Mid-Semester Examination

Space Flight Mechanics (AE313 / AE613), IIST 2026

Duration	2 hours
Maximum marks	30
Part A — objective (Q1–Q4)	15 marks
Part B — numerical and derivation (Q5–Q6)	15 marks

Instructions

1. All questions are compulsory.
2. The whole paper is one conversation between Johannes Kepler and Isaac Newton. In Part A they leave sentences unfinished. Each gap is a blank such as (c) [1] ________, with its marks in brackets. Write the question number, the blank label and your answer in the answer book. Do not copy the dialogue.
3. “Answering as Johannes” means using geometry, observation and his three laws only: no forces and no calculus. “Answering as Isaac” allows any mathematics.
4. Use the data table unless a question says otherwise. Planetary orbits are circular and coplanar, and engine burns are impulsive, unless a question says otherwise.
5. Diagrams and write-up that are not legible will not be graded.

Data

Quantity	Value
Earth: gravitational parameter \(\mu_\oplus\); radius \(R_\oplus\)	\(3.986\times10^{5}\ \mathrm{km^3\,s^{-2}}\); \(6378\) km
Standard gravity \(g_0\)	\(9.81\ \mathrm{m\,s^{-2}}\)
Geostationary orbit	radius \(42\,164\) km; period (one sidereal day) \(23.934\) h
Sun: gravitational parameter \(\mu_\odot\); radius \(R_\odot\)	\(1.327\times10^{11}\ \mathrm{km^3\,s^{-2}}\); \(6.96\times10^{5}\) km
Astronomical unit	\(1\ \mathrm{AU} = 1.496\times10^{8}\) km
Year; Earth’s orbital speed	\(365.25\) d; \(v_\oplus = 29.78\) km/s
Moon	orbit radius \(384\,400\) km; sidereal period \(27.32\) d
Venus	orbit radius \(0.723\) AU
Mars	orbit radius \(1.524\) AU; sidereal period \(686.98\) d
Jupiter	orbit radius \(5.203\) AU; \(m_J/M_\odot = 1/1047.6\)
Saturn	orbit radius \(9.537\) AU

Prologue: two visitors

This paper looks at the first half of the course from two viewpoints: pure geometry (Kepler’s view) and dynamics (Newton’s view). It shows how far geometry and a few assumptions can take us, and where dynamics is really needed. To set this up, I have written a dialogue between Johannes and Isaac, as if they had been transported to 2026. Each arrived knowing only what he knew on the day he died.

To avoid ambiguity, here is what each of them knew on his deathbed.

• Johannes Kepler (1571–1630)
  • Had Tycho Brahe’s observations of Mars, the geometry of conics, and logarithms.
  • Assumed that the orbits were elliptical, and knew his three laws (1609, 1619).
  • Believed that the Sun drives the planets with a force that weakens as \(1/r\), and that comets move in straight lines.
• Isaac Newton (1643–1727)
  • Discovered the laws of motion and the law of gravitation, and invented fluxions (calculus).
  • Knew Kepler’s three laws.
  • Never learned the value of the gravitational constant \(G\) (Cavendish, 1798), and never saw Halley’s comet return (1758).

Since July, both have been sitting in the back row of AE313. Isaac takes careful notes; Johannes argues with the slides. It is the night before the mid-semester examination, and they are revising together. In Part A, you finish the lines they leave unfinished. In Part B, they set each other problems.

Part A — The revision (15 marks)

Q1. The crowd [2.5 marks]

Scene: the hostel common room, nine in the evening. Isaac opens his notebook.

Isaac: The course began where I began: every body attracts every other. Treat the Sun, the planets and the moons as \(n\) point masses \(m_i\) at positions \(\mathbf R_i\) in an inertial frame. In the Principia I reckoned that, at conjunction, Jupiter pulls on Saturn with about \(1/211\) of the Sun’s pull (Figure 1, left panel). With today’s data the ratio is \(1/\)(a) [1.5] __________.

Figure 1: Left: the Sun, Jupiter and Saturn in line, with Jupiter between. Right: the Moon at new moon, pulled towards the Sun and towards the Earth. Arrow lengths are schematic.

