1  SFM 2025 End Semester Exam Questions-Answers

1. We resume our review of the course from where we stopped. After studying two-dimensional planetary motion, we extended the analysis to three dimensions. In doing so, we introduced three additional orbital elements \(i\), \(\Omega\), and \(\omega\), called (a) - 0.5 marks ——, (b) - 0.5 marks ——, and (c) - 0.5 marks ——, respectively. A satellite of Earth with (d) - 0.5 marks —— (condition written in terms of the orbital elements) has retrograde motion.
   
   We then studied orbital maneuvers, focusing on impulsive maneuvers rather than continuous-thrust models to simplify the analysis. This was justified by the argument that (e) - 2 marks ——.
   
   We showed that (f) - 1 mark —— (number) transfer orbits exist for transferring a satellite from one elliptical orbit to another. An external criteria was used to choose a particular orbit. The most important criteria we found to be minimization of fuel consumption. This is equivalent to minimizing the total \(\Delta v\) of the transfer. We found that it is always advantageous to perform impulsive maneuvers at the perigee of the first orbit. A heuristic argument in support of this statement is: (g) - 2 marks ——.
   
   We also observed that the difference in total mechanical energy \(\left(\tfrac{1}{2}v^2 - \tfrac{\mu}{r}\right)\) between the initial and final orbits does not, by itself, determine the \(\Delta v\) required for transfer. An example of two planar orbits with the same total mechanical energy but a nonzero \(\Delta v\) required for transfer is: (h) - 1 mark ——.
   
   Next, we explored rendezvous and chase maneuvers between two satellites. We consider a fixed transfer time maneuver. Let the initial state vectors of the chaser and target satellites be \((\mathbf{r}_c,\mathbf{v}_c)\) and \((\mathbf{r}_t,\mathbf{v}_t)\), respectively. We first determine the position of the target at the rendezvous time. The target state at rendezvous, \((\mathbf{r}^+_t,\mathbf{v}^+_t)\), is obtained by solving (j) - 1 mark —— equations (name of the equations is expected) for the elliptical and parabolic cases. The chaser state at rendezvous, \((\mathbf{r}^+_c,\mathbf{v}^+_c)\), is then found by noting that (k) - 1 mark —— for a rendezvous, while (l) - 1 mark —— for a chase maneuver. Once the terminal states of both target and chaser are known, we solve (m) - 1 mark —— to determine the transfer orbit.
   
   We also studied orbit transfers in 3D. To change the inclination of a satellite by \(45^\circ\) (while keeping all other orbital parameters fixed), we apply a \(\Delta v\) of (i) - 1 mark —— at the ascending node.
   
   We then considered interplanetary transfers. To avoid the full multi-body problem, we introduced the concept of the sphere of influence (SOI) for each body. These two ideas require further clarification.
   
   As an example, consider a spacecraft moving under the gravitational attraction of both Earth and Sun. Using Newton’s law of gravitation, we can write the equations of motion of the spacecraft as (n) - 2 marks ——. These equations can be integrated numerically (e.g. by RK4) to obtain the trajectory for given initial conditions. With modern computational resources (unlike during the Apollo era), solving these equations is quite feasible. Nevertheless, we still prefer to work with the two-body problem instead of the full multi-body problem because (o) - 1 mark ——.
   
   Next, we examined the sphere of influence more carefully. The gravitational force due to the Sun is smaller than that due to Earth for any point with \(r < r_{\text{SOI}}\), where \(r_{\text{SOI}}\) is the radius of Earth’s sphere of influence. (p) - 1 mark True / False ——.
   
   Finally, we returned to Earth-bound ballistic trajectories. We assumed (q) - 1 mark —— to make the ballistic trajectories symmetric. In ballistic trajectory problems, we are typically given the launch site, the target site, and the missile velocity at burnout altitude, and are asked to determine (r) - 1 mark ——.
   
   This completes our brief tour of the course. To conclude, recall that our model of orbital mechanics is a simplified representation of reality. List any four ways in which this model must be modified to better match the real world: (s) - 2 marks ——.

