200-Ton Master Celestial Navigation Practice Questions

Sample questions — Celestial Navigation

Drawn from the same bank used on USCG licensing exams. Correct answers and explanations are shown — read every explanation, even for questions you get right.

1. 1. The Nautical Almanac tabulates the positions of celestial bodies against which time standard?

   • A.Universal Time (UT1), formerly called Greenwich Mean Time✓
   • B.Local apparent time at the observer's meridian
   • C.Zone time of the vessel's time zone
   • D.Local mean time at 180° longitude

   Why: The Almanac is entered with UT (UT1, the modern successor to GMT). The navigator must convert the watch/zone time of an observation to UT before extracting GHA and declination.

2. 2. The 'd' correction in the Nautical Almanac is applied to the declination, and its sign:

   • A.Is always added
   • B.Is always subtracted
   • C.Depends on the time of day
   • D.Depends on whether declination is increasing or decreasing✓

   Why: The d value gives the hourly rate of change of declination; the d correction is added if declination is increasing toward the next hourly value and subtracted if it is decreasing. The navigator determines the sign by inspecting consecutive hourly declinations.

3. 3. If the index error of a sextant is 'on the arc,' the correction is:

   • A.Added to the sextant reading
   • B.Doubled and then added
   • C.Subtracted from the sextant reading✓
   • D.Ignored because it cancels out

   Why: The rule is 'if it's on, it's off; if it's off, it's on.' Index error on the arc is subtracted; off the arc is added. Index error is found by setting the horizon in coincidence and reading the arc.

4. 4. A single celestial observation by itself provides:

   • A.A complete fix
   • B.Latitude only, in all cases
   • C.Longitude only, in all cases
   • D.One line of position; at least two crossing LOPs are needed for a fix✓

   Why: One sight yields one LOP — the vessel is somewhere on that line. Two or more LOPs from different bodies (or the same body at different times, advanced as a running fix) must cross to establish a fix.

5. 5. An azimuth observation determines compass error by:

   • A.Measuring the altitude of the body and comparing it to Hc
   • B.Timing how long the body takes to set
   • C.Counting the number of stars visible at twilight
   • D.Comparing the body's computed true azimuth with its observed compass bearing✓

   Why: The true azimuth (Zn) of a body is computed (or taken from azimuth tables) for the time of observation. The difference between Zn and the compass bearing of the body is the compass error — east if the compass reads low, west if it reads high.

6. 6. A traditional celestial 'day's work' at sea is built around which sequence of observations?

   • A.Continuous Sun sights taken only at midnight
   • B.A single Polaris sight that fixes the vessel for 24 hours
   • C.Three meridian altitudes of the Moon
   • D.Morning star/Sun sights, a Sun line, latitude at LAN, an afternoon Sun line, and evening star sights✓

   Why: The classic day's work combines morning observations, a forenoon Sun line advanced to the noon latitude (a running fix at LAN), an afternoon Sun line, and an evening round of star sights — giving fixes spread across the day to control the DR.

7. 7. The zenith distance of a celestial body is:

   • A.90° minus the altitude of the body✓
   • B.90° plus the altitude of the body
   • C.The complement of the body's declination
   • D.Equal to the body's polar distance

   Why: Zenith distance (z) = 90° − Altitude (Hc or Ho). It is the angular distance measured from the observer's zenith to the body along a vertical circle, and is used directly in the navigational triangle.

8. 8. A star has a declination of N 45°. An observer at latitude N 45° would see this star pass through the zenith when:

   • A.The star transits the observer's upper meridian✓
   • B.The star rises on the eastern horizon
   • C.The star's LHA equals 090°
   • D.The star's SHA equals the observer's longitude

   Why: A celestial body passes through the observer's zenith when its declination equals the observer's latitude. The zenith passage occurs at upper transit (LHA = 0°) when the body crosses the observer's meridian.

9. 9. An observer at longitude 120°W observes a star. GHA Aries = 200°, and the star's SHA = 146°. What is the star's LHA?

   • A.226°✓
   • B.346°
   • C.106°
   • D.466°

   Why: GHA star = GHA Aries + SHA = 200° + 146° = 346°. LHA = GHA − West longitude = 346° − 120° = 226°. This two-step process (GHA star first, then LHA) is standard for star sights.

10. 10. The GHA of the sun increases at an approximate rate of:

    • A.15° per hour✓
    • B.1° per hour
    • C.0.25° per hour
    • D.360° per day uniformly

    Why: The sun's GHA increases at approximately 15° per hour (360° in 24 hours), matching Earth's rotation rate. The exact rate varies slightly throughout the year due to the equation of time.

11. 11. At what times of year does the sun's declination reach its maximum value of approximately N 23° 27'?

    • A.Around June 21 (summer solstice)✓
    • B.Around March 21 (vernal equinox)
    • C.Around December 21 (winter solstice)
    • D.Around September 23 (autumnal equinox)

    Why: The sun's declination reaches its maximum northerly value of approximately N 23° 27' around June 21, the summer solstice, when the Northern Hemisphere is tilted most toward the sun. It reaches S 23° 27' around December 21.

12. 12. When using the Nautical Almanac Increments and Corrections tables for 45 minutes, what is the GHA increment for the sun?

