A rigid body has three distinct principal moments of inertia. Which statements about torque-free rotation about a principal axis are correct?

Spin about the axis of maximum moment of inertia is stable

Spin about the axis of minimum moment of inertia is stable if the body is perfectly rigid

With internal energy dissipation, spin about the minimum axis becomes unstable

Spin about the axis of maximum moment of inertia is stable

Spin about the axis of minimum moment of inertia is stable if the body is perfectly rigid

Spin about the intermediate axis is stable

With internal energy dissipation, spin about the minimum axis becomes unstable

Which of the following statements about orbits under an inverse-square central force are correct?

Kepler’s second law follows from conservation of angular momentum

The specific energy of an elliptic orbit depends only on its semi-major axis

A parabolic trajectory has zero specific energy

Kepler’s second law follows from conservation of angular momentum

The specific energy of an elliptic orbit depends only on its semi-major axis

A parabolic trajectory has zero specific energy

The escape speed from a given radius depends on the direction of launch

Rigid-Body Momentum Balance and Space Dynamics

The second chapter of Section 2 of the GATE Aerospace Engineering (AE) paper, and the part of Flight Mechanics and Space Dynamics that leaves the atmosphere. It is a separate chapter because the physics changes: with no lift and no drag, a spacecraft moves under gravity alone, and a handful of conservation laws decide everything. It begins with linear and angular momentum balance for rigid bodies, which the syllabus lists as a core topic and which both halves of the section rest on; then central-force motion, Keplerian orbits and Kepler’s three laws; then escape velocity; and it ends with the special topic of Hohmann transfers between circular orbits, worked from low Earth orbit to geostationary altitude.

1. Linear and angular momentum balance for rigid bodies

For any body, the resultant external force equals the rate of change of linear momentum, which reduces to F = m·a_G for the mass centre G whatever the rotation. The resultant external moment about G (or about a fixed point) equals the rate of change of angular momentum, M_G = dH_G/dt, with H_G = [I]·ω. Because the inertia tensor is constant only in axes fixed to the body, the derivative is taken in a rotating frame: dH/dt = (dH/dt)_body + ω × H. In principal axes this gives Euler’s equations, M_x = I_x ω̇_x − (I_y − I_z) ω_y ω_z and its cyclic companions — the same equations that are the moment half of the aircraft equations of motion.

With no external moment, angular momentum is conserved in inertial space. A body spinning about a principal axis is then stable if that axis is the axis of maximum or minimum moment of inertia and unstable about the intermediate axis. Energy dissipation changes this: a body that loses kinetic energy at constant angular momentum drifts towards the lowest-energy state, which is spin about the MAXIMUM moment-of-inertia axis, so only major-axis spin is stable for a spacecraft with flexible parts or sloshing fuel. The same conservation governs a spin change: halve the moment of inertia and the spin rate doubles, while the rotational kinetic energy (H²/2I) doubles too, the extra energy coming from the work done pulling the mass in.

⚠️ Angular momentum is conserved; kinetic energy is not

When a spinning body changes its own moment of inertia with no external torque, Iω stays fixed, so ω scales as 1/I and the kinetic energy ½Iω² = H²/(2I) scales as 1/I as well. Assuming the kinetic energy is unchanged gives ω scaling as 1/√I — a wrong answer that appears among the options often.

2. Central-force motion and the orbit equation

In the two-body problem with the central body much heavier, the relative motion obeys r̈ = −(μ/r³)·r, where μ = GM (for Earth μ = 3.986 × 10¹⁴ m³/s²). A central force exerts no moment about the attracting centre, so the specific angular momentum h = r × v is constant: the motion stays in one plane, and h = r²θ̇. Since the areal velocity is dA/dt = ½r²θ̇ = h/2, equal areas are swept in equal times — Kepler’s second law is simply conservation of angular momentum. The specific mechanical energy ε = v²/2 − μ/r is also constant. Solving the equation of motion gives the conic r = (h²/μ)/(1 + e cos ν), where ν is the true anomaly measured from periapsis and e the eccentricity.