Solution

• 1. Inclination
• 2. Longitude of ascending node
• 3. Argument of perigee
• 4. \(\pi/2 < i < \pi\)
• 5. Because the duration of application of propulsive forces is much smaller than the overall trajectory
• 6. infinite
• 7. The KE of orbit is \(\frac{v^2}{2}\) and the \(\Delta KE = v\Delta v\). The change in energy is highest when the \(\Delta v\) is applied when the velocity vector is collinear with the velocity vector. Also, the the change is proportional to the magnitude of \(v\). Hence, it should be applied at the highest velocity, that is, the periapse point.
• 8. There are infinite such orbits. One example are two elliptical orbits with same \((a, e)\) but different apse lines.
• 1. \(\Delta v = 2\frac{v}{\sin \frac{45}{2}} = \frac{2v}{\sin 22.5} \approx 0.765 v\)
• 10. Kepler’s equation for Elliptical orbits and Barker’s equation for parabolic orbits.
• 11. \(r_c^+=r_t^+\), \(v_c^+=v_t^+\) (Rendezvous)
• 12. \(r_c^+=r_t^+\), \(v_c^+\) is free (Chase)
• 13. Lambert’s problem

• 14. \(\ddot{\mathbf{r}} = -\mu_e \frac{\mathbf{r}-\mathbf{r}_e}{\vert\vert\mathbf{r}-\mathbf{r}_e\vert\vert^3} - \mu_s\frac{\mathbf{r}-\mathbf{r}_s}{\vert\vert\mathbf{r}-\mathbf{r}_s\vert\vert^3}\)
• 15. System is extremely sensitive to the initial conditions, i.e. it is chaotic.
• 16. False.

Here is a python code to calculate the acceleration due to Sun and Earth at a point in the \(r_{SOI}\) of Earth.

import math

# -----------------------------
# 1. Physical constants
# -----------------------------

# Gravitational parameters (m^3/s^2)
mu_earth = 3.986004418e14      # μ⊕
mu_sun   = 1.32712440018e20    # μ☉

# Earth's orbital semimajor axis (1 AU) in meters
AU = 1.495978707e11            # a

# -----------------------------
# 2. Earth sphere of influence
# -----------------------------
# Laplace SOI radius:
# r_SOI = a * (μ_earth / μ_sun)^(2/5)

r_SOI = AU * (mu_earth / mu_sun)**(2.0 / 5.0)

print(f"Earth sphere of influence radius:")
print(f"  r_SOI = {r_SOI:.3e} m  ≈ {r_SOI/1000:.1f} km\n")

# -----------------------------
# 3. Choose a point inside r_SOI
# -----------------------------
# Pick a radius as a fraction of r_SOI.
# Any fraction between 0 and 1 is allowed.
# We'll pick 0.5 (i.e., r = 0.5 r_SOI), where a_sun > a_earth.

fraction = 0.5
r = fraction * r_SOI   # distance from Earth's center (m)

print(f"Point considered:")
print(f"  r = {fraction:.3f} * r_SOI = {r:.3e} m ≈ {r/1000:.1f} km from Earth's center\n")

# -----------------------------
# 4. Distances to Sun and accelerations
# -----------------------------
# Assume the point is on the line between Earth and Sun,
# on the Sun-facing side: distance to Sun = AU - r.
# (For r << AU this difference is small, but we include it for correctness.)

R_sun = AU - r  # distance from point to Sun (m)

# Accelerations
# a_earth = μ_earth / r^2
# a_sun   = μ_sun   / R_sun^2

a_earth = mu_earth / r**2
a_sun   = mu_sun   / R_sun**2

print("Gravitational accelerations at that point:")
print(f"  a_earth = {a_earth:.6e} m/s^2")
print(f"  a_sun   = {a_sun:.6e} m/s^2")
print(f"  a_sun / a_earth = {a_sun / a_earth:.3f}\n")

Earth sphere of influence radius:
  r_SOI = 9.246e+08 m  ≈ 924646.8 km

Point considered:
  r = 0.500 * r_SOI = 4.623e+08 m ≈ 462323.4 km from Earth's center