Johannes (pointing at the right panel): Then look at the Moon. At new moon the Sun pulls it one way and the Earth pulls it the other, and the Sun pulls more than twice as hard. So the Moon is really the Sun’s satellite, and any two-body model of the Earth and the Moon must fail.

(b) [1] True or false? Justify in one line. ____________________

Answers to Q1

Blank	Answer	Basis
(a)	216	\(\dfrac{m_J}{M_\odot}\left(\dfrac{9.537}{9.537 - 5.203}\right)^2 = \dfrac{4.842}{1047.6}\)
(b)	F	The Earth–Moon relative motion feels only the difference between the Sun’s pulls on the Moon and on the Earth, the tidal term \(\approx 2\mu_\odot r_{EM}/d^3 = 3.0\times10^{-5}\ \mathrm{m\,s^{-2}}\). That is about 1.1% of the Earth’s pull (\(2.70\times10^{-3}\ \mathrm{m\,s^{-2}}\)).

Q2. The pair [4.5 marks]

Johannes: Enough of crowds. Give me two bodies and nothing else. My ellipse! Before I found the ellipse, I believed that a planet’s speed is inversely proportional to its distance from the Sun.

Isaac: Not the speed. In Figure 2, the quantity that is exactly proportional to \(1/r\) at every point of the orbit is (a) [1] ________ [(A) the speed \(v\) (B) the radial velocity \(v_r\) (C) the transverse velocity \(v_\theta\) (D) the areal velocity].

Figure 2: Velocity at a point of an ellipse: its radial and transverse components, and the flight-path angle \(\gamma\) measured from the local horizontal.

Johannes: Look at the same figure. My path is most steeply inclined to the local horizontal where the planet is at its mean distance, \(r = a\).

(b) [1.5] True or false? Justify, and give the greatest flight-path angle in terms of \(e\). ____________________

Isaac: I did not trust the inverse square until I had tried it on the Moon. The Moon’s centripetal acceleration is (c) [1] ________ \(\mathrm{m\,s^{-2}}\), and \(g_0\) divided by it is ________. Compare this with \((r_{\rm Moon}/R_\oplus)^2 = 3632\).

Isaac: Also, I used Huygens’ moon of Saturn to weigh Saturn against the Sun. Titan circles Saturn at \(1\,221\,870\) km, once in \(15.945\) days. Compared with the Earth’s orbit, this gives \(m_{\rm Saturn}/M_\odot\) closest to (d) [1] ________ [(A) 1/1050 (B) 1/3500 (C) 1/19 400 (D) 1/333 000].

Answers to Q2

Blank	Answer	Basis
(a)	C	\(v_\theta = r\dot f = h/r\)
(b)	T; \(\ \sin\gamma_{\max} = e\)	\(\tan\gamma = \dfrac{e\sin f}{1+e\cos f}\) peaks at \(\cos f = -e\), where \(r = p/(1-e^2) = a\) (the ends of the minor axis)
(c)	\(2.72\times10^{-3}\); about 3600	\(4\pi^2 r/T^2\) with \(r = 3.844\times10^8\) m, \(T = 27.32\) d
(d)	B	\(\left(\dfrac{1.2219\times10^6}{1.496\times10^8}\right)^3\left(\dfrac{365.25}{15.945}\right)^2 = \dfrac{1}{3498}\) (the Principia gave 1/3021)

Q3. The ferry [5 marks]

Scene: one in the morning, a mission-control video playing on a laptop.

Johannes: In 1610 I wrote to Galileo that once there are ships with sails fit for the heavenly breezes, men will be found to brave even that emptiness. And there on the screen are the ships. This Hohmann ferry of 1925 is merely half of one of my ellipses. It touches the inner circle at perihelion and the outer circle at aphelion (Figure 3).

Figure 3: The Hohmann transfer ellipse between circles of radii \(r_1\) and \(r_2\). The planet is at the focus F, and C is the centre of the transfer ellipse.

(a) [2] Match each length marked in Figure 3 with a mean of \(r_1\) and \(r_2\). One entry in the right-hand column is not used.

