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GELE practice questions

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Question 1Geodesy and Geodetic Surveying

In the definition of a reference ellipsoid, the flattening f is expressed as:

  1. (a − b) / a Correct answer.
  2. (a − b) / b
  3. (a + b) / a
  4. b / (a − b)

Explanation

Flattening is the ratio of the difference between the semimajor axis a and semiminor axis b to the semimajor axis, so f = (a − b)/a. It is a dimensionless number of the order of 1/300 for the Earth.

Question 2Geodesy and Geodetic Surveying

The standard width in longitude of a UTM zone is:

  1. 8°
  2. 3°
  3. 12°
  4. 6° Correct answer.

Explanation

The Universal Transverse Mercator system divides the globe into zones that are 6° of longitude wide, each with its own central meridian. This keeps the scale distortion within acceptable limits for mapping and survey work.

Question 3Geodesy and Geodetic Surveying

The geoid is best described as:

  1. the reference ellipsoid surface adopted for a datum
  2. the mean topographic surface of the land mass
  3. a sphere whose radius equals the mean radius of the Earth
  4. an equipotential surface of gravity near mean sea level Correct answer.

Explanation

The geoid is the particular equipotential surface of the Earth's gravity field that most closely coincides with mean sea level, extended hypothetically under the continents. All orthometric heights are referred to it.

Question 4Geodesy and Geodetic Surveying

Orthometric height is the vertical distance measured:

  1. from the ellipsoid along the ellipsoid normal
  2. from the geoid along the curved plumb line Correct answer.
  3. from mean sea level along the plumb line at a tide gauge
  4. from the geoid along the ellipsoid normal

Explanation

Orthometric height is the distance from the geoid up to the point measured along the curved plumb line, that is, along the direction of gravity. Ellipsoidal height, by contrast, is measured along the ellipsoid normal.

Question 5Geodesy and Geodetic Surveying

Geodetic latitude at a point on the ellipsoid is the angle between:

  1. the plumb line at the point and the equatorial plane
  2. the radius vector from the geocentre and the equatorial plane
  3. the ellipsoid normal at the point and the equatorial plane Correct answer.
  4. the normal to the ellipsoid and the polar axis

Explanation

Geodetic latitude is defined by the direction of the ellipsoid normal at the point measured from the equatorial plane. It differs from astronomic latitude, which uses the direction of gravity, and from geocentric latitude, which uses the radius vector.

Question 6Geodesy and Geodetic Surveying

Which segment of the GPS as originally designed contains the master control station and the monitor stations?

  1. control segment Correct answer.
  2. space segment
  3. signal segment
  4. user segment

Explanation

The control segment comprises the master control station, monitor stations and upload antennas, which track the satellites and update their broadcast ephemerides and clocks. The space segment is the satellite constellation and the user segment is the receivers.

Question 7Geodesy and Geodetic Surveying

In spirit leveling, the instrument's line of sight is made horizontal by:

  1. setting the vertical circle to zero
  2. measuring the height of the instrument above the peg
  3. rotating the telescope until the crosshair bisects the target
  4. centering the bubble of the level vial in its run Correct answer.

Explanation

A spirit level works because the bubble settles at the highest point of a curved vial; when the bubble is centred in its run, the axis of the vial and hence the line of sight are horizontal. The vertical circle and the target bisection do not control the horizontality of the sight line.

Question 8Geodesy and Geodetic Surveying

A geodetic azimuth is measured:

  1. counter-clockwise from south, 0° to 360°
  2. clockwise from east, 0° to 360°
  3. from the magnetic meridian, clockwise
  4. clockwise from the north, 0° to 360° Correct answer.

Explanation

Azimuth is a horizontal direction reckoned clockwise from a reference meridian. A geodetic azimuth uses the geodetic (or grid) north direction and runs clockwise through 360°.

Question 9Geodesy and Geodetic Surveying

The scale factor on the central meridian of a UTM zone is:

  1. 1.0004
  2. 1.0000
  3. 0.9990
  4. 0.9996 Correct answer.

Explanation

UTM applies a central-meridian scale factor of 0.9996 so that distortion is spread across the zone rather than accumulating; scale equals exactly 1 along two lines either side of the central meridian. This keeps the maximum scale error inside the zone within about one part in a thousand.

Question 10Geodesy and Geodetic Surveying

The deflection of the vertical at a point is the angle between:

  1. two successive equipotential surfaces
  2. the direction of gravity and the ellipsoid normal Correct answer.
  3. the ellipsoid normal and the equatorial plane
  4. the local horizon and the celestial equator

Explanation

The deflection of the vertical expresses how far the actual gravity direction (the plumb line) departs from the ellipsoid normal at the point. It has a north-south and an east-west component and links astronomic and geodetic coordinates.