Gravitational accelerations at that point:
  a_earth = 1.864859e-03 m/s^2
  a_sun   = 5.966907e-03 m/s^2
  a_sun / a_earth = 3.200

• 17. \(h_{bo} = h_{re}\) where \(h_{bo}\) is the height of burnout and \(h_{re}\) is the height of re-entry.
• 18. \(\phi_{bo}\) where \(\phi_{bo}\) is the flight path angle at burnout.
• 1. Gravitational perturbations of non-spherical celestial bodies
  2. Three-body effects
  3. Atmospheric drag
  4. Solar radiation pressure
  5. Impulsive approximation of propulsive forces
  6. Move from point mass approximation of satellites to 3D objects thereby introduction attitude dynamics
  7. Remove assumption of \(i=0\) for all the planets
  8. Remove assumption of \(e=0\) for inner planets
  9. Introduce gravitational perturbations of the moons of the planets
  10. Include relativistic effects for Mercury and Venus

2. Draw the diagram of a hyperbolic orbit. Show, the asymptote, apse line, true anomaly (\(\theta\)) aiming radius (\(\Delta\)) and turn angle (\(\delta\)). Also derive the expression for aiming radius in terms of \(a\) and \(e\) from geometry.

Solution

Derivation:

We want the aiming radius (impact parameter) \(\Delta\) in terms of hyperbolic elements \(a, e\).

For a hyperbolic orbit:

• Semimajor axis \(a<0\) (or we use \(|a|\) as its magnitude),
• Eccentricity \(e>1\),
• Specific energy \[ \varepsilon = \frac{v_\infty^2}{2} = -\frac{\mu}{2a}, \] hence \(v_\infty^2 = \dfrac{\mu}{|a|}\) if \((|a| = -a)\).

Let \(b\) be the aiming radius / impact parameter: the perpendicular distance between the asymptotic incoming velocity vector and the center (in the far field). Then:

• Specific angular momentum \[ h = b.v_\infty. \]

Also from the conic relations:

\[ e = \sqrt{1 + \frac{2\varepsilon h^2}{\mu^2}}. \]

Substitute \(\varepsilon = v_\infty^2/2\) and \(h = b v_\infty\):

\[ \begin{aligned} e^2 &= 1 + \frac{2\left(\tfrac{v_\infty^2}{2}\right)(b^2 v_\infty^2)}{\mu^2} = 1 + \frac{b^2 v_\infty^4}{\mu^2}. \end{aligned} \]

But \(v_\infty^2 = \mu/|a|\), hence

\[ v_\infty^4 = \frac{\mu^2}{|a|^2}. \]

So

\[ e^2 = 1 + \frac{b^2}{|a|^2} \quad\Rightarrow\quad b^2 = |a|^2\left(e^2 - 1\right). \]

Thus the aiming radius is

\[ \boxed{\Delta = b = |a|.\sqrt{e^2 - 1}}. \]

If you use a convention where \(a>0\) is taken as the magnitude of the semimajor axis even for hyperbola, you can simply write

\[ \boxed{\Delta = a.\sqrt{e^2 - 1}}. \]

The diagram should show:

• focus at the attracting body,
• hyperbolic branch,
• line of apsides (pericentre direction),
• asymptote lines meeting at angle \(\delta\) (turn angle),
• true anomaly \(f\) from periapsis to current position,
• aiming radius \(\Delta\) as the perpendicular distance from the asymptotic line to the focus.

3. Calculate all the six orbital parameters of the satellite given the state vector \(\mathbf{r} = -6045 \mathbf{\hat{i}} - 3490 \mathbf{\hat{j}} + 2500 \mathbf{\hat{k}}\) km and \(\mathbf{v} = -3.457 \mathbf{\hat{i}} + 6.618 \mathbf{\hat{j}} + 2.533 \mathbf{\hat{k}}\) km/s.

Solution

Given (all in km, km/s):

\[ \mathbf r = (-6045,,-3490,;2500), \quad \mathbf v = (-3.457,;6.618,;2.533), \] and Earth’s (= 398600 ^3/2).

We’ll compute the classical orbital elements (\(a, e, i, \Omega, \omega, f\)).

1. Magnitudes

\[ r = |\mathbf r| = \sqrt{(-6045)^2 + (-3490)^2 + 2500^2} \approx 7414.32\ \text{km}. \]

\[ v = |\mathbf v| = \sqrt{(-3.457)^2 + 6.618^2 + 2.533^2} \approx 7.88447\ \text{km/s}. \]

2. Specific angular momentum

\[ \mathbf h = \mathbf r \times \mathbf v = \begin{vmatrix} \hat{\mathbf i} & \hat{\mathbf j} & \hat{\mathbf k}\ -6045 & -3490 & 2500\ -3.457 & 6.618 & 2.533 \end{vmatrix} \]

Compute components:

• \(h_x = (-3490)(2.533) - (2500)(6.618) \approx -25385.17\),
• \(h_y = (2500)(-3.457) - (-6045)(2.533) \approx 6669.485\),
• \(h_z = (-6045)(6.618) - (-3490)(-3.457) \approx -52070.74\).

So

\[ \mathbf h \approx (-25385.17,\ 6669.49,\ -52070.74)\ \text{km}^2/\text{s}, \] \[ h = |\mathbf h| \approx 58311.67\ \text{km}^2/\text{s}. \]

3. Eccentricity vector

\[ \mathbf e = \frac{\mathbf v \times \mathbf h}{\mu} - \frac{\mathbf r}{r}. \]

Compute \(\mathbf v \times \mathbf h\) numerically:

\[ \mathbf e \approx (-0.0916,\ -0.1422,\ 0.0264), \]

so

\[ e = |\mathbf e| \approx 0.1712. \]

This is a moderately eccentric ellipse.

4. Specific mechanical energy and semimajor axis

Specific energy:

\[ \varepsilon = \frac{v^2}{2} - \frac{\mu}{r} = \frac{(7.88447)^2}{2} - \frac{398600}{7414.32} \approx 31.084 - 53.758 \approx -22.674\ \text{km}^2/\text{s}^2. \]

Semimajor axis:

\[ a = -\frac{\mu}{2\varepsilon} = -\frac{398600}{2(-22.674)} \approx 8788.10\ \text{km}. \]

5. Inclination

Let \(\hat{\mathbf k} = (0,0,1)\). Then

\[ \cos i = \frac{h_z}{h} = \frac{-52070.74}{58311.67} \approx -0.8916, \] \[ i = \arccos(-0.8916) \approx 153.25^\circ. \]

So the orbit is retrograde (inclination > \(90^\circ\)).

6. Node vector and RAAN

Node vector:

\[ \mathbf n = \hat{\mathbf k} \times \mathbf h = (0,0,1)\times(h_x,h_y,h_z) = (-h_y,;h_x,;0). \]

So

\[ \mathbf n \approx (-6669.49,\ -25385.17,\ 0), \] \[ n = |\mathbf n| \approx 26247.9. \]

RAAN \(\Omega\) defined via

\[ \cos\Omega = \frac{n_x}{n},\qquad \sin\Omega = \frac{n_y}{n}. \]

Here

\[ \cos\Omega = \frac{-6669.49}{26247.9} \approx -0.2541, \quad \sin\Omega = \frac{-25385.17}{26247.9} \approx -0.9672. \]

Both sine and cosine are negative ⇒ \(\Omega\) is in the third quadrant:

\[ \Omega \approx 255.28^\circ. \]

7. Argument of perigee \(\omega\)

Use

\[ \cos\omega = \frac{\mathbf n\cdot\mathbf e}{ne}. \]

Compute scalar product (numerically):

\[ \mathbf n\cdot\mathbf e \approx (-6669.49)(-0.0916) +(-25385.17)(-0.1422) + 0\times 0.0264 \approx 5201.2. \]

Hence

\[ \cos\omega = \frac{5201.2}{(26247.9)(0.1712)} \approx 0.9391 \Rightarrow \omega \approx 20.07^\circ. \]

Since \(e_z > 0\), the usual quadrant tests keep \(\omega\) in the first quadrant, consistent with this value.

