Orbital Mechanics & Astrodynamics

Q1. (3 marks)

• 1st law: Every planet moves in an ellipse with the Sun at one focus (orbits are conic sections). (1)
• 2nd law: The line joining a planet to the Sun sweeps equal areas in equal times. (1)
• 3rd law: The square of the orbital period is proportional to the cube of the semi-major axis, $T^2\propto a^3$. (1)

Q2. (6 marks) — ½ mark definition + ½ mark meaning each.

• $a$ (semi-major axis): sets orbit size/energy.
• $e$ (eccentricity): sets orbit shape (circle→hyperbola).
• $i$ (inclination): tilt of orbit plane vs reference (equatorial) plane.
• $\Omega$ (RAAN): angle from reference direction to ascending node — orients the plane.
• $\omega$ (argument of perigee): angle from ascending node to perigee — orients ellipse in plane.
• $\nu$ (true anomaly): angle from perigee to the body — locates it on the orbit at epoch.

Q3. (4 marks) — 1 mark each.

• $e=0$: circle, $\varepsilon<0$ (bound).
• $e=0.4$: ellipse, $\varepsilon<0$ (bound).
• $e=1$: parabola, $\varepsilon=0$ (escape/marginal).
• $e=1.5$: hyperbola, $\varepsilon>0$ (unbound).

Q4. (5 marks) (a) $v=\sqrt{GM/r}=\sqrt{3.986\times10^5/7000}=\sqrt{56.94}=7.546\ \mathrm{km/s}$. (2) (b) $T=2\pi\sqrt{r^3/GM}=2\pi\sqrt{(7000)^3/3.986\times10^5}=2\pi\sqrt{8.606\times10^5}=2\pi(927.7)=5829\ \mathrm{s}=97.2\ \mathrm{min}$. (2) (c) $\varepsilon=-GM/2r=-3.986\times10^5/14000=-28.47\ \mathrm{km^2/s^2}$. (1)

Q5. (4 marks)

• Semi-major axis: $a=(r_p+r_a)/2=(7000+42000)/2=24500\ \mathrm{km}$. (1)
• Vis-viva: $v^2=GM(2/r-1/a)$. (1)
• At perigee: $v_p^2=3.986\times10^5(2/7000-1/24500)=3.986\times10^5(2.857\times10^{-4}-4.082\times10^{-5})=3.986\times10^5(2.449\times10^{-4})=97.63$. (1)
• $v_p=9.88\ \mathrm{km/s}$. (1)

Q6. (4 marks)

• Angular momentum per unit mass $h=r^2\dot\theta=$ const (central force → torque zero). (1)
• Area swept by radius vector: $dA=\tfrac12 r\,(r\,d\theta)=\tfrac12 r^2\,d\theta$. (1)
• Therefore $\dfrac{dA}{dt}=\tfrac12 r^2\dot\theta=\tfrac12 h$. (1)
• Since $h$ is constant, $dA/dt$ is constant → equal areas in equal times (Kepler's 2nd law). (1)

Q7. (6 marks) (a) $a_t=(r_1+r_2)/2=(7000+42000)/2=24500\ \mathrm{km}$. (1) Circular speeds: $v_{c1}=\sqrt{GM/r_1}=7.546\ \mathrm{km/s}$; $v_{c2}=\sqrt{GM/r_2}=\sqrt{3.986\times10^5/42000}=3.081\ \mathrm{km/s}$. (b) Perigee speed of transfer: $v_p=\sqrt{GM(2/r_1-1/a_t)}=9.882\ \mathrm{km/s}$. $\Delta v_1=v_p-v_{c1}=9.882-7.546=2.336\ \mathrm{km/s}$. (2) (c) Apogee speed: $v_a=\sqrt{GM(2/r_2-1/a_t)}=\sqrt{3.986\times10^5(4.762\times10^{-5}-4.082\times10^{-5})}=\sqrt{3.986\times10^5(6.803\times10^{-6})}=\sqrt{2.712}=1.647\ \mathrm{km/s}$. $\Delta v_2=v_{c2}-v_a=3.081-1.647=1.434\ \mathrm{km/s}$. (2) (d) $\Delta v_{tot}=2.336+1.434=3.77\ \mathrm{km/s}$. (1)

Q8. (4 marks)

• Kepler's equation: $M=E-e\sin E$. (1)
• Newton–Raphson: $E_{n+1}=E_n-\dfrac{E_n-e\sin E_n-M}{1-e\cos E_n}$. (1)
• With $E_0=0.5$, $e=0.2$, $M=0.5$: $f=0.5-0.2\sin(0.5)-0.5=-0.2(0.4794)=-0.09589$. $f'=1-0.2\cos(0.5)=1-0.2(0.8776)=0.8245$. $E_1=0.5-(-0.09589/0.8245)=0.5+0.1163=0.6163$ rad. (2)

Q9. (3 marks)

• Vis-viva: $v^2=GM\left(\dfrac{2}{r}-\dfrac{1}{a}\right)$. (1)
• Specific energy: $\varepsilon=-\dfrac{GM}{2a}$. (1)
• $\varepsilon<0$ means total energy is negative → kinetic energy insufficient to reach infinity; the body is gravitationally bound (finite, closed orbit, $a>0$). (1)

[
  {"claim":"Circular orbit speed at r=7000 km ≈ 7.546 km/s","code":"GM=3.986e5; r=7000; v=sqrt(GM/r); result = abs(float(v)-7.546)<0.01"},
  {"claim":"Circular orbit period at r=7000 km ≈ 5829 s","code":"import sympy as sp; GM=3.986e5; r=7000; T=2*sp.pi*sqrt(r**3/GM); result = abs(float(T)-5829)<5"},
  {"claim":"Hohmann total dv (7000->42000 km) ≈ 3.77 km/s","code":"GM=3.986e5; r1=7000.0; r2=42000.0; a=(r1+r2)/2; vc1=sqrt(GM/r1); vc2=sqrt(GM/r2); vp=sqrt(GM*(2/r1-1/a)); va=sqrt(GM*(2/r2-1/a)); dv=(vp-vc1)+(vc2-va); result = abs(float(dv)-3.77)<0.02"},
  {"claim":"One Newton-Raphson step for Kepler eqn (e=0.2, M=0.5, E0=0.5) gives E≈0.6163","code":"e=0.2; M=0.5; E0=0.5; f=E0-e*sin(E0)-M; fp=1-e*cos(E0); E1=E0-f/fp; result = abs(float(E1)-0.6163)<0.001"}
]