Length	Mean of \(r_1\) and \(r_2\)
P. \(a\)	1. geometric mean
Q. \(b\)	2. harmonic mean
R. \(p\)	3. arithmetic mean
S. \(ae\)	4. half the difference, \((r_2 - r_1)/2\)
	5. root-mean-square

Johannes: With my third law alone, I can tell you how long the ferry takes from the Earth’s orbit to Venus’s orbit: (b) [1] ________ days.

Isaac: Going outwards, one can also overshoot and fall back. Figure 4 compares three plans, all starting from a circular orbit of radius \(r_1\):

• a Hohmann transfer to a circular orbit of radius \(r_2 = R\,r_1\);
• a bi-elliptic transfer to the same orbit, with its intermediate apoapsis pushed out to infinity;
• a single burn that escapes from \(r_1\).

Figure 4: Total speed change, in units of the initial circular speed, for three ways of going outwards from a circular orbit of radius \(r_1\).

(c) [1] Identify curves 1, 2 and 3. ________

Isaac: Between coplanar circular orbits, the Hohmann transfer is cheaper than every bi-elliptic transfer whenever \(r_2/r_1\) is below about (d) [1] ________ [(A) 3.3 (B) 11.94 (C) 15.58 (D) 100].

Answers to Q3

Blank	Answer	Basis
(a)	P–3, Q–1, R–2, S–4	\(b^2 = r_pr_a = r_1r_2\); \(p = b^2/a\); ½ mark each
(b)	146 d	\(\tfrac12\left(\tfrac{1.723}{2}\right)^{3/2}\) yr
(c)	1: Hohmann; 2: bi-elliptic with \(r_b\to\infty\) (bi-parabolic); 3: escape, \((\sqrt2-1)v_{c1}\)	Only the Hohmann cost vanishes at \(R = 1\)
(d)	B	Curves 1 and 2 cross at \(R = 11.94\); curve 1 peaks at 15.58

Q4. The door [3 marks]

Scene: three in the morning, under the stars.

Johannes: This Voyager 1 (not visible even to Johannes) is now about 170 AU from the Sun and still moves at about 17.0 km/s. Will it ever come back?

Isaac: Ignoring the planets, the local escape speed from the Sun at 170 AU is (a) [1] ________ km/s, and Voyager’s hyperbolic excess speed is ________ km/s.

Johannes: In 2025 astronomers found 3I/ATLAS, a visitor from another star that crosses the solar system once and will never return. Here at last is my straight-line comet!

Isaac: Nearly, Johannes. It moves on a hyperbola about the Sun. For 3I/ATLAS, with \(e = 6.14\) \(\delta =\) (b) [1] ________ degrees. The first visitor, 1I/ʻOumuamua (2017), had \(e \approx 1.20\) and was turned through ________ degrees. Your straight lines, Johannes, are the limit of my hyperbolas as (c) [1] ________.

Answers to Q4

Blank	Answer	Basis
(a)	3.23 km/s; 16.7 km/s	\(\sqrt{2\mu_\odot/(170\ \mathrm{AU})}\); \(\ v_\infty = \sqrt{v^2 - v_{\rm esc}^2}\)
(b)	\(18.7^\circ\); \(113^\circ\)	\(\delta = 2\arcsin(1/e)\), from \(\delta = 2f_\infty - \pi\) and \(\cos f_\infty = -1/e\)
(c)	\(e \to \infty\)	\(\delta \to 0\): the Sun no longer turns the path

Part B — Problems they set each other (15 marks)

Q5. Who needs whom? [8 marks]

Johannes: Isaac, I think people make far too much of your achievement in finding the law of gravitation. As I shall show, I can derive your law from my three laws alone.

(a) Write Johannes’ derivation. Starting from Kepler’s three laws, show that each planet has an acceleration directed towards the Sun, of magnitude \(\mu/r^2\), with the same constant \(\mu\) for every planet. What must Isaac add to turn this into his law of universal gravitation? [3]

Isaac: I agree. I think your work was far harder. There was a great deal of trial and error and speculation, because you did not know why the heavens moved. Your third law fascinates me most. You stated \(T^2 = k a^3\), with \(k\) a constant. To arrive at it, you needed the ratios \(a_i/a_j\) for planets \(i\) and \(j\). How did you find them?

