Chapter 9 Quiz — Orbit Types and Their Uses
Twenty quick checks on the orbit catalog and the reasoning behind it. Mix of multiple choice, true/false-with-justification, and short answer. Target: 16 / 20. Below that, reread the flagged sections in the "Topics to review" map before moving on to Chapter 10. Answer from memory first; a justification you cannot state is a fact you have not yet learned.
Multiple choice
1. Low Earth orbit is usually taken to span altitudes of roughly: - (a) $50$–$160\ \text{km}$ - (b) $160$–$2{,}000\ \text{km}$ - (c) $2{,}000$–$20{,}000\ \text{km}$ - (d) $20{,}000$–$36{,}000\ \text{km}$
2. A GPS satellite completes one orbit in about: - (a) $90\ \text{minutes}$ - (b) $6\ \text{hours}$ - (c) $12\ \text{hours}$ - (d) $24\ \text{hours}$
3. The geostationary altitude (height above the equator) is closest to: - (a) $20{,}200\ \text{km}$ - (b) $35{,}786\ \text{km}$ - (c) $42{,}164\ \text{km}$ - (d) $384{,}000\ \text{km}$
4. A satellite must be seen from a fixed ground dish that never moves. The correct orbit is: - (a) a $500\ \text{km}$ polar orbit - (b) a Molniya orbit - (c) a geostationary orbit - (d) a sun-synchronous orbit
5. The geostationary period is set equal to one sidereal day ($86{,}164\ \text{s}$) rather than the $86{,}400\ \text{s}$ solar day because: - (a) the satellite must keep pace with Earth's rotation relative to the stars, not the Sun - (b) the solar day is not exactly $24$ hours on any given date - (c) atmospheric drag slows the satellite by four minutes per day - (d) the sidereal day gives a rounder altitude number
6. The classic Molniya orbit uses an inclination of $63.4^\circ$ because at that inclination: - (a) the orbit becomes sun-synchronous - (b) Earth's oblateness stops rotating the orbit's apogee away from the north - (c) the period becomes exactly $24\ \text{hours}$ - (d) the satellite passes directly over the North Pole
7. A sun-synchronous orbit typically has an inclination near: - (a) $0^\circ$ - (b) $28^\circ$ - (c) $55^\circ$ - (d) $98^\circ$
8. Fly a camera to a higher orbit and, all else equal, you gain __ and lose ____: - (a) resolution; coverage - (b) coverage; resolution - (c) coverage; radiation tolerance - (d) delta-v; period
9. Satellite-navigation systems (GPS, Galileo) are placed in MEO rather than LEO mainly because MEO: - (a) is closer to the ground, improving accuracy - (b) lets a small constellation keep four-plus satellites in view everywhere - (c) is below the radiation belts - (d) has a $24$-hour period
10. A geostationary transfer orbit (GTO) is best described as: - (a) a circular orbit at half the GEO altitude - (b) an ellipse with perigee in LEO and apogee at GEO altitude - (c) a polar orbit used to reach GEO - (d) the graveyard orbit above GEO
11. At end of life, a geostationary satellite is usually disposed of by: - (a) de-orbiting it into the atmosphere - (b) leaving it in its operational slot - (c) boosting it a few hundred km into a graveyard orbit above GEO - (d) lowering it into MEO
12. Which quantity is the same for a GPS satellite and a Molniya satellite? - (a) eccentricity - (b) inclination - (c) semi-major axis (and hence period) - (d) apogee altitude
True / false — justify in one sentence
13. A geostationary orbit can have any inclination and still keep the satellite fixed in the sky.
14. A higher orbit always means a faster-moving satellite.
15. Any polar orbit is automatically sun-synchronous.
16. A satellite in MEO generally faces a harsher radiation environment than one in LEO.
17. Three geostationary satellites spaced $120^\circ$ apart give complete coverage of the entire Earth, poles included.
18. The delta-v to raise a retiring satellite from GEO into its graveyard orbit is only on the order of tens of meters per second.
Short answer
19. In one or two sentences, explain why your phone needs to receive signals from four GPS satellites — not three — to compute its position.
20. State, in one sentence each, the essential difference between a polar orbit and a sun-synchronous orbit — what does each one buy the mission?
Answer key
| # | Answer | One-line rationale |
|---|---|---|
| 1 | b | LEO runs from ~160 km (above which drag is survivable) to ~2,000 km (below the inner belt). |
| 2 | c | GPS is semi-synchronous: half a sidereal day, ~11 h 58 min. |
| 3 | b | Altitude $35{,}786\ \text{km}$; the $42{,}164\ \text{km}$ in (c) is the radius. |
| 4 | c | Only a geostationary satellite appears motionless, so a fixed dish can be aimed once. |
| 5 | a | The satellite must match Earth's true rotation (against the stars); the sidereal day is that rotation. |
| 6 | b | $63.4^\circ$ is the critical inclination where J2 apsidal precession vanishes, freezing the apogee. |
| 7 | d | Sun-synchronous orbits are slightly retrograde, ~$98^\circ$, to precess eastward with the Sun. |
| 8 | b | Higher = wider footprint (more coverage) but greater distance (worse resolution). |
| 9 | b | MEO's altitude lets ~24 satellites guarantee 4-in-view globally; LEO would need hundreds. |
| 10 | b | GTO is the transfer ellipse: low perigee, GEO-altitude apogee. |
| 11 | c | GEO is too high to de-orbit cheaply, so retiring satellites climb to a graveyard orbit. |
| 12 | c | Same period (half a sidereal day) means the same $a \approx 26{,}560\ \text{km}$; their shapes differ. |
| 13 | False | Only an equatorial ($i=0^\circ$), circular geosynchronous orbit is truly stationary; a tilted one traces a figure-eight. |
| 14 | False | Higher is slower ($v_{\text{circ}}=\sqrt{\mu/r}$ falls with $r$); GEO's $3.07$ km/s is well under LEO's $7.67$. |
| 15 | False | Polar means $i\approx90^\circ$ (global reach); sun-synchronous additionally requires the plane to precess once/year (constant lighting). |
| 16 | True | MEO spends its life among the Van Allen belts; LEO sits mostly beneath them. |
| 17 | False | GEO reaches only to ~$81^\circ$ latitude; the poles are left out — the gap Molniya orbits fill. |
| 18 | True | Raising the orbit ~300 km costs only ~$11\ \text{m/s}$, a small reserved final burn. |
| 19 | — | Three distances fix a point if the clock is perfect; the receiver's clock error is a fourth unknown, so a fourth satellite is needed to solve for it. |
| 20 | — | Polar buys where (near-$90^\circ$ inclination → global coverage); sun-synchronous buys when/under what light (plane precesses once/year → constant local solar time). |
Topics to review by question
| If you missed… | Review |
|---|---|
| 1, 8 | §9.1 Low Earth Orbit (altitude range; coverage-resolution trade) |
| 2, 9, 16 | §9.2 Medium Earth Orbit and navigation |
| 3, 4, 5, 10, 11, 18 | §9.3 Geostationary orbit, GTO, and the graveyard orbit |
| 6, 12, 17 | §9.4 Highly elliptical and Molniya orbits |
| 7, 15, 20 | §9.5 Sun-synchronous and polar orbits |
| 13, 14 | §9.3 and Chapter 6 ("higher is slower") |
| 19 | §9.2 (the four-satellite fix) and Chapter 26 |
If you scored $16$ or better, you can reason from a mission to its orbit — you are ready to learn to move between orbits in Chapter 10. If not, the map above points you to the two or three sections that will close the gap fastest.