GELE Geodesy Reviewer
12 Geodesy practice questions for the Geodetic Engineer Licensure Examination, each with the correct answer and an explanation of why it's right.
Geodesy Practice Questions with Answers
- 1easy
Which of the following BEST defines the geoid?
- A.The mathematical ellipsoid that approximates the shape of the Earth
- B.The equipotential surface of Earth's gravity field that best fits global mean sea level
- C.The physical surface of the Earth including mountains and ocean floors
- D.The surface generated by rotating a meridian ellipse around the Earth's axis
Show answer & explanation
Answer: B. The equipotential surface of Earth's gravity field that best fits global mean sea level
Step 1: Recall the three surfaces in geodesy — ellipsoid, geoid, and terrain. Step 2: The ellipsoid is a purely mathematical surface (e.g., WGS84, PRS92). Step 3: The terrain is the physical ground surface. Step 4: The geoid is defined by gravity — specifically, it is the equipotential surface of Earth's gravity field that coincides with mean sea level worldwide. Step 5: Options A and D describe the ellipsoid; Option C describes the terrain. Only Option B correctly defines the geoid.
- 2easy
In geodesy, the relationship among ellipsoidal height (h), orthometric height (H), and geoid undulation (N) is expressed as:
- A.H = h + N
- B.N = h + H
- C.h = H + N
- D.h = H − N
Show answer & explanation
Answer: C. h = H + N
Step 1: The three heights in geodesy are measured from different reference surfaces. Step 2: h (ellipsoidal height) is measured from the ellipsoid upward to the point. Step 3: N (geoid undulation) is the height of the geoid above the ellipsoid. Step 4: H (orthometric height) is measured from the geoid upward to the point. Step 5: Geometrically, h = H + N because the ellipsoidal height equals the orthometric height plus the height of the geoid above the ellipsoid. Rearranging gives H = h − N, which is the working formula for GNSS-to-elevation conversion.
- 3easy
A GNSS receiver directly provides which type of height?
- A.Orthometric height (H)
- B.Dynamic height
- C.Ellipsoidal height (h)
- D.Geopotential number (C)
Show answer & explanation
Answer: C. Ellipsoidal height (h)
Step 1: GNSS (e.g., GPS, GLONASS, Galileo) computes positions using satellite ranging referenced to a mathematical ellipsoid (WGS84 globally; PRS92 for the Philippines). Step 2: The height component output by GNSS processing is therefore the ellipsoidal height h — the vertical distance from the ellipsoidal surface to the point. Step 3: Orthometric height H (the engineering 'elevation above MSL') requires a geoid model to convert from h using H = h − N. Step 4: Dynamic heights and geopotential numbers are derived from spirit leveling plus gravity observations — not directly from GNSS. Step 5: Always remember: raw GNSS height ≠ usable elevation without a geoid model.
- 4easy
A GNSS survey yields an ellipsoidal height h = 52.30 m. The geoid undulation at that point is N = −30.10 m. What is the orthometric height H?
- A.22.20 m
- B.82.40 m
- C.−82.40 m
- D.30.10 m
Show answer & explanation
Answer: B. 82.40 m
Step 1: Write the formula: H = h − N. Step 2: Substitute the given values: H = 52.30 m − (−30.10 m). Step 3: Subtracting a negative number means adding: H = 52.30 + 30.10. Step 4: H = 82.40 m above mean sea level. Step 5: The negative N means the geoid lies below the ellipsoid at this point, so the orthometric height is larger than the ellipsoidal height — a common scenario in parts of the Philippines where N is negative. Watch the double-negative sign carefully on the board exam.
- 5easy
A benchmark has a levelled orthometric height H = 112.20 m and a GNSS ellipsoidal height h = 124.50 m. What is the geoid undulation N at this point?
- A.−12.30 m
- B.236.70 m
- C.12.30 m
- D.112.20 m
Show answer & explanation
Answer: C. 12.30 m
Step 1: Start from h = H + N and rearrange: N = h − H. Step 2: Substitute: N = 124.50 − 112.20. Step 3: N = 12.30 m. Step 4: A positive N means the geoid lies 12.30 m above the ellipsoid at this location. Step 5: This method — using co-located GNSS and leveling benchmarks — is how geoid models are validated in the field. In the Philippines, NAMRIA uses this approach in combination with gravimetry to refine the Philippine Geoid Model.
- 6easy
How does the magnitude of gravity (g) vary from the equator to the poles?
- A.Gravity is constant everywhere on Earth's surface
- B.Gravity decreases from equator to poles
- C.Gravity increases from equator to poles
- D.Gravity increases from equator to 45° latitude, then decreases
Show answer & explanation
Answer: C. Gravity increases from equator to poles
Step 1: Earth is an oblate spheroid — it is flattened at the poles and bulges at the equator. Step 2: At the equator, the distance from the Earth's center to the surface is greater, so gravity is weaker (≈ 9.78 m/s²). Step 3: At the poles, the distance is shorter, so gravity is stronger (≈ 9.83 m/s²). Step 4: Additionally, the centrifugal effect of Earth's rotation reduces the effective gravity at the equator more than at the poles. Step 5: Both effects (shorter radius + less centrifugal reduction) make gravity increase monotonically from equator to poles.
- 7easy
The geoid undulates relative to the ellipsoid primarily because of:
- A.Variations in Earth's mass density distribution
- B.Errors in satellite orbit determination
- C.Tidal forces from the Moon only
- D.Atmospheric pressure differences
Show answer & explanation
Answer: A. Variations in Earth's mass density distribution
Step 1: The geoid is an equipotential surface of gravity. Step 2: Gravity is generated by Earth's mass; where mass concentrations exist (e.g., dense rock, mountain ranges), gravity is stronger and the geoid bulges upward toward those masses. Step 3: Where there is a mass deficiency (e.g., ocean trenches, less-dense crust), the geoid dips. Step 4: The smooth ellipsoid does not account for these internal mass variations, so the geoid undulates ±100 m relative to it. Step 5: Satellite errors, tides, and atmospheric pressure are secondary or unrelated — mass density heterogeneity is the primary cause.
- 8easy
Spirit (differential) leveling directly produces which type of height differences?
- A.Ellipsoidal height differences
- B.Geoid undulation differences
- C.Orthometric height differences (elevations above MSL)
- D.Dynamic height differences only after gravity correction
Show answer & explanation
Answer: C. Orthometric height differences (elevations above MSL)
Step 1: Spirit leveling measures height differences along the equipotential gravity field — the instrument level bubble aligns with the local equipotential surface. Step 2: When referenced to a tide gauge (mean sea level), leveled heights become orthometric heights H — what engineers call 'elevation above MSL.' Step 3: Ellipsoidal heights come from GNSS, not leveling. Step 4: Strictly, raw leveled differences are 'geopotential differences,' and rigorous orthometric heights require a small gravity correction — but for board-exam purposes, spirit leveling gives orthometric heights. Step 5: This is why leveling networks (like NAMRIA's vertical control network) define the national orthometric height datum.
- 9easy
If the geoid lies BELOW the ellipsoid at a given point, the geoid undulation N is:
- A.Positive
- B.Zero
- C.Negative
- D.Indeterminate without additional data
Show answer & explanation
Answer: C. Negative
Step 1: Geoid undulation N is defined as the height of the geoid above the ellipsoid. Step 2: 'Height above' implies measured upward from the ellipsoid to the geoid. Step 3: If the geoid is BELOW the ellipsoid, the geoid is at a negative vertical distance from the ellipsoid — therefore N < 0. Step 4: In the formula H = h − N: if N is negative, then H = h − (negative) = h + |N|, making H larger than h. Step 5: This is the most common board-exam sign trap. Many parts of the Philippines have negative N values — always check the sign before computing H.
- 10easy
In engineering and construction projects in the Philippines, the practical reference surface for 'elevation' is the:
- A.WGS84 ellipsoid
- B.PRS92 ellipsoid
- C.Geoid (mean sea level)
- D.Physical terrain surface
Show answer & explanation
Answer: C. Geoid (mean sea level)
Step 1: Engineering projects require a height reference that is physically meaningful — water flows downhill from higher to lower elevations. Step 2: This behavior is governed by gravity: water flows along equipotential surfaces toward lower potential. Step 3: The geoid, as an equipotential surface approximating mean sea level, is the natural reference for 'elevation above MSL' used in construction, flood mapping, and infrastructure design. Step 4: Ellipsoidal heights (from WGS84 or PRS92) do not reflect gravity-driven flow — two points at the same ellipsoidal height can have water flowing between them if the geoid tilts. Step 5: NAMRIA establishes the Philippine vertical datum using tide gauge observations and leveling networks referenced to the geoid.
- 11easy
What is the minimum number of GNSS satellites required to obtain a three-dimensional position fix?
- A.3
- B.4
- C.5
- D.6
Show answer & explanation
Answer: B. 4
Step 1 – Identify the unknowns: A 3-D GNSS position has four unknowns: the three Cartesian coordinates (X, Y, Z) of the receiver, plus the receiver clock bias (Δt). Step 2 – Equation count: Each satellite provides one pseudorange equation. To solve a system of four unknowns, you need at least four independent equations. Step 3 – Therefore, a minimum of 4 satellites is required. Step 4 – Why not 3? Three satellites would yield three equations, which cannot uniquely solve four unknowns. The clock bias is the critical fourth unknown that students often overlook. Step 5 – Why not more? More satellites improve geometry and accuracy (lower DOP) but 4 is the mathematical minimum.
- 12easy
A GPS signal takes 0.067 s to travel from a satellite to a receiver. What is the pseudorange? (c = 299,792,458 m/s)
- A.20,086,095 m
- B.18,500,000 m
- C.25,000,000 m
- D.4,474,514 m
Show answer & explanation
Answer: A. 20,086,095 m
Step 1 – Recall the pseudorange formula: ρ = c × Δt, where c is the speed of light and Δt is the signal travel time. Step 2 – Substitute values: ρ = 299,792,458 m/s × 0.067 s. Step 3 – Compute: ρ = 299,792,458 × 0.067 = 20,086,094.7 m ≈ 20,086,095 m (~20,086 km). Step 4 – Sanity check: GPS satellites orbit at approximately 20,200 km altitude, so a pseudorange near 20,086 km is physically consistent. Step 5 – Common mistake: Students sometimes use c = 3 × 10⁸ m/s (approximate), which gives 20,100,000 m. Always use the precise value (299,792,458 m/s) in board exam calculations.
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