Johannes: I am glad you asked. It was a clever trick of geometry. I had Tycho’s data. Every 687 days Mars returns to exactly the same place in its orbit, while the Earth is at a different place in its own. Look at Figure 5, where I show a simplified case with Mars and the Earth orbiting in the same plane. At each return of Mars I know the angle \(\alpha_i\) at the Sun between the directions to the Earth and to Mars, and I measure the angle \(\beta_i\) at the Earth between the Sun and Mars.

Figure 5: Kepler’s triangulation. Mars (\(M\)) returns to the same heliocentric position every \(T_M \approx 687\) days, while the Earth is seen at a different point \(P_i\) of its orbit each time. Idealized coplanar model; the Earth’s eccentricity is exaggerated.

(b) Write down the formula Johannes uses to obtain the ratio \(SP_i/SM\) of the Earth–Sun and Mars–Sun distances from the angles \(\alpha_i\) and \(\beta_i\). [2]

Once we have these ratios at enough points, spread over many years, we can map out the whole orbit of the Earth in units of \(SM\), and hence find the ratio \(a_E/a_M\).

Isaac: That is clever indeed. But remember that I never knew the value of the universal gravitational constant either. Only last night I read about the experiment Mr Cavendish performed in 1798, seventy years after my death. Figure 6 shows his apparatus, seen from above.

Figure 6: Cavendish torsion balance, top view (schematic, twist exaggerated). A light rod of half-length \(a\) carrying two small spheres \(m\) hangs from a thin wire of torsion constant \(\kappa\). Two large spheres \(M\) pull the small ones sideways, so the rod turns by \(\theta\). A mirror on the wire reflects a light beam onto a scale at distance \(L\); the spot moves by \(s\).

Isaac (continuing): The difficulty is that the attraction between two lumps of lead is absurdly small, far too small for any balance that weighs things. So Cavendish did not try to measure the force; he measured an angle. He hung a light rod, with a small lead ball of mass \(m\) at each end, from a long, thin wire. When the rod turns, the wire twists and pushes back with a torque proportional to the twist, \(\kappa\theta\). He then brought two great lead balls of mass \(M\) close to the small ones, one on each side, so that both attractions turn the rod the same way. The rod turns until the twist of the wire balances the pull of the great balls. The angle is tiny, so he fixed a small mirror to the wire and read the deflection of a reflected beam of light on a distant scale. He did not know the stiffness \(\kappa\) of his wire either. Instead, he let the rod swing freely and timed one full oscillation, \(T_{\text{osc}} = 2\pi\sqrt{I/\kappa}\), where \(I\) is the moment of inertia of the system. That period gave him \(\kappa\). So from a length, a distance, a mass, an angle and a time, he found my constant.

(c) Derive an expression for \(G\) in terms of \(M\), \(a\), \(r\), \(\theta\) and \(T_{\text{osc}}\). Treat the rod as massless and the spheres as point masses, and neglect the attraction of each large sphere on the far small sphere. Also relate \(\theta\) to the scale reading \(s\) and the distance \(L\). [3]

Solution

(a) Second law ⇒ central force. Equal areas in equal times means \(\tfrac12\,|\mathbf r \times \dot{\mathbf r}| = \text{const}\). The orbit lies in a fixed plane, so \(\mathbf h = \mathbf r \times \dot{\mathbf r}\) is constant. Then \(\dot{\mathbf h} = \mathbf r \times \ddot{\mathbf r} = \mathbf 0\), so the acceleration is parallel to \(\mathbf r\): it points along the line to the Sun. In polar coordinates, \(h = r^2\dot f\).

First law ⇒ inverse square. The radial acceleration is \(a_r = \ddot r - r\dot f^2\). With \(u = 1/r\) and \(\dot f = h u^2\), \[ \dot r = -h\,\frac{du}{df}, \qquad \ddot r = -h^2u^2\,\frac{d^2u}{df^2} \quad\Rightarrow\quad a_r = -h^2u^2\left(\frac{d^2u}{df^2} + u\right). \] For the ellipse with the Sun at a focus, \(u = (1 + e\cos f)/p\). Hence \(u'' + u = 1/p\) and \[ a_r = -\frac{h^2}{p}\,\frac{1}{r^2}. \] On each orbit, the acceleration is directed towards the Sun and varies as \(1/r^2\).

Third law ⇒ the same constant for all planets. The areal rate times the period equals the area of the ellipse, \(\tfrac12 hT = \pi ab\), and \(p = b^2/a\). Hence \[ \frac{h^2}{p} = \frac{4\pi^2a^2b^2}{T^2}\cdot\frac{a}{b^2} = \frac{4\pi^2a^3}{T^2} = \frac{4\pi^2}{k} \equiv \mu, \] which is the same for every planet. So \(\ddot{\mathbf r} = -\mu\,\mathbf r/r^3\), with a single \(\mu\) belonging to the Sun.

Where geometry stops. Kepler’s laws give only the acceleration and the one number \(\mu\). Turning this into a force \(F = m\mu/r^2\) needs \(F = ma\). Writing \(\mu = GM_\odot\), with the force proportional to both masses, needs Newton’s third law (the planet pulls the Sun as hard as the Sun pulls the planet). It also needs the claim that the same law holds between any two bodies. Johannes cannot supply these dynamical steps.

(b) In triangle \(S P_i M\) the third angle, at Mars, is \(\gamma_i = \pi - \alpha_i - \beta_i\). By the sine rule, \(SP_i/\sin\gamma_i = SM/\sin\beta_i\), so \[ \frac{r_E}{r_M} = \frac{SP_i}{SM} = \frac{\sin\gamma_i}{\sin\beta_i} = \frac{\sin(\alpha_i + \beta_i)}{\sin\beta_i}. \] Here \(\alpha_i\) is the difference between the heliocentric longitudes of the Earth and Mars. The Earth’s heliocentric longitude is the Sun’s observed longitude plus \(180^\circ\), and \(\beta_i\) is the observed elongation of Mars from the Sun.

(c) The force on one small sphere is \(F = GMm/r^2\). Both spheres turn the rod in the same sense, so at equilibrium \[ \kappa\theta = 2Fa = \frac{2GMma}{r^2}. \tag{1} \] For the free oscillation, \(I\ddot\phi = -\kappa\phi\) with \(I = 2ma^2\), so \[ T_{\text{osc}} = 2\pi\sqrt{I/\kappa} \quad\Rightarrow\quad \kappa = \frac{8\pi^2ma^2}{T_{\text{osc}}^2}. \tag{2} \] Substituting (2) into (1) and cancelling \(m\) and \(a\): \[ \boxed{G = \frac{4\pi^2\,a\,r^2\,\theta}{M\,T_{\text{osc}}^2}} \] The small mass \(m\) cancels, because it appears in both the gravitational torque and the moment of inertia. For the optical lever: turning the mirror by \(\theta\) turns the reflected beam by \(2\theta\). So for small angles \(s = 2L\theta\), i.e. \(\theta = s/(2L)\).

Q6. One push or two? [7 marks]

Johannes: Here is a circle, and here is one of my ellipses, crossing it. Surely a single push at the crossing is the cheapest way from one to the other: no waiting, no second burn.

Isaac: Perhaps. Let us calculate before we judge.

A spacecraft is in a circular Earth orbit of radius \(r_1 = 10\,000\) km. It must be transferred to a coplanar elliptical orbit of eccentricity \(e = 0.3\), whose semi-latus rectum equals the radius of the circle, \(p = r_1\) (Figure 7).

Figure 7: The circle of radius \(r_1\) (blue) and the target ellipse (black), which cross at X and X\('\). Dashed: the two tangential transfer routes (i) and (ii). Stars mark their burn points.