8. True anomaly \(f\) at the given position \(\mathbf r\)

\[ \cos f = \frac{\mathbf e\cdot\mathbf r}{er}. \]

Compute \(\mathbf e\cdot\mathbf r\) (numerically):

\[ \mathbf e\cdot\mathbf r \approx (-0.0916)(-6045) + (-0.1422)(-3490) + (0.0264)(2500) \approx 362.2 + 496.3 + 66.1 \approx 924.6. \]

Then

\[ \cos f = \frac{924.6}{(0.1712)(7414.32)} \approx 0.8801, \] so

\[ f \approx \arccos(0.8801) \approx 28.45^\circ. \]

Check the sign using \(\mathbf r\cdot \mathbf v\): here \(\mathbf r\cdot\mathbf v > 0\) (radial component positive), meaning we are after perigee with \(0<f<180^\circ\), so \(f\approx 28.45^\circ\) is consistent.

Final set of six classical orbital elements:

\[ \boxed{ \begin{aligned} a \approx 8788\ \text{km},\ e \approx 0.171,\ i \approx 153.25^\circ,\ \Omega \approx 255.28^\circ,\ \omega \approx 20.07^\circ,\ f \approx 28.45^\circ. \end{aligned}} \]

4. A spacecraft at a radius \(r\) has a speed \(v\) and a flight path angle \(\gamma\). What is \(\sigma\) in the following expression in terms of \(r\), \(v\) and \(\gamma\).

\[e = \sqrt{1 + \sigma (\sigma-2)\cos^2 \gamma}\]

Solution

Given

\[ e = \sqrt{1 + \sigma(\sigma - 2)\cos^2\gamma}, \]

we must express \(\sigma\) in terms of \(r, v, \gamma\).

From basic orbital relations:

• radial distance \(r\),
• speed \(v\),
• flight-path angle \(\gamma\) (angle between velocity and local transverse direction).

Decompose velocity:

\[ v_r = v\sin\gamma,\qquad v_t = v\cos\gamma, \] specific angular momentum \[ h = r v_t = r v\cos\gamma, \] specific mechanical energy \[ \varepsilon = \frac{v^2}{2} - \frac{\mu}{r}. \]

For any Keplerian orbit:

\[ e^2 = 1 + \frac{2\varepsilon h^2}{\mu^2}. \]

Substitute \(\varepsilon\) and \(h\):

\[ \begin{aligned} e^2 &= 1 + \frac{2\left(\dfrac{v^2}{2} - \dfrac{\mu}{r}\right)(r^2 v^2\cos^2\gamma)}{\mu^2} &= 1 + \frac{(v^2 - 2\mu/r) r^2 v^2\cos^2\gamma}{\mu^2}. \end{aligned} \]

Factor:

\[ e^2 = 1 + \left[ \frac{r v^2}{\mu}\left(\frac{r v^2}{\mu} - 2\right)\right]\cos^2\gamma. \]

So if we define

\[ \boxed{\displaystyle \sigma = \frac{r v^2}{\mu}}, \]

then

\[ e^2 = 1 + \sigma(\sigma - 2)\cos^2\gamma, \]

which matches the given form.

So the answer is:

\[ \boxed{\sigma = \dfrac{r v^2}{\mu}}. \]

5. Find the total \(\Delta v\) requirement for bielliptic Hohmann transfer from a geocentric circular orbit of 7000 km radius to one of 105,000 km radius. Let the apogee of the first ellipse be 210,000 km. Compare the \(\Delta v\) schedule and total flight time with that for an ordinary single Hohmann transfer ellipse.

Orbits

Solution

Let Earth’s \(\mu = 398600\ \text{km}^3/\text{s}^2\).

Denote

• initial circular radius \(r_1 = 7000\ \text{km}\),
• final circular radius \(r_3 = 105000\ \text{km}\),
• intermediate apogee radius \(r_2 = 210000\ \text{km}\).

