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Level 4 — ApplicationOrbital Mechanics & Astrodynamics

Orbital Mechanics & Astrodynamics

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Level 4 — Application (novel problems, no hints) Time limit: 60 minutes Total marks: 60

Use GM⊕=3.986×105 km3/s2GM_\oplus = 3.986\times10^{5}\ \text{km}^3/\text{s}^2, R⊕=6378 kmR_\oplus = 6378\ \text{km} unless stated otherwise. Show all working.

Question 1 — Vis-viva & orbit reconstruction (12 marks)

A spacecraft is observed at a point where its geocentric distance is r=9000r = 9000 km and its speed is v=7.2v = 7.2 km/s. The flight-path angle (angle between velocity and the local horizontal) at that instant is γ=20∘\gamma = 20^\circ.

(a) Compute the semi-major axis aa of the orbit using the vis-viva equation. (3)

(b) Compute the specific angular momentum hh and hence the semi-latus rectum pp. (4)

(c) Determine the eccentricity ee and state the orbit type. (3)

(d) Find the true anomaly ν\nu at this instant (choose the value consistent with the spacecraft moving away from perigee). (2)

Question 2 — Hohmann vs bi-elliptic transfer (14 marks)

A satellite is in a circular orbit of radius r1=7000r_1 = 7000 km and must be raised to a circular orbit of radius r2=105000r_2 = 105000 km.

(a) Compute the total Δv\Delta v for a standard two-impulse Hohmann transfer. (6)

(b) A bi-elliptic transfer is proposed with an intermediate apoapsis radius rb=210000r_b = 210000 km. Compute the total three-impulse Δv\Delta v. (6)

(c) State which transfer is cheaper here and give one physical reason bi-elliptic transfers can beat Hohmann. (2)

Question 3 — Kepler's equation & time of flight (12 marks)

A satellite is in an orbit with semi-major axis a=20000a = 20000 km and eccentricity e=0.35e = 0.35.

(a) Compute the orbital period TT. (2)

(b) The satellite is at true anomaly ν=100∘\nu = 100^\circ. Find the eccentric anomaly EE. (3)

(c) Using Kepler's equation, find the mean anomaly MM and hence the time elapsed since perigee passage. (4)

(d) Set up (do NOT iterate to convergence) one Newton–Raphson step to solve Kepler's equation for EE given M=1.0M = 1.0 rad, using initial guess E0=ME_0 = M. Report the updated estimate E1E_1. (3)

Question 4 — J2 nodal precession & Sun-synchronous design (12 marks)

The J2J_2 nodal precession rate is Ω˙=−32 J2(R⊕p)2ncos⁡i,\dot\Omega = -\frac{3}{2}\,J_2\left(\frac{R_\oplus}{p}\right)^2 n\cos i, where n=GM/a3n=\sqrt{GM/a^3} and p=a(1−e2)p=a(1-e^2). Take J2=1.0826×10−3J_2 = 1.0826\times10^{-3}.

(a) A satellite is in a circular orbit (e=0e=0) at altitude 700 km. Compute its mean motion nn and its natural period. (3)

(b) For a Sun-synchronous orbit, Ω˙\dot\Omega must equal 360∘360^\circ per year (i.e. +1.991×10−7+1.991\times10^{-7} rad/s). Determine the required inclination ii for the 700 km circular orbit. (6)

(c) Explain physically why Sun-synchronous orbits are retrograde (i>90∘i>90^\circ). (3)

Question 5 — Clohessy–Wiltshire rendezvous (10 marks)

A chaser spacecraft operates in the LVLH frame of a target in a circular orbit of mean motion nn. The in-plane CW equations are x¨−2ny˙−3n2x=0,y¨+2nx˙=0.\ddot x - 2n\dot y - 3n^2 x = 0,\qquad \ddot y + 2n\dot x = 0.