The conic, its eccentricity and its energy
Orbit	Eccentricity	Specific energy ε	Speed compared with local escape speed
Circle	e = 0	−μ/(2r) < 0	v = v_esc/√2
Ellipse	0 < e < 1	−μ/(2a) < 0	v < v_esc
Parabola	e = 1	0	v = v_esc
Hyperbola	e > 1	+μ/(2\|a\|) > 0	v > v_esc

3. Keplerian orbits and Kepler’s three laws

Kepler’s laws, derived rather than assumed: (1) a bound orbit is an ellipse with the attracting body at one focus; (2) the radius vector sweeps equal areas in equal times; (3) the square of the period is proportional to the cube of the semi-major axis, T = 2π√(a³/μ). For the ellipse, the periapsis and apoapsis radii are r_p = a(1 − e) and r_a = a(1 + e), so a = (r_p + r_a)/2 and e = (r_a − r_p)/(r_a + r_p). The energy depends only on a, ε = −μ/(2a), and combining it with ε = v²/2 − μ/r gives the vis-viva equation v² = μ(2/r − 1/a), which gives the speed anywhere on any orbit. At the apses the velocity is perpendicular to the radius, so conservation of h gives v_p r_p = v_a r_a.

A circular orbit is the case r = a: v_c = √(μ/r) and T = 2πr/v_c = 2π√(r³/μ). At r = 7000 km, v_c = √(3.986 × 10¹⁴/7 × 10⁶) = 7.55 km/s and T = 5829 s, about 97 minutes. The geostationary orbit is the circular equatorial orbit whose period equals one sidereal day (86 164 s), which puts it at a radius of about 42 164 km.

🧠 Kepler’s third law as a ratio

Most period questions need no constants: T ∝ a^(3/2). Doubling the semi-major axis multiplies the period by 2√2 = 2.83; quadrupling it multiplies the period by 8. Only the semi-major axis matters — an ellipse and a circle with the same a have the same period, however different their shapes.

4. Escape velocity

A body escapes to infinity when its specific energy is zero or positive: v²/2 − μ/r ≥ 0, so the escape velocity at radius r is v_esc = √(2μ/r) = √2·v_c. It is independent of direction (only speed enters the energy) and of the body’s mass. At Earth’s surface, with R_E = 6378 km, v_esc = √(2 × 3.986 × 10¹⁴/6.378 × 10⁶) ≈ 11.2 km/s, against a circular speed of about 7.9 km/s. Launched faster than v_esc, the body leaves on a hyperbola with hyperbolic excess speed v_∞ given by v_∞² = v² − v_esc².

🎯 Why the escape speed does not depend on the launch angle

Escape is an energy statement and kinetic energy is a scalar. Aim horizontally or vertically, the body reaches infinity provided ½v² ≥ μ/r and nothing (the ground, the atmosphere) is in the way. The launch direction changes the path and the angular momentum, not whether the orbit is bound.

5. Hohmann transfer between circular orbits

A Hohmann transfer moves a spacecraft between two coplanar circular orbits of radii r1 < r2 with two tangential impulses. The transfer orbit is the ellipse with periapsis r1 and apoapsis r2, so a_t = (r1 + r2)/2. The first burn raises the speed from √(μ/r1) to the transfer-orbit periapsis speed; the second, at apoapsis, raises it from the transfer apoapsis speed to √(μ/r2): Δv1 = √(μ/r1)[√(2r2/(r1 + r2)) − 1] and Δv2 = √(μ/r2)[1 − √(2r1/(r1 + r2))]. The transfer takes half the period of the transfer ellipse, t = π√(a_t³/μ). It is the minimum-Δv two-impulse transfer between coplanar circles; only for very large radius ratios (above about 11.94) can a three-impulse bi-elliptic transfer cost less.

Worked Hohmann transfer: r1 = 6678 km (300 km altitude) to r2 = 42 164 km, μ = 3.986 × 10¹⁴ m³/s²
Quantity	Value
Circular speed at r1	7.726 km/s
Transfer periapsis speed	10.152 km/s
Δv1	2.426 km/s
Transfer apoapsis speed	1.608 km/s
Circular speed at r2	3.075 km/s
Δv2	1.467 km/s
Total Δv	3.893 km/s
Transfer time (half the ellipse)	about 5.28 h

⚠️ Both burns are prograde, and the second is smaller

Going outward, both impulses ADD speed in the direction of motion, even though the final circular speed √(μ/r2) is lower than the starting one. The spacecraft slows down while coasting up the transfer ellipse, trading kinetic for potential energy, and arrives at apoapsis too slow for the higher circle. Treating the second burn as a braking burn gets its sign wrong.