1. Single Hohmann transfer \(r_1 \to r_3\)

Semimajor axis of Hohmann ellipse:

\[ a_{13} = \frac{r_1 + r_3}{2} = \frac{7000 + 105000}{2} = 56000\ \text{km}. \]

Circular speeds:

\[ v_{c1} = \sqrt{\frac{\mu}{r_1}} \approx \sqrt{\frac{398600}{7000}} \approx 7.546\ \text{km/s}, \] \[ v_{c3} = \sqrt{\frac{\mu}{r_3}} \approx \sqrt{\frac{398600}{105000}} \approx 1.948\ \text{km/s}. \]

Speeds on transfer ellipse (vis-viva):

At perigee \((r=r_1)\):

\[ v_{p,13} = \sqrt{\mu\Big(\frac{2}{r_1} - \frac{1}{a_{13}}\Big)} \approx 10.333\ \text{km/s}, \]

At apogee \((r=r_3)\):

\[ v_{a,13} = \sqrt{\mu\Big(\frac{2}{r_3} - \frac{1}{a_{13}}\Big)} \approx 0.689\ \text{km/s}. \]

So the impulses:

\[ \Delta v_{1,H} = v_{p,13} - v_{c1} \approx 10.333 - 7.546 \approx 2.787\ \text{km/s}, \]

\[ \Delta v_{2,H} = v_{c3} - v_{a,13} \approx 1.948 - 0.689 \approx 1.260\ \text{km/s}. \]

Total:

\[ \boxed{\Delta v_H \approx 4.05\ \text{km/s}}. \]

Time of flight (half ellipse):

\[ T_H = \pi\sqrt{\frac{a_{13}^3}{\mu}} \approx 6.59\times 10^4\ \text{s} \approx 18.3\ \text{h}. \]

---

2. Bi-elliptic transfer \(r_1 \to r_2 \to r_3\)

First ellipse: between \(r_1\) and \(r_2\),

\[ a_{12} = \frac{r_1 + r_2}{2} = \frac{7000 + 210000}{2} = 108500\ \text{km}. \]

Second ellipse: between \(r_2\) and \(r_3\),

\[ a_{23} = \frac{r_2 + r_3}{2} = \frac{210000 + 105000}{2} = 157500\ \text{km}. \]

First burn at \(r_1\)

Speed on ellipse 1 at perigee \((r_1)\):

\[ v_{p,12} = \sqrt{\mu\Big(\frac{2}{r_1} - \frac{1}{a_{12}}\Big)} \approx 10.498\ \text{km/s}. \]

So

\[ \Delta v_{1,B} = v_{p,12} - v_{c1} \approx 10.498 - 7.546 \approx 2.95\ \text{km/s}. \]

Second burn at \(r_2\)

Speed on ellipse 1 at apogee \((r_2)\):

\[ v_{a,12} = \sqrt{\mu\Big(\frac{2}{r_2} - \frac{1}{a_{12}}\Big)} \approx 0.350\ \text{km/s}. \]

Speed on ellipse 2 at apogee \((r_2)\):

\[ v_{a,23} = \sqrt{\mu\Big(\frac{2}{r_2} - \frac{1}{a_{23}}\Big)} \approx 1.125\ \text{km/s}. \]

So

\[ \Delta v_{2,B} = v_{a,23} - v_{a,12} \approx 1.125 - 0.350 \approx 0.775\ \text{km/s}. \]

Third burn at \(r_3\)

Speed on ellipse 2 at perigee ((r_3)):

\[ v_{p,23} = \sqrt{\mu\Big(\frac{2}{r_3} - \frac{1}{a_{23}}\Big)} \approx 2.551\ \text{km/s}. \]

Circular speed at \(r_3\) was \(v_{c3}\approx 1.948\ \text{km/s}\). To circularize:

\[ \Delta v_{3,B} = v_{c3} - v_{p,23} \approx 1.948 - 2.551 \approx -0.301\ \text{km/s}. \]

Magnitude \(|\Delta v_{3,B}| \approx 0.301\ \text{km/s}\).