(a) The chaser starts at rest relative to the target at position x0x_0 (radial), y0=0y_0=0. With no thrust, show that the motion is bounded and find the amplitude of the resulting relative oscillation in xx. (5)

(b) Explain, using the secular term in the yy-solution, why a purely radial offset does not drift but an along-track velocity offset does. (5)

Answer keyMark scheme & solutions

Question 1

(a) Vis-viva: v2=GM(2r−1a)v^2 = GM\left(\dfrac{2}{r}-\dfrac{1}{a}\right). v2=51.84v^2 = 51.84, 2GM/r=2(3.986×105)/9000=88.5782GM/r = 2(3.986\times10^5)/9000 = 88.578 km²/s². 1a=1GM(2GMr−v2)=88.578−51.843.986×105=9.217×10−5\dfrac{1}{a} = \dfrac{1}{GM}\left(\dfrac{2GM}{r}-v^2\right) = \dfrac{88.578-51.84}{3.986\times10^5} = 9.217\times10^{-5}. a=10850a = 10850 km (≈ 1.085×1041.085\times10^4 km). (3)

(b) h=r vcos⁡γ=9000×7.2×cos⁡20∘=9000×7.2×0.9397=60,890h = r\,v\cos\gamma = 9000\times7.2\times\cos20^\circ = 9000\times7.2\times0.9397 = 60{,}890 km²/s. (2) p=h2/GM=(6.089×104)2/3.986×105=3.708×109/3.986×105=9302p = h^2/GM = (6.089\times10^4)^2/3.986\times10^5 = 3.708\times10^9/3.986\times10^5 = 9302 km. (2)

(c) p=a(1−e2)⇒e=1−p/a=1−9302/10850=0.14268=0.378p = a(1-e^2)\Rightarrow e = \sqrt{1-p/a} = \sqrt{1-9302/10850} = \sqrt{0.14268} = 0.378. (2) 0<e<10<e<1 ⇒ ellipse. (1)

(d) Orbit equation r=p1+ecos⁡ν⇒cos⁡ν=p/r−1e=9302/9000−10.378=0.033560.378=0.0888r = \dfrac{p}{1+e\cos\nu}\Rightarrow \cos\nu = \dfrac{p/r - 1}{e} = \dfrac{9302/9000 - 1}{0.378} = \dfrac{0.03356}{0.378} = 0.0888. ν=±84.9∘\nu = \pm 84.9^\circ; since γ>0\gamma>0 (moving away from perigee, radius increasing), take ν=+84.9∘\nu = +84.9^\circ. (2)

Question 2

GM=3.986×105GM=3.986\times10^5.

(a) Hohmann. Transfer ellipse: at=(r1+r2)/2=(7000+105000)/2=56000a_t = (r_1+r_2)/2 = (7000+105000)/2 = 56000 km.

  • vc1=GM/r1=3.986×105/7000=56.94=7.546v_{c1} = \sqrt{GM/r_1} = \sqrt{3.986\times10^5/7000} = \sqrt{56.94} = 7.546 km/s.
  • vp=GM(2/r1−1/at)=3.986×105(2/7000−1/56000)v_{p} = \sqrt{GM(2/r_1 - 1/a_t)} = \sqrt{3.986\times10^5(2/7000 - 1/56000)} =3.986×105(2.8571×10−4−1.7857×10−5)=3.986×105×2.6786×10−4=106.77=10.333= \sqrt{3.986\times10^5(2.8571\times10^{-4}-1.7857\times10^{-5})} = \sqrt{3.986\times10^5\times2.6786\times10^{-4}} = \sqrt{106.77}=10.333.
  • Δv1=10.333−7.546=2.787\Delta v_1 = 10.333 - 7.546 = 2.787 km/s.
  • vc2=GM/r2=3.986×105/105000=3.796=1.948v_{c2} = \sqrt{GM/r_2} = \sqrt{3.986\times10^5/105000} = \sqrt{3.796}=1.948 km/s.
  • va=GM(2/r2−1/at)=3.986×105(1.9048×10−5−1.7857×10−5)=3.986×105×1.1905×10−6=0.4745=0.6889v_a = \sqrt{GM(2/r_2 - 1/a_t)} = \sqrt{3.986\times10^5(1.9048\times10^{-5}-1.7857\times10^{-5})} = \sqrt{3.986\times10^5\times1.1905\times10^{-6}} = \sqrt{0.4745}=0.6889.
  • Δv2=1.948−0.6889=1.259\Delta v_2 = 1.948 - 0.6889 = 1.259 km/s.
  • ΔvH=2.787+1.259=4.046 km/s\boxed{\Delta v_{H} = 2.787+1.259 = 4.046\ \text{km/s}} (6)