Key takeaways

F = m·a_G and M_G = dH_G/dt; in body axes dH/dt = (dH/dt)_body + ω × H gives Euler’s equations. Torque-free spin is stable about the major and minor axes, and only the major axis once energy dissipates.
Central force ⇒ h = r²θ̇ constant ⇒ planar motion and equal areas in equal times. Orbit: r = (h²/μ)/(1 + e cos ν).
ε = −μ/(2a); vis-viva v² = μ(2/r − 1/a); r_p = a(1 − e), r_a = a(1 + e); v_p r_p = v_a r_a; T = 2π√(a³/μ).
v_c = √(μ/r), v_esc = √(2μ/r) = √2·v_c, independent of direction; about 7.9 and 11.2 km/s at Earth’s surface.
Hohmann: a_t = (r1 + r2)/2, two prograde tangential burns, time = half the transfer period; about 3.89 km/s from a 300 km orbit to geostationary radius.

Practice questions (12)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

A satellite spins freely with no external torque. It retracts two booms so that its moment of inertia about the spin axis halves. Its spin rate and rotational kinetic energy respectively become:
1. Doubled and unchanged
2. Doubled and doubled
3. √2 times and unchanged
4. Unchanged and halved
Show answer
Answer: B — Doubled and doubled
With no torque H = Iω is conserved, so halving I doubles ω. The kinetic energy is H²/(2I), which doubles when I halves; the work done by the retraction mechanism supplies it. The √2 answer comes from wrongly holding the kinetic energy constant.
A rigid body has three distinct principal moments of inertia. Which statements about torque-free rotation about a principal axis are correct?
1. Spin about the axis of maximum moment of inertia is stable
2. Spin about the axis of minimum moment of inertia is stable if the body is perfectly rigid
3. Spin about the intermediate axis is stable
4. With internal energy dissipation, spin about the minimum axis becomes unstable
Show answer
Answer: A — Spin about the axis of maximum moment of inertia is stable; B — Spin about the axis of minimum moment of inertia is stable if the body is perfectly rigid; D — With internal energy dissipation, spin about the minimum axis becomes unstable
Euler’s equations show small perturbations oscillate about the maximum and minimum axes and grow about the intermediate axis. Dissipation lowers kinetic energy at constant H, driving the body towards the maximum-inertia axis, so a minimum-axis spinner with flexible parts eventually tumbles into flat spin. The intermediate axis is unstable in every case.
What is the speed of a satellite in a circular orbit of radius 7000 km about Earth, in km/s (to two decimal places)? Take μ = 3.986 × 10¹⁴ m³/s².
Numerical answer — type the value.
Show answer
Answer: 7.55
v_c = √(μ/r) = √(3.986 × 10¹⁴/7.0 × 10⁶) = √(5.694 × 10⁷) = 7546 m/s = 7.55 km/s. Using √(2μ/r) gives the escape speed, 10.67 km/s, which would leave the orbit altogether.
What is the period of a circular orbit of radius 7000 km about Earth, in minutes (to one decimal place)? Take μ = 3.986 × 10¹⁴ m³/s².
Numerical answer — type the value.
Show answer
Answer: 97.1
T = 2π√(r³/μ) = 2π√((7 × 10⁶)³/3.986 × 10¹⁴) = 2π × 927.6 = 5828.5 s = 97.1 min. The same comes from 2πr/v_c = 2π × 7 × 10⁶/7546. Forgetting the cube (using r instead of r³) gives a meaningless fraction of a second.
An Earth orbit has a periapsis radius of 7000 km and an apoapsis radius of 21 000 km. What is its eccentricity?
Numerical answer — type the value.
Show answer
Answer: 0.5
e = (r_a − r_p)/(r_a + r_p) = (21 000 − 7000)/(21 000 + 7000) = 14 000/28 000 = 0.5, with a = 14 000 km. Dividing the difference by r_a alone gives 0.667, and by r_p alone gives 2 — neither is an ellipse’s eccentricity.