Totals

Total \(\Delta v\):

\[ \Delta v_B \approx 2.95 + 0.78 + 0.30 \approx 4.03\ \text{km/s}. \]

Two half-ellipses in time:

\[ T_{12} = \pi\sqrt{\frac{a_{12}^3}{\mu}} \approx 1.78\times 10^5\ \text{s}, \]

\[ T_{23} = \pi\sqrt{\frac{a_{23}^3}{\mu}} \approx 3.11\times 10^5\ \text{s}. \]

Total:

\[ T_B = T_{12} + T_{23} \approx 4.89\times 10^5\ \text{s} \approx 135.8\ \text{h} \approx 5.66\ \text{days}. \]

---

Comparison

• Ordinary Hohmann: \[ \Delta v_H \approx 4.05\ \text{km/s},\quad T_H \approx 18.3\ \text{h}. \]
• Bi-elliptic with \(r_2 = 210000\ \text{km}\): \[ \Delta v_B \approx 4.03\ \text{km/s},\quad T_B \approx 5.66\ \text{days}. \]

So the bi-elliptic saves only about \(0.02\ \text{km/s}\) in \(\Delta v\) (a tiny reduction) but increases the transfer time by a factor of \(\approx 7.4\).

6. During the test firing of a ballistic missile, the following measurements were made: \(h_{bo}\) = 1/5 DU, \(v_{bo}\) = 2/3 DU/TU, \(h_{apogee}\) = 0.5 DU. Assuming a symmetrical trajectory, what was the free-flight range of the missile during this test? (1 DU= 6378 km,1 DU/TU = 7.905 km/s)

Solution

From given data \[ \begin{aligned} DU &= 6378.0 \ \text{km} \\ DU/TU &= 7.905 \ \text{km/s} \\ h_{bo} &= 1/5 DU = 1275.6 \ \text{km} \\ v_{bo} &= 2/3 DU/TU = 5.2702 \ \text{km/s} \\ h_{apogee} &= 0.5 DU = 3189.1 \ \text{km} \\ \end{aligned} \]

and \(\mu = 398600 \ \text{km}^3/\text{s}^2\) and \(r_{earth} = 6378\) km.

First, computing the orbital energy and velocity

\[ \begin{aligned} E &= \frac{v_{bo}^2}{2} - \frac{\mu}{r_{bo}} \\ &= \frac{5.2702^2}{2} - \frac{398600}{6378+1275.6} = -38.1914 \ \text{km}^2/\text{s}^2 \\ v_a &= \sqrt{2\left(\frac{\mu}{r_a}+E\right)} \\ &= \sqrt{2\left(\frac{398600}{6378+3189.1}-38.1914\right)} = 2.6351 \ \text{km/s} \end{aligned} \]

Then the angular momentum \(h\)

\[ h = r_a . v_a = (6378+3189.1).3.6351 = 25,221.91 \ \text{km}^2/\text{s} \]

Comparing above expression with \(h = r_{bo} v_{bo}\cos\phi_{bo}\) gives

\[ \cos\phi_{bo} = 0.625 \]

and

\[ Q_{bo} = \frac{v_{bo}^2r_{bo}}{\mu} = 0.533 \]

Now, using the free-flight range equation

\[ \cos{\frac{\psi}{2}} = \frac{1-Q_{bo}\cos^2\phi_{bo}}{\sqrt{1+Q_{bo}(Q_{bo}-2)\cos^2\phi_{bo}}} = 0.95 \]

So, \(\psi = 36.4^\circ\)

and the free-flight range

\[ R_{ff} = 36.4\times 111.12 = 4043.6 \ \text{km} \]

where, \(111.12 \ \text{km/deg}\) is the arc-length per degree of the greater circle on the spherical earth.

7. Find the total delta‑v requirement for bielliptic Hohmann transfer from a geocentric circular orbit of 7000 km radius to one of 105,000 km radius. Let the apogee of the first ellipse be 210,000 km. Compare the delta‑v schedule and total flight time with that for an ordinary single Hohmann transfer ellipse.

