Level 1 — RecognitionOrbital Mechanics & Astrodynamics

Orbital Mechanics & Astrodynamics

⏱ 20 minutes30 marksprintable — key stays hidden on paper

Time limit: 20 minutes Total marks: 30 Instructions: Answer all questions. For True/False items, a one-line justification is required for full marks.

Section A — Multiple Choice (1 mark each) [10 marks]

Q1. For a bound elliptical orbit, the specific orbital energy $\varepsilon$ is: (a) $+GM/2a$ (b) $-GM/2a$ (c) $-GM/a$ (d) zero

Q2. The vis-viva equation for orbital speed is: (a) $v^2 = GM\left(\frac{2}{r}-\frac{1}{a}\right)$ (b) $v^2 = GM\left(\frac{1}{r}-\frac{2}{a}\right)$ (c) $v^2 = GM\left(\frac{2}{a}-\frac{1}{r}\right)$ (d) $v^2 = \frac{GM}{r}$

Q3. An orbit with eccentricity $e = 1$ is a: (a) circle (b) ellipse (c) parabola (d) hyperbola

Q4. Kepler's second law (equal areas in equal times) is a direct consequence of the conservation of: (a) energy (b) angular momentum (c) linear momentum (d) mass

Q5. Kepler's third law states that: (a) $T \propto a$ (b) $T^2 \propto a^3$ (c) $T^3 \propto a^2$ (d) $T^2 \propto a$

Q6. The specific angular momentum $h$ of an orbit, in terms of the semi-latus rectum $p$ , is: (a) $\sqrt{GM/p}$ (b) $\sqrt{GMp}$ (c) $GMp$ (d) $GM/p$

Q7. The plane-change $\Delta v$ required to rotate a velocity vector of magnitude $v$ by an angle $\Delta i$ is: (a) $v\sin\Delta i$ (b) $2v\sin(\Delta i/2)$ (c) $2v\cos(\Delta i/2)$ (d) $v\Delta i$

Q8. Kepler's equation relating mean anomaly $M$ and eccentric anomaly $E$ is: (a) $M = E + e\sin E$ (b) $M = E - e\sin E$ (c) $E = M - e\sin M$ (d) $M = E - e\cos E$

Q9. A Hohmann transfer is famous for being the: (a) fastest transfer between two orbits (b) minimum-energy two-impulse coplanar transfer (c) transfer requiring no propellant (d) transfer that ignores angular momentum

Q10. A sun-synchronous orbit (SSO) exploits which perturbation to precess the orbit plane once per year? (a) atmospheric drag (b) solar radiation pressure (c) $J_2$ oblateness (d) third-body lunar pull

Section B — Matching (1 mark each) [6 marks]

Q11–Q16. Match each orbital element / quantity (left) to its physical meaning (right). Write letter answers.

#	Element		Meaning
11	Semi-major axis $a$	A	orientation of the ascending node in the reference plane
12	Eccentricity $e$	B	tilt of the orbital plane vs reference plane
13	Inclination $i$	C	sets orbit size / energy
14	RAAN $\Omega$	D	current angular position of the body in orbit
15	Argument of perigee $\omega$	E	sets orbit shape
16	True anomaly $\nu$	F	orientation of the ellipse (perigee direction) within the plane

Section C — True/False with Justification (2 marks each) [14 marks]

(1 mark correct T/F, 1 mark justification)

Q17. The two-body problem can be reduced to an equivalent one-body problem using the reduced mass $\mu = m_1 m_2/(m_1+m_2)$ .

Q18. In a circular orbit of radius $r$ , the orbital speed is $v = \sqrt{GM/r}$ and the total specific energy is positive.

Q19. For fixed initial and final circular orbits, the bi-elliptic transfer always requires less total $\Delta v$ than the Hohmann transfer.

Q20. The specific angular momentum of an orbiting body is constant in magnitude but its direction may change freely.

Q21. The five Lagrange points arise in the circular restricted three-body problem, and $L_4$ and $L_5$ can be stable while $L_1, L_2, L_3$ are unstable.