(b) Bi-elliptic, rb=210000r_b = 210000.

  • Ellipse 1: at1=(r1+rb)/2=(7000+210000)/2=108500a_{t1}=(r_1+r_b)/2 = (7000+210000)/2 = 108500. vp1=GM(2/r1−1/at1)=3.986×105(2.8571×10−4−9.2166×10−6)=3.986×105×2.7649×10−4=110.21=10.498v_{p1}=\sqrt{GM(2/r_1-1/a_{t1})}=\sqrt{3.986\times10^5(2.8571\times10^{-4}-9.2166\times10^{-6})}=\sqrt{3.986\times10^5\times2.7649\times10^{-4}}=\sqrt{110.21}=10.498. Δv1=10.498−7.546=2.952\Delta v_1 = 10.498-7.546 = 2.952.
  • At rbr_b on ellipse 1: va1=GM(2/rb−1/at1)=3.986×105(9.5238×10−6−9.2166×10−6)=3.986×105×3.072×10−7=0.12246=0.3499v_{a1}=\sqrt{GM(2/r_b-1/a_{t1})}=\sqrt{3.986\times10^5(9.5238\times10^{-6}-9.2166\times10^{-6})}=\sqrt{3.986\times10^5\times3.072\times10^{-7}}=\sqrt{0.12246}=0.3499.
  • Ellipse 2: at2=(r2+rb)/2=(105000+210000)/2=157500a_{t2}=(r_2+r_b)/2=(105000+210000)/2=157500. va2=GM(2/rb−1/at2)=3.986×105(9.5238×10−6−6.3492×10−6)=3.986×105×3.1746×10−6=1.2654=1.1250v_{a2}=\sqrt{GM(2/r_b-1/a_{t2})}=\sqrt{3.986\times10^5(9.5238\times10^{-6}-6.3492\times10^{-6})}=\sqrt{3.986\times10^5\times3.1746\times10^{-6}}=\sqrt{1.2654}=1.1250. Δv2=1.1250−0.3499=0.7751\Delta v_2 = 1.1250 - 0.3499 = 0.7751 (burn at apoapsis raises periapsis to r2r_2).
  • At r2r_2 on ellipse 2: vp2=GM(2/r2−1/at2)=3.986×105(1.9048×10−5−6.3492×10−6)=3.986×105×1.2698×10−5=5.062=2.2499v_{p2}=\sqrt{GM(2/r_2-1/a_{t2})}=\sqrt{3.986\times10^5(1.9048\times10^{-5}-6.3492\times10^{-6})}=\sqrt{3.986\times10^5\times1.2698\times10^{-5}}=\sqrt{5.062}=2.2499. Δv3=2.2499−1.948=0.3019\Delta v_3 = 2.2499 - 1.948 = 0.3019 (deceleration to circularize).
  • ΔvBE=2.952+0.7751+0.3019=4.029 km/s\boxed{\Delta v_{BE}=2.952+0.7751+0.3019=4.029\ \text{km/s}} (6)

(c) Bi-elliptic (4.0294.029) is marginally cheaper than Hohmann (4.0464.046) for this ratio r2/r1=15r_2/r_1=15 (bi-elliptic wins when the ratio ≳11.94\gtrsim 11.94). Physical reason: raising apoapsis very high means the periapsis-raising burn occurs where orbital speed is small (Oberth effect is weak there), so the plane/energy change costs less; the extra energy is recovered efficiently. (2)

Question 3

(a) T=2πa3/GM=2π(2×104)3/3.986×105=2π8×1012/3.986×105=2π2.0070×107=2π(4480.5)=28,152T = 2\pi\sqrt{a^3/GM} = 2\pi\sqrt{(2\times10^4)^3/3.986\times10^5}=2\pi\sqrt{8\times10^{12}/3.986\times10^5}=2\pi\sqrt{2.0070\times10^7}=2\pi(4480.5)=28{,}152 s ≈ 7.82 h. (2)

(b) tan⁡E2=1−e1+etan⁡ν2\tan\dfrac{E}{2}=\sqrt{\dfrac{1-e}{1+e}}\tan\dfrac{\nu}{2}. (1−0.35)/(1+0.35)=0.65/1.35=0.4815=0.6939\sqrt{(1-0.35)/(1+0.35)}=\sqrt{0.65/1.35}=\sqrt{0.4815}=0.6939. tan⁡(50∘)=1.19175\tan(50^\circ)=1.19175; product =0.8270=0.8270; E/2=arctan⁡(0.8270)=39.58∘E/2 = \arctan(0.8270)=39.58^\circ; E=79.16∘=1.3816E=79.16^\circ = 1.3816 rad. (3)

(c) Kepler: M=E−esin⁡E=1.3816−0.35sin⁡(1.3816)=1.3816−0.35(0.98218)=1.3816−0.3438=1.0378M = E - e\sin E = 1.3816 - 0.35\sin(1.3816) = 1.3816 - 0.35(0.98218)=1.3816-0.3438=1.0378 rad. (2) t=M/nt = M/n, n=2π/T=2π/28152=2.2318×10−4n = 2\pi/T = 2\pi/28152 = 2.2318\times10^{-4} rad/s. t=1.0378/2.2318×10−4=4650t = 1.0378/2.2318\times10^{-4}=4650 s (≈ 1.29 h since perigee). (2)

(d) f(E)=E−esin⁡E−Mf(E)=E-e\sin E - M, f′(E)=1−ecos⁡Ef'(E)=1-e\cos E. With M=1.0M=1.0, e=0.35e=0.35, E0=1.0E_0=1.0: f(E0)=1.0−0.35sin⁡1.0−1.0=−0.35(0.84147)=−0.29452f(E_0)=1.0-0.35\sin1.0-1.0=-0.35(0.84147)=-0.29452. f′(E0)=1−0.35cos⁡1.0=1−0.35(0.54030)=0.81090f'(E_0)=1-0.35\cos1.0=1-0.35(0.54030)=0.81090. E1=E0−f/f′=1.0−(−0.29452/0.81090)=1.0+0.36321=1.3632E_1 = E_0 - f/f' = 1.0 -(-0.29452/0.81090)=1.0+0.36321=1.3632 rad. (3)

Question 4

(a) a=R⊕+700=7078a = R_\oplus + 700 = 7078 km. n=GM/a3=3.986×105/(7078)3=3.986×105/3.5457×1011=1.1242×10−6=1.0603×10−3n=\sqrt{GM/a^3}=\sqrt{3.986\times10^5/(7078)^3}=\sqrt{3.986\times10^5/3.5457\times10^{11}}=\sqrt{1.1242\times10^{-6}}=1.0603\times10^{-3} rad/s. (2) T=2π/n=5926T=2\pi/n = 5926 s ≈ 98.8 min. (1)

(b) For e=0e=0, p=a=7078p=a=7078 km. Required Ω˙=+1.991×10−7\dot\Omega = +1.991\times10^{-7} rad/s. cos⁡i=−Ω˙32J2(R⊕/a)2n.\cos i = -\frac{\dot\Omega}{\tfrac32 J_2 (R_\oplus/a)^2 n}. Compute denominator: 32J2=1.6239×10−3\tfrac32 J_2 = 1.6239\times10^{-3}; (R⊕/a)2=(6378/7078)2=(0.90110)2=0.81198(R_\oplus/a)^2=(6378/7078)^2=(0.90110)^2=0.81198; D=1.6239×10−3×0.81198×1.0603×10−3=1.3981×10−6D = 1.6239\times10^{-3}\times0.81198\times1.0603\times10^{-3}=1.3981\times10^{-6}. (3) cos⁡i=−1.991×10−7/1.3981×10−6=−0.14241\cos i = -1.991\times10^{-7}/1.3981\times10^{-6}=-0.14241. i=arccos⁡(−0.14241)=98.19∘i = \arccos(-0.14241)=98.19^\circ. (3)

(c) The natural (uncontrolled) precession from Earth's oblateness is retrograde for prograde orbits (cos⁡i>0\cos i>0 gives Ω˙<0\dot\Omega<0). To make the node advance eastward to follow the Sun (+360∘360^\circ/yr), we need Ω˙>0\dot\Omega>0, which requires cos⁡i<0\cos i<0, i.e. i>90∘i>90^\circ (retrograde). The equatorial bulge torque then precesses the node in the required direction. (3)

Question 5

(a) With x(0)=x0x(0)=x_0, y(0)=0y(0)=0, x˙(0)=y˙(0)=0\dot x(0)=\dot y(0)=0. Standard CW solution: x(t)=(4−3cos⁡nt)x0+x˙0nsin⁡nt+2y˙0n(1−cos⁡nt).x(t)=(4-3\cos nt)x_0 + \frac{\dot x_0}{n}\sin nt + \frac{2\dot y_0}{n}(1-\cos nt). With x˙0=y˙0=0\dot x_0=\dot y_0=0: x(t)=x0(4−3cos⁡nt)x(t)=x_0(4-3\cos nt), which oscillates between x0x_0 (at cos⁡=1\cos=1) and x0(4−3(−1))=7x0x_0(4-3(-1))=7x_0...

Correcting: x(t)=x0(4−3cos⁡nt)x(t)=x_0(4-3\cos nt) ranges from xmin⁡=x0x_{\min}=x_0 to xmax⁡=7x0x_{\max}=7x_0; mean =4x0=4x_0, amplitude =3x0=3x_0 about the drifted center 4x04x_0. Motion is bounded (no secular term because the drift condition y˙0=−2nx0\dot y_0 = -2n x_0 is... here y˙0=0≠−2nx0\dot y_0=0\ne -2nx_0, so along-track drifts — see (b)). The radial motion itself is bounded with amplitude 3∣x0∣3|x_0|. (5)

(b) Along-track solution: y(t)=y0+(2x˙0n)(...)−(3y˙0t+6nx0t)  ⇒  y(t)∋−(6nx0+3y˙0) t.y(t)=y_0 + \left(\frac{2\dot x_0}{n}\right)(... ) - \left(3\dot y_0 t + 6n x_0 t\right)\frac{}{} \;\Rightarrow\; y(t)\ni -(6n x_0 + 3\dot y_0)\,t. The secular (linearly growing) term is −(6nx0+3y˙0)t-(6n x_0 + 3\dot y_0)t.

  • A purely radial offset with matched conditions vanishes only if 6nx0+3y˙0=06nx_0+3\dot y_0=0; but the bounded/no-drift condition on the periodic part actually comes from the xx-drift term −2y˙0n-\tfrac{2\dot y_0}{n}... The key point: the secular drift coefficient is −(6nx0+3y˙0)-(6nx_0+3\dot y_0). An initial along-track velocity offset y˙0\dot y_0 directly feeds this secular term, producing unbounded along-track drift, whereas a purely radial position offset produces bounded ep