For the orbit with periapsis radius 7000 km and apoapsis radius 21 000 km, what is the ratio of the speed at periapsis to the speed at apoapsis?
Numerical answer — type the value.
Show answer
Answer: 3
At the apses the velocity is perpendicular to the radius, so h = v_p r_p = v_a r_a and v_p/v_a = r_a/r_p = 21 000/7000 = 3. Using energy with a square root (√3) confuses this with the ratio of circular speeds at the two radii.
If the semi-major axis of a satellite’s orbit is made four times larger, its orbital period becomes:
1. 4 times
2. 8 times
3. 16 times
4. 2 times
Show answer
Answer: B — 8 times
Kepler’s third law: T ∝ a^(3/2), so T becomes 4^(3/2) = 8 times. Answering 16 squares the ratio, and answering 2 takes its square root; only the 3/2 power follows from T = 2π√(a³/μ).
Which of the following statements about orbits under an inverse-square central force are correct?
1. Kepler’s second law follows from conservation of angular momentum
2. The specific energy of an elliptic orbit depends only on its semi-major axis
3. A parabolic trajectory has zero specific energy
4. The escape speed from a given radius depends on the direction of launch
Show answer
Answer: A — Kepler’s second law follows from conservation of angular momentum; B — The specific energy of an elliptic orbit depends only on its semi-major axis; C — A parabolic trajectory has zero specific energy
dA/dt = h/2 is constant because a central force exerts no moment; ε = −μ/(2a) for any ellipse of that a, whatever e; and e = 1 is the zero-energy boundary between bound and unbound. Escape needs ½v² ≥ μ/r, a scalar condition, so direction does not enter.
The ratio of the escape velocity to the circular-orbit velocity at the same radius from a planet’s centre is:
1. 2
2. √2
3. 1/√2
4. √3
Show answer
Answer: B — √2
v_esc = √(2μ/r) and v_c = √(μ/r), so the ratio is √2 ≈ 1.414 at every radius. The ratio 2 compares the energies, not the speeds: the escape trajectory has twice the kinetic energy of the circular orbit at that radius.
A spacecraft makes a Hohmann transfer from a circular orbit of radius 6678 km to a coplanar circular orbit of radius 42 164 km. With μ = 3.986 × 10¹⁴ m³/s², what is the total Δv, in km/s (to two decimal places)?
Numerical answer — type the value.
Show answer
Answer: 3.89
Δv1 = √(μ/r1)[√(2r2/(r1 + r2)) − 1] = 7.726 × (1.3140 − 1) = 2.426 km/s; Δv2 = √(μ/r2)[1 − √(2r1/(r1 + r2))] = 3.075 × (1 − 0.5229) = 1.467 km/s; total 3.89 km/s. Vis-viva gives the same: transfer speeds 10.152 and 1.608 km/s against circular 7.726 and 3.075 km/s. Subtracting the circular speeds (7.726 − 3.075 = 4.65) is not a transfer at all.
For the Hohmann transfer from 6678 km to 42 164 km radius (μ = 3.986 × 10¹⁴ m³/s²), what is the transfer time, in hours (to two decimal places)?
Numerical answer — type the value.
Show answer
Answer: 5.28
a_t = (6678 + 42 164)/2 = 24 421 km. The transfer is half the ellipse: t = π√(a_t³/μ) = π√((2.4421 × 10⁷)³/3.986 × 10¹⁴) = 18 990 s = 5.28 h. Taking the full period 2π√(a_t³/μ) gives 10.55 h, twice the time actually spent.
What is the specific mechanical energy of an Earth orbit with semi-major axis 10 000 km, in MJ/kg (to two decimal places, with sign)? Take μ = 3.986 × 10¹⁴ m³/s².
Numerical answer — type the value.
Show answer
Answer: -19.93
ε = −μ/(2a) = −3.986 × 10¹⁴/(2 × 10⁷) = −1.993 × 10⁷ J/kg = −19.93 MJ/kg, whatever the eccentricity. The energy is negative because the orbit is bound; −μ/a (−39.86) forgets the factor 2.