Self-Assessment Quiz: Orbital Maneuvers

Twenty questions to check your grasp of impulsive maneuvers, the Hohmann and bi-elliptic transfers, plane changes, phasing, rendezvous, and combined burns. Answer each before opening the key. Aim for 16 or more. Use $\mu_\oplus = 3.986\times10^{5}\ \text{km}^3/\text{s}^2$.

Question 1

An impulsive maneuver is idealized as:

A) a long, slow burn spread over many orbits B) an instantaneous velocity change at a single point, with position unchanged C) a change of position with velocity unchanged D) a maneuver that uses no propellant

Question 2

A Hohmann transfer between two circular orbits uses:

A) one burn B) two burns, along an ellipse tangent to both circles C) three burns via a high intermediate apoapsis D) a continuous spiral

Question 3

In a Hohmann transfer that raises an orbit, the two burns are:

A) both retrograde B) both prograde C) one prograde, one radial D) one normal, one prograde

Question 4

The LEO-to-GEO Hohmann transfer costs about:

A) $0.9\ \text{km/s}$ B) $3.9\ \text{km/s}$ C) $9.4\ \text{km/s}$ D) $11.2\ \text{km/s}$

Question 5

The Hohmann transfer time (one way) is:

A) $2\pi\sqrt{a_t^3/\mu}$ (a full period) B) $\pi\sqrt{a_t^3/\mu}$ (half a period) C) independent of the orbit radii D) always exactly one hour

Question 6

A bi-elliptic transfer can beat a Hohmann transfer only when the ratio $r_2/r_1$ is:

A) less than $2$ B) roughly $1$ C) larger than about $11.94$ D) exactly $\pi$

Question 7

The delta-v cost of a pure plane change of angle $\Delta i$ at orbital speed $v$ is:

A) $v\,\Delta i$ exactly B) $2v\sin(\Delta i/2)$ C) $v/\sin(\Delta i)$ D) independent of $v$

Question 8

Plane changes are cheapest when performed:

A) as low and fast as possible B) as high and slow as possible (e.g., at apoapsis) C) exactly at the equator crossing, regardless of speed D) during the launch ascent only

Question 9

A $60^\circ$ plane change costs a delta-v equal to:

A) half the orbital speed B) the full orbital speed $v$ C) twice the orbital speed D) the escape velocity

Question 10

To catch up with a target ahead of you in the same circular orbit, you should first burn:

A) prograde (toward the target) B) retrograde (dropping to a lower, faster orbit) C) radially outward D) normal to the plane

Question 11

A phasing orbit is used to change a spacecraft's:

A) orbital plane B) angular position (phase) relative to a target C) mass ratio D) specific impulse

Question 12

Two impulsive burns done separately cost $|\Delta\mathbf{v}_1| + |\Delta\mathbf{v}_2|$; done as one combined burn they cost:

A) more, always B) $|\Delta\mathbf{v}_1 + \Delta\mathbf{v}_2|$, which is never larger C) exactly the same D) zero

Question 13

In an impulsive prograde burn at perigee, the point of the orbit that rises is:

A) the perigee (the burn point itself) B) the apogee (the point opposite the burn) C) the ascending node D) no point rises

Question 14

The combined speed-change-plus-plane-change burn is computed with:

A) a simple scalar sum B) the law of cosines, $\sqrt{v_a^2 + v_{c2}^2 - 2v_av_{c2}\cos\Delta i}$ C) the rocket equation D) Kepler's third law

Question 15 (True/False, justify)

"Making a rocket launch on a precise azimuth and time is mostly about tradition." True or false? Justify using the cost of plane changes.

Question 16 (True/False, justify)

"A bi-elliptic transfer always saves a lot of delta-v compared with a Hohmann transfer." True or false? Explain.

Question 17 (True/False, justify)

"During the coasting phase of a Hohmann transfer, the spacecraft's specific orbital energy stays constant." True or false? Explain briefly.

Question 18 (Short answer)

A satellite in a $400\ \text{km}$ circular orbit ($v = 7.673\ \text{km/s}$) needs a $10^\circ$ plane change. Compute its cost, and say in one sentence why this is expensive.

Question 19 (Short answer)

Explain, in two sentences, why the Hohmann transfer needs exactly two burns and not one.

Question 20 (Short answer)

State the "combine and defer" rule for the plane change of a GEO satellite, and say roughly how much delta-v it saves compared with doing the circularization and a $28.5^\circ$ plane change separately.

Answer Key

Q Ans Note
1 B Impulsive = instantaneous velocity kick at a point; position unchanged.
2 B Two burns along an ellipse tangent to both circular orbits.
3 B Both burns are prograde (speed-ups) when raising the orbit.
4 B $2.40 + 1.46 \approx 3.86\ \text{km/s}$.
5 B Half the transfer ellipse's period; depends on $a_t = (r_1+r_2)/2$.
6 C Below $\approx 11.94$, Hohmann is always optimal; bi-elliptic needs a large ratio.
7 B $\Delta v = 2v\sin(\Delta i/2)$ from the isosceles velocity triangle.
8 B Cost $\propto v$, so change plane where $v$ is smallest (high, slow).
9 B $2v\sin 30^\circ = v$ — the whole orbital velocity.
10 B Drop to a lower, faster, shorter-period orbit to gain, then re-raise.
11 B A phasing orbit changes the along-track phase relative to a target.
12 B Vectors add as a triangle; the combined magnitude is never larger.
13 B A burn moves the point opposite it; prograde at perigee raises apogee.
14 B Law of cosines on the two velocity vectors separated by $\Delta i$.
15 False It is physics, not tradition: fixing the plane later costs $2v\sin(\Delta i/2)$, huge in LEO, so you must be born into the right plane.
16 False It beats Hohmann only for large ratios ($\gtrsim 11.94$), and even then by little, at a big cost in time.
17 True Energy changes only when the engine fires; during the coast $\varepsilon = -\mu/(2a_t)$ is constant while speed varies.
18 — $\Delta v = 2(7.673)\sin 5^\circ = 15.346 \times 0.08716 = 1.34\ \text{km/s}$. Expensive because the cost scales with the full LEO orbital speed.
19 — One burn only moves the orbit point opposite it: the first burn raises apogee to the target radius but leaves perigee low (an ellipse); a second burn at the new apogee raises perigee to match, rounding it into the target circle.
20 — Defer the plane change to slow-moving GTO apogee and fold it into the circularization burn (law of cosines): $\approx 1.83\ \text{km/s}$ combined versus $1.46 + 1.51 = 2.97\ \text{km/s}$ separate — a saving of about $1.15\ \text{km/s}$.

Topics to review by question

Questions Topic Section
1, 13 Impulsive maneuvers §10.1
2, 3, 4, 5, 19 The Hohmann transfer §10.2
6, 16 Bi-elliptic transfers §10.3
7, 8, 9, 15, 18 Plane changes §10.4
10, 11, 17 Phasing & rendezvous §10.5, §10.2
12, 14, 20 Combined maneuvers §10.6