(a) Find the semi-major axis, the periapsis radius and the apoapsis radius of the target ellipse. Show that the circle meets the ellipse at true anomaly \(f = \pm 90^\circ\). [2]

(b) Single impulse. At the crossing point X, find the radial and transverse velocity components on the ellipse. Show that the insertion burn is purely radial, and that its magnitude is \[ \Delta v_{1} = e\,v_c, \qquad v_c = \sqrt{\mu / r_1}. \] [2]

(c) Show that, in units of \(v_c\), \[ \begin{aligned} \Delta v_{\text{(i)}} &= \sqrt{2(2-e)} - (2-e), \\ \Delta v_{\text{(ii)}} &= (2+e) - \sqrt{2(2+e)}, \end{aligned} \] and hence prove that both two-impulse transfers are cheaper than the single impulse for every \(0 < e < 1\). [2]

(d) Give one mission reason for choosing it anyway. [1]

Solution

(a) \(a = p/(1-e^2) = 10\,000/0.91 = 10\,989\) km, \(\;r_p = p/(1+e) = 7\,692\) km, \(\;r_a = p/(1-e) = 14\,286\) km. The circle meets the ellipse where \(r = p/(1+e\cos f) = p\), i.e. \(\cos f = 0\), so \(f = \pm 90^\circ\). Because \(r_p < r_1 < r_a\), the orbits cross; they are not tangent.

(b) On the ellipse, \[ v_r = \sqrt{\mu/p}\; e\sin f, \qquad v_\theta = \sqrt{\mu/p}\,(1 + e\cos f). \] At \(f = 90^\circ\): \(v_r = e\sqrt{\mu/p}\) and \(v_\theta = \sqrt{\mu/p} = v_c\). The transverse component already equals the circular speed, so only the radial component must be added: \[ \Delta v_1 = e\,v_c = 0.3 \times 6.313 = 1.894 \text{ km/s}. \] The flight-path angle after the burn is \(\gamma = \tan^{-1} e = 16.7^\circ\).

(c) Work with \(\mu = p = 1\), so that \(v_c = 1\), \(r_a = 1/(1-e)\) and \(r_p = 1/(1+e)\).

(i) The transfer ellipse runs from \(1\) to \(r_a\). Its periapsis speed is \(\sqrt{2r_a/(1+r_a)} = \sqrt{2/(2-e)}\), and its apoapsis speed is \((1-e)\sqrt{2/(2-e)}\) (conservation of angular momentum). The target’s apoapsis speed is \(1-e\). Hence \[ \Delta v_{\text{(i)}} = \Big(\sqrt{\tfrac{2}{2-e}} - 1\Big)\big[1 + (1-e)\big] = \sqrt{2(2-e)} - (2-e). \] (ii) The transfer ellipse runs from \(r_p\) to \(1\). Its apoapsis speed is \(\sqrt{2/(2+e)}\), and its periapsis speed is \((1+e)\sqrt{2/(2+e)}\). The target’s periapsis speed is \(1+e\). Hence \[ \Delta v_{\text{(ii)}} = \Big(1 - \sqrt{\tfrac{2}{2+e}}\Big)\big[1 + (1+e)\big] = (2+e) - \sqrt{2(2+e)}. \] Comparing with \(\Delta v_1 = e\): \[ \Delta v_{\text{(i)}} < e \iff \sqrt{2(2-e)} < 2 \iff e > 0, \qquad \Delta v_{\text{(ii)}} < e \iff \sqrt{2(2+e)} > 2 \iff e > 0. \] Both inequalities hold for every \(0 < e < 1\). For small \(e\) both costs are \(\approx e/2\), half the single-impulse cost. As \(e \to 1\), \(\Delta v_{\text{(i)}} \to \sqrt 2 - 1 \approx 0.414\), while \(\Delta v_1 \to 1\). Transfer (i) is always the cheaper of the two.

(d) A radial burn is perpendicular to the velocity, so to first order it does no work. It changes the specific energy \(v^2/2\) only through the \((\Delta v)^2\) term, and it leaves the angular momentum \(r v_\theta\) unchanged. It spends its \(\Delta v\) turning the velocity vector rather than resizing it. A tangential burn acts along \(\mathbf v\) and changes the energy by \(v\,\Delta v\), so every km/s goes into changing the orbit’s energy. Reasons to accept the single impulse anyway:

• it is immediate, with no 69–111 min coast;
• it needs only one engine restart;
• it arrives at a specific point on the target orbit, which helps with phasing or rendezvous timing.

Epilogue

Johannes: So the heavenly breezes were never needed: only the right push, at the right place, on the right ellipse.

Isaac: Your ellipse, Johannes. I only explained it.

End of paper.