Question 11Geodesy and Geodetic Surveying

For a total station sighting a target at a distance D in kilometres, with instrument and target at the same height, the combined correction for curvature and atmospheric refraction applied to the observed height difference is approximately:

  1. 0.0675 D² m Correct answer.
  2. 0.0675 D m
  3. 0.132 D² m
  4. 0.0785 D² m

Explanation

Curvature alone lowers the line of sight by about 0.0785 D² metres, and standard refraction raises it by roughly one-seventh of that, leaving a net correction near 0.0675 D² metres. The correction grows with the square of the distance, not linearly.

Question 12Geodesy and Geodetic Surveying

In relative GNSS positioning, differencing the observations of two receivers at the same epoch primarily removes:

  1. satellite clock errors common to both receivers Correct answer.
  2. the geometric range between satellite and receiver
  3. multipath at the rover receiver only
  4. the effect of the ionosphere entirely

Explanation

A between-receiver difference cancels any error that is common to both stations for the same satellite, and the satellite clock offset is exactly such an error. Errors that depend on the individual signal path, such as multipath and the bulk of the ionospheric delay, are not removed.

Question 13Geodesy and Geodetic Surveying

A map projection described as conformal preserves:

  1. areas in correct proportion
  2. angles, so small shapes stay undistorted Correct answer.
  3. distances along the central meridian only
  4. great circle paths as straight lines

Explanation

A conformal projection keeps the scale the same in all directions at a point, so angles are reproduced correctly and small figures keep their shape. The price is that scale varies from place to place, so areas are not preserved.

Question 14Geodesy and Geodetic Surveying

Within a UTM zone, the grid scale factor at a point depends mainly on:

  1. the distance of the point from the central meridian Correct answer.
  2. the latitude of the point only
  3. the elevation of the point above the geoid only
  4. the convergence of the meridians

Explanation

In the transverse Mercator projection the scale grows with the square of the distance from the central meridian, so the point's easting determines the grid scale factor. Elevation introduces a separate scale effect handled by the sea-level or combination factor.

Question 15Geodesy and Geodetic Surveying

A geocentric datum such as WGS 84 has its origin at:

  1. the centre of an ellipsoid aligned to a national origin point
  2. the mean sea surface of the Atlantic Ocean
  3. the mass centre of the Earth Correct answer.
  4. a point on the equator at the Greenwich meridian at sea level

Explanation

A geocentric datum places the origin of its Cartesian coordinate system at the Earth's centre of mass, with the Z axis along the conventional terrestrial pole. Local or regional datums, by contrast, are often optimised to fit the ellipsoid to a particular area.

Question 16Geodesy and Geodetic Surveying

In a national vertical control network, the published elevation of a benchmark is normally a:

  1. depth below the terrain surface from a total station
  2. geoid undulation from a global model
  3. orthometric height on a tide-gauge datum Correct answer.
  4. ellipsoidal height derived from GNSS

Explanation

Levelling networks propagate heights from a reference tide gauge and publish orthometric heights referred to that datum, realised through benchmarks. Ellipsoidal heights require GNSS plus a geoid model to be converted to orthometric heights.

Question 17Geodesy and Geodetic Surveying

The astronomical latitude of a station is determined from the direction of:

  1. the station's radius vector from the Earth's centre
  2. the direction of gravity relative to the celestial pole Correct answer.
  3. the magnetic needle at the station
  4. the ellipsoid normal at the station

Explanation

Astronomical latitude comes from observing celestial bodies whose positions are known, referred to the local direction of gravity as defined by a spirit level or meridian instrument. It therefore reflects the geoid rather than the ellipsoid.

Question 18Geodesy and Geodetic Surveying

In precise geodetic triangulation, the spherical excess of a triangle is used to:

  1. convert a geodetic azimuth to a grid azimuth
  2. convert a measured slope distance into a horizontal distance
  3. reduce a measured distance to the ellipsoid surface
  4. close the triangle by adjusting for the expected angle sum Correct answer.

Explanation

On a curved reference surface the three angles of a triangle sum to more than 180° by an amount equal to the spherical excess. Knowing this quantity lets the surveyor test and adjust the observed angles before computing the network.

Question 19Geodesy and Geodetic Surveying

Which statement about the prime vertical radius of curvature N and the meridian radius of curvature M at latitude φ is correct?

  1. N is constant over the ellipsoid and equal to the semimajor axis a
  2. N equals M at the equator and is smaller towards the poles
  3. N = a(1 − e² sin²φ)^−1/2, which exceeds M except at the poles Correct answer.
  4. M is always greater than N between the equator and the pole

Explanation

The prime vertical radius is N = a(1 − e² sin²φ)^−1/2, while the meridian radius is M = a(1 − e²)(1 − e² sin²φ)^−3/2, so M is smaller than N at every latitude except the poles, where both equal a/(1 − e²)^1/2. This difference matters in reducing geodetic observations.

Question 20Geodesy and Geodetic Surveying

A three-parameter datum transformation between two geocentric datums is defined by:

  1. seven parameters including rotations and scale
  2. a new central meridian and standard parallels
  3. three origin translations and no rotations Correct answer.
  4. three rotations plus a scale factor

Explanation

The three-parameter (geocentric translation) transformation simply shifts the origin by ΔX, ΔY, ΔZ between two datums that are already assumed to be parallel and of the same size. Adding three rotations and a scale factor produces the seven-parameter Helmert or Bursa-Wolf transformation.