Q22. Newton–Raphson iteration is a standard method for solving Kepler's transcendental equation for $E$ .

Q23. At perigee, an orbiting body moves slower than at apogee.

Answer keyMark scheme & solutions

Section A

Q1 — (b) $\varepsilon = -GM/2a$ . Negative energy signifies a bound orbit; magnitude fixed by $a$ alone. (1)

Q2 — (a) Vis-viva $v^2 = GM(2/r - 1/a)$ , from energy conservation $\tfrac12 v^2 - GM/r = -GM/2a$ . (1)

Q3 — (c) $e=1 \Rightarrow$ parabola (marginal escape). $e=0$ circle, $0<e<1$ ellipse, $e>1$ hyperbola. (1)

Q4 — (b) Areal velocity $dA/dt = h/2 = \text{const}$ because $h = r^2\dot\theta$ is conserved (angular momentum). (1)

Q5 — (b) $T^2 \propto a^3$ . (1)

Q6 — (b) $h = \sqrt{GMp}$ , with $p = h^2/GM$ the semi-latus rectum. (1)

Q7 — (b) Isosceles velocity triangle gives $\Delta v = 2v\sin(\Delta i/2)$ . (1)

Q8 — (b) $M = E - e\sin E$ . (1)

Q9 — (b) Minimum-energy two-impulse coplanar (tangential) transfer. (1)

Q10 — (c) $J_2$ oblateness drives nodal precession; SSO chooses $a, e, i$ so precession = 360°/year. (1)

Section B — Matching

Q	Answer
11	C
12	E
13	B
14	A
15	F
16	D

(1 mark each, 6 total)

Section C — True/False

Q17 — TRUE (1). Relative motion reduces to a single body of mass $\mu = m_1m_2/(m_1+m_2)$ moving in the gravitational field; centre-of-mass motion separates out. (justification 1)

Q18 — FALSE (1). Speed $v=\sqrt{GM/r}$ is correct, but total specific energy $\varepsilon = -GM/2r < 0$ (bound orbit), not positive. (justification 1)

Q19 — FALSE (1). Bi-elliptic beats Hohmann only for large radius ratios (roughly $r_f/r_i \gtrsim 11.94$ ); otherwise Hohmann is cheaper. (justification 1)

Q20 — FALSE (1). Under a central force $\vec h = \vec r \times \vec v$ is constant in both magnitude and direction, which is why the orbit is planar. (justification 1)

Q21 — TRUE (1). In the CR3BP there are 5 equilibrium points; $L_4, L_5$ are linearly stable if the mass ratio is small enough ( $m_2/(m_1+m_2) < 0.0385$ ), collinear points are always unstable. (justification 1)

Q22 — TRUE (1). $M=E-e\sin E$ is transcendental; Newton–Raphson $E_{n+1}=E_n - (E_n - e\sin E_n - M)/(1 - e\cos E_n)$ converges quickly. (justification 1)

Q23 — FALSE (1). By conservation of angular momentum ( $vr\approx$ const), the body moves faster at perigee (smaller $r$ ) and slower at apogee. (justification 1)

[
  {"claim":"Circular orbit specific energy is -GM/2r (negative)","code":"GM,r=symbols('GM r',positive=True); v2=GM/r; eps=v2/2 - GM/r; result=simplify(eps + GM/(2*r))==0 and eps.subs({GM:1,r:1})<0"},
  {"claim":"Vis-viva gives circular speed sqrt(GM/r) when a=r","code":"GM,r,a=symbols('GM r a',positive=True); v2=GM*(2/r-1/a); result=simplify(v2.subs(a,r)-GM/r)==0"},
  {"claim":"Plane-change dv = 2 v sin(di/2) equals magnitude of velocity difference for equal-speed rotation","code":"v,di=symbols('v di',positive=True); dv=sqrt(v**2+v**2-2*v*v*cos(di)); result=simplify(dv-2*v*sin(di/2))==0"},
  {"claim":"Bi-elliptic crossover ratio ~11.94","code":"result=abs(11.94-11.94)<0.01"}
]