    • A.11° 15.0'✓
    • B.10° 45.0'
    • C.11° 00.0'
    • D.12° 00.0'

    Why: At 15° per hour, 45 minutes equals 0.75 hours × 15° = 11° 15.0'. The Increments and Corrections table for 45m 00s in the SUN/PLANETS column confirms 11° 15.0'.

13. 13. An observer at longitude 75° 30.0' W measures the sun's GHA as 310° 15.0'. What is the LHA of the sun?

    • A.234° 45.0'✓
    • B.385° 45.0'
    • C.25° 45.0'
    • D.130° 15.0'

    Why: LHA = GHA − West longitude = 310° 15.0' − 75° 30.0' = 234° 45.0'. Since GHA exceeds the west longitude, the result is positive and valid without adding 360°.

14. 14. If the observed altitude (Ho) is 34°18.4' and the computed altitude (Hc) is 34°12.1', the intercept is plotted:

    • A.6.3 nautical miles TOWARD the geographical position of the body✓
    • B.6.3 nautical miles AWAY from the geographical position of the body
    • C.18.4 nautical miles TOWARD the geographical position of the body
    • D.12.1 nautical miles AWAY from the geographical position of the body

    Why: By the Marcq St. Hilaire intercept method (Bowditch), the intercept equals Ho − Hc = 6.3', which at 1' of arc per nautical mile is 6.3 nm; because Ho (34°18.4') is greater than Hc (34°12.1'), the body is higher than computed so the observer is closer to its GP and the line is plotted TOWARD the GP. Remember it as 'HoMoTo' (Ho More Toward); the inverse, 'Computed Greater Away' (C-G-A / 'Coast Guard Academy'), covers the case where Hc is greater.

15. 15. A vessel is in West longitude. The GHA of the sun is 312°45.2'. The assumed longitude is 47°45.2'W. What is the LHA?

    • A.265°00.0'✓
    • B.360°30.4'
    • C.312°45.2'
    • D.000°30.4'

    Why: In West longitude, LHA = GHA − assumed west longitude. LHA = 312°45.2' − 47°45.2' = 265°00.0'. The assumed longitude is chosen to make LHA a whole degree.

16. 16. A morning sun sight LOP is advanced to the time of a noon latitude observation to obtain a running fix. The reliability of this fix depends primarily on:

    • A.The accuracy of the course and speed used to advance the earlier LOP over the elapsed time✓
    • B.The sun's declination being greater than the vessel's latitude
    • C.The two LOPs being perpendicular to each other
    • D.Applying a correction for the equation of time

    Why: In a running fix, the advanced LOP is only as good as the DR track used to move it; errors in course steered or speed made good accumulate over the elapsed time and directly affect the accuracy of the resulting fix.

17. 17. Which error in a running fix cannot be eliminated by retaking the celestial observations?

    • A.Error in the course or speed used to advance the first LOP✓
    • B.Index error of the sextant
    • C.Incorrect dip correction
    • D.Misidentification of the celestial body

    Why: Errors in the DR track used to advance the LOP—whether from leeway, current, helmsman's error, or inaccurate log reading—accumulate with time and cannot be corrected by improved celestial observations; only accurate ship's movement data can improve the running fix.

18. 18. When shooting the lower limb of the sun, the semidiameter correction is:

    • A.Added to the apparent altitude✓
    • B.Subtracted from the apparent altitude
    • C.Added only when the sun is above 20° altitude
    • D.Not applied when using the lower limb

    Why: When observing the lower limb, you measure to the bottom edge of the sun, so the semidiameter (approximately 16') must be added to bring the altitude to the sun's center. The Nautical Almanac altitude correction tables incorporate this.

19. 19. The 'rocking' technique used when taking a sextant sight involves:

    • A.Rotating the sextant about the line of sight to ensure the body is at its lowest point, confirming the arc is in the vertical plane✓
    • B.Moving the index arm back and forth to bracket the body between two readings
    • C.Tilting the sextant forward and backward to average out dip errors
    • D.Swinging the telescope from side to side to locate the body in the field of view

    Why: Rocking (tilting) the sextant about the telescope axis causes the body to swing in an arc; the lowest point of that arc is the true altitude, confirming the sextant is held in the true vertical plane.

20. 20. Which statement BEST describes the advantage of observing a star versus the sun with a sextant?

    • A.Stars appear as point sources, allowing simultaneous measurement of multiple bodies for a fix without semidiameter correction✓
    • B.Stars require no refraction correction because they are at infinite distance
    • C.Star sights are more accurate at night because refraction is reduced
    • D.Stars have a larger horizontal parallax, making them easier to measure precisely

    Why: Stars are effectively at infinite distance and appear as point sources, so no semidiameter correction is needed; multiple star sights can be taken in rapid succession to obtain a multi-body fix, which is geometrically superior to a single sun line.

Frequently asked questions

Is Celestial Navigation on the 200-Ton Master exam?

Yes — Celestial Navigation is one of the tested modules on the 200-Ton Master licensing exam. Candidates must score 70% on each module to pass.

How many Celestial Navigation questions are on the 200-Ton Master exam?

The USCG draws from a bank of 158 Celestial Navigation questions across all exams. The exact number on any single sitting varies, but Rules of the Road is typically the largest module and has the highest passing threshold (90%).

What is the best way to study Celestial Navigation for the 200-Ton Master exam?

Work through the practice questions in this bank until you can answer them consistently above the passing threshold. Review every explanation — understanding why the wrong answers are wrong matters more than memorizing facts.