Question 21Geodesy and Geodetic Surveying

A horizontal distance measured at mean terrain elevation H is reduced to the ellipsoid by multiplying it by approximately:

  1. H / R
  2. (R + H) / R
  3. R / H
  4. R / (R + H) Correct answer.

Explanation

Distances at height H above the ellipsoid subtend a larger arc than the corresponding ellipsoidal chord, so the reduction factor is R/(R + H), where R is a mean radius of curvature. The correction is small but significant for precise work at high elevation.

Question 22Geodesy and Geodetic Surveying

In Stokes' formula, the geoid undulation N at a point is computed by integrating:

  1. satellite clock offsets relative to a reference station
  2. gravity anomalies over the whole Earth Correct answer.
  3. deflections of the vertical over the whole Earth
  4. terrain elevations within a limited cap around the point

Explanation

Stokes' integral combines gravity anomalies distributed over the entire sphere with the Stokes function, which depends on the spherical distance from the computation point. Deflections of the vertical are obtained instead from the Vening Meinesz integral of the same anomalies.

Question 23Geodesy and Geodetic Surveying

Flattening f of an ellipsoid of revolution with semi-major axis a and semi-minor axis b is defined as:

  1. (a − b)/b
  2. b/a
  3. (a + b)/a
  4. (a − b)/a Correct answer.

Explanation

Flattening is the fractional difference between the equatorial and polar semi-axes, f = (a − b)/a. For the WGS 84 ellipsoid it is about 1/298.257223563.

Question 24Geodesy and Geodetic Surveying

A station has an ellipsoidal height h of 120.0 m and lies where the geoid undulation N is −25.0 m. Its orthometric height H is:

  1. 145.0 m Correct answer.
  2. 95.0 m
  3. −95.0 m
  4. 120.0 m

Explanation

The three heights are related by h = H + N, so H = h − N. Substituting gives 120.0 − (−25.0) = 145.0 m.

Question 25Geodesy and Geodetic Surveying

The longitudinal width of a standard UTM zone is:

  1. 4°
  2. 8°
  3. 6° Correct answer.
  4. 3°

Explanation

UTM divides the Earth into 60 zones, each 6° of longitude wide, numbered from west to east starting at 180°.

Question 26Geodesy and Geodetic Surveying

To keep northings positive in the southern hemisphere, each southern UTM zone is given a false northing of:

  1. 1 000 000 m
  2. 10 000 000 m Correct answer.
  3. 0 m
  4. 500 000 m

Explanation

A false northing of 10 000 000 m is added at the equator in southern zones, so northings increase southward from that value. The 500 000 m false easting handles eastings at the central meridian.

Question 27Geodesy and Geodetic Surveying

The nominal orbital period of a GPS satellite in the operational constellation is about:

  1. 6 sidereal hours
  2. 12 sidereal hours Correct answer.
  3. 36 sidereal hours
  4. 24 sidereal hours

Explanation

GPS satellites are placed in semi-synchronous orbits with a period of roughly 12 sidereal hours, so the ground track repeats about twice per sidereal day. This geometry gives repeated visibility over the same ground area.

Question 28Geodesy and Geodetic Surveying

International Atomic Time (TAI) and Coordinated Universal Time (UTC) differ by:

  1. a constant 0.9 s
  2. the DUT1 value in seconds
  3. an integer number of leap seconds Correct answer.
  4. a fixed 32.184 s

Explanation

UTC is kept within 0.9 s of UT1 by inserting or removing whole leap seconds, so TAI − UTC is always an integer number of seconds (37 s at present). The 32.184 s offset belongs to the relation between TT and TAI.

Question 29Geodesy and Geodetic Surveying

Geometric Dilution of Precision (GDOP) describes:

  1. geometry's effect on position uncertainty Correct answer.
  2. the number of satellites above the elevation mask
  3. the residual ionospheric delay after differencing
  4. the received signal strength at the antenna

Explanation

GDOP is a dimensionless factor derived from the geometry matrix of the satellite constellation; a larger GDOP magnifies the effect of measurement errors on the computed position. Good geometry, with satellites spread widely in the sky, gives a small GDOP.

Question 30Geodesy and Geodetic Surveying

On a Mercator projection, the meridians are drawn as:

  1. concentric circular arcs
  2. equally spaced vertical straight lines Correct answer.
  3. converging straight lines
  4. curved lines meeting at the poles

Explanation

Mercator is a conformal cylindrical projection; meridians map to equally spaced parallel straight lines and parallels to straight horizontal lines. This makes rhumb lines straight but greatly exaggerates scale near the poles.

Question 31Geodesy and Geodetic Surveying

A seven-parameter Helmert transformation between two geocentric datums consists of three translations, three rