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GELE Geodesy Reviewer 2026

12 Geodesy practice questions for the Geodetic Engineer Licensure Examination, each with the correct answer and an explanation of why it's right.

269 Geodesy questions in the bank

Geodesy Practice Questions with Answers

  1. 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.

  2. 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.

  3. 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.

  4. 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.

  5. 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.

  6. 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.

  7. 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.

  8. 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.

  9. 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.

  10. 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.

  11. 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.

  12. 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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