ⇒ hyperbola. Analysis of Specific Keplerian Orbits Circular Orbit (e = 0) Radius constant r = a = p = h 2 / μ r=a=p=h^2/\mu r = a = p = h 2 / μ . Orbital speed: v c = μ / r . v_c = \sqrt{\mu/r}. v c = μ / r . Period: T c = 2 π r 3 / μ . T_c = 2\pi\sqrt{r^{3}/\mu}. T c = 2 π r 3 / μ . Elliptical Orbit (0 < e < 1) Periapsis (closest): (\nu=0^\circ) ⇒ r p = p 1 + e = a ( 1 − e ) . r_p = \dfrac{p}{1+e} = a(1-e). r p = 1 + e p = a ( 1 − e ) . Apoapsis (farthest): (\nu=180^\circ) ⇒ r a = p 1 − e = a ( 1 + e ) . r_a = \dfrac{p}{1-e} = a(1+e). r a = 1 − e p = a ( 1 + e ) . Velocity at an arbitrary (r): v 2 = μ ( 2 r − 1 a ) . v^2 = \mu\left(\dfrac{2}{r}-\dfrac{1}{a}\right). v 2 = μ ( r 2 − a 1 ) . Period from Kepler III: T = 2 π a 3 / μ . T = 2\pi\sqrt{a^{3}/\mu}. T = 2 π a 3 / μ . Parabolic Orbit (e = 1) Specific energy = 0. Radius relation: r = p 1 + cos ν = h 2 / μ 1 + cos ν . r=\dfrac{p}{1+\cos\nu} = \dfrac{h^2/\mu}{1+\cos\nu}. r = 1 + cos ν p = 1 + cos ν h 2 / μ . Periapsis distance: r p = p / 2. r_p = p/2. r p = p /2. Hyperbolic Orbit (e > 1) Open trajectory; excess specific energy (>0). Approach asymptote defined by eccentricity and semi-major axis (a < 0). Activity 1 (Practice Problems) Circular Earth orbit with speed v = 7 km/s v=7\,\text{km/s} v = 7 km/s .Use v < e m > c = μ < / e m > E / ( R E + h ) v<em>c=\sqrt{\mu</em>E/(R_E+h)} v < e m > c = μ < / e m > E / ( R E + h ) . Solve for altitude (h). ((\muE = 3.986\times10^{14}\,\text{m}^3\,\text{s}^{-2}); (R E=6378\,\text{km})). Mars elliptical orbit, major axis 2 a = 40 , 000 km ⇒ a = 20 , 000 km 2a=40,000\,\text{km}\Rightarrow a=20,000\,\text{km} 2 a = 40 , 000 km ⇒ a = 20 , 000 km .Period: T = 2 π a 3 / μ < e m > M a r s T=2\pi\sqrt{a^{3}/\mu<em>{Mars}} T = 2 π a 3 / μ < e m > M a r s , with μ < / e m > M a r s = 4.282 × 10 13 m 3 s − 2 \mu</em>{Mars}=4.282\times10^{13}\,\text{m}^3\,\text{s}^{-2} μ < / e m > M a r s = 4.282 × 1 0 13 m 3 s − 2 . Derive velocities:Start from general equation r = h 2 / μ 1 + e cos ν r=\dfrac{h^2/\mu}{1+e\cos\nu} r = 1 + e cos ν h 2 / μ along with energy v 2 = μ ( 2 r − 1 a ) . v^2=\mu\left(\dfrac{2}{r}-\dfrac{1}{a}\right). v 2 = μ ( r 2 − a 1 ) . Specialize to:
• Circle: set e = 0 , a = r e=0,\;a=r e = 0 , a = r ⇒ v c = μ / r v_c=\sqrt{\mu/r} v c = μ / r .
• Ellipse: retain e e e , use vis-viva result above. Reference Constants G = 6.6743 × 10 − 11 m 3 kg − 1 s − 2 G = 6.6743\times10^{-11}\,\text{m}^3\,\text{kg}^{-1}\,\text{s}^{-2} G = 6.6743 × 1 0 − 11 m 3 kg − 1 s − 2 .Earth radius: R E = 6378 km R_E = 6378\,\text{km} R E = 6378 km . Earth mass: M E = 5.9722 × 10 24 kg M_E = 5.9722\times10^{24}\,\text{kg} M E = 5.9722 × 1 0 24 kg . Sun mass: M ⊙ = 1.9891 × 10 30 kg M_\odot = 1.9891\times10^{30}\,\text{kg} M ⊙ = 1.9891 × 1 0 30 kg . Mars mass: M M a r s = 6.41693 × 10 23 kg M_{Mars}=6.41693\times10^{23}\,\text{kg} M M a r s = 6.41693 × 1 0 23 kg . Standard gravitational parameters: μ < e m > E = G M < / e m > E , μ < e m > M a r s = G M < / e m > M a r s . \mu<em>E = G M</em>E, \; \mu<em>{Mars}=G M</em>{Mars}. μ < e m > E = GM < / e m > E , μ < e m > M a r s = GM < / e m > M a r s . Next Topic Detailed derivation of the general equation of motion for a body in a Keplerian orbit (start from Newton’s second law in polar coordinates, apply conservation of angular momentum, integrate).