Solution

Given

\[ r_A = 7000~\text{km}, \qquad r_B = 210{,}000~\text{km}, \qquad r_C = r_D = 105{,}000~\text{km} \]

Thus,

\[ \frac{r_B}{r_A} = 30, \qquad \frac{r_C}{r_A} = 15. \]

Orbit 1

Since this is a circular orbit,

\[ v_{A})_1 = \sqrt{\frac{\mu}{r_A}} = \sqrt{\frac{398{,}600}{7000}} = 7.546~\text{km/s}. \]

Orbit 2

\[ h_2 = \sqrt{ 2\mu \frac{r_A r_B}{r_A + r_B} } = 73{,}487~\text{km}^2/\text{s}. \]

\[ v_{A})_2 = \frac{h_2}{r_A} = \frac{73{,}487}{7000} = 10.498~\text{km/s}. \]

\[ v_{B})_2 = \frac{h_2}{r_B} = \frac{73{,}487}{210{,}000} = 0.34994~\text{km/s}. \]

Orbit 3

\[ h_3 = \sqrt{ 2\mu \frac{105{,}000 \cdot 210{,}000}{105{,}000 + 210{,}000} } = 236{,}230~\text{km}^2/\text{s}. \]

\[ v_{B})_3 = \frac{h_3}{r_B} = \frac{236{,}230}{210{,}000} = 1.1249~\text{km/s}. \]

\[ v_{C})_3 = \frac{h_3}{r_C} = \frac{236{,}230}{105{,}000} = 2.2498~\text{km/s}. \]

Orbit 4 (Circular)

\[ v_{C})_4 = v_{D})_4 = \sqrt{\frac{398{,}600}{105{,}000}} = 1.9484~\text{km/s}. \]

Total Δv — Bielliptic

\[ \Delta v_{\text{bielliptical}} = |v_{A})_2 - v_{A})_1| + |v_{B})_3 - v_{B})_2| + |v_{C})_4 - v_{C})_3| \]

\[ \Delta v_{\text{bielliptical}} = |10.498 - 7.546| + |1.1249 - 0.34994| + |1.9484 - 2.2498| = 4.0285~\text{km/s}. \]

Time of Flight — Bielliptic

\[ a_2 = \tfrac{1}{2}(7000 + 210{,}000) = 108{,}500~\text{km}, \]

\[ a_3 = \tfrac{1}{2}(105{,}000 + 210{,}000) = 157{,}500~\text{km}. \]

\[ t_{\text{bielliptical}} = \frac{1}{2}\left(\frac{2\pi}{\sqrt{\mu}}a_2^{3/2} + \frac{2\pi}{\sqrt{\mu}}a_3^{3/2}\right) = 488{,}870~\text{s} = 5.66~\text{days}. \]

Hohmann Transfer

\[ h_5 = \sqrt{ 2\mu \frac{7000 \cdot 105{,}000}{7000 + 105{,}000} } = 72{,}330~\text{km}^2/\text{s}. \]

\[ v_{A})_5 = \frac{h_5}{r_A} = \frac{72{,}330}{7000} = 10.333~\text{km/s}. \]

\[ v_{D})_5 = \frac{h_5}{r_D} = \frac{72{,}330}{105{,}000} = 0.68886~\text{km/s}. \]

\[ \Delta v_{\text{Hohmann}} = (10.333 - 7.546) + (1.9484 - 0.68886) = 4.0463~\text{km/s}. \]

Time of Flight — Hohmann

\[ a_5 = \tfrac{1}{2}(7000 + 105{,}000) = 56{,}000~\text{km}, \] \[ t_{\text{Hohmann}} = \tfrac{1}{2}\frac{2\pi}{\sqrt{\mu}}a_5^{3/2} = 65{,}942~\text{s} = 0.763~\text{days}. \]

• Hohmann Δv: 4.0463 km/s
• Bielliptic Δv: 4.0285 km/s
• Bielliptic is ~0.44% more efficient
• But bielliptic flight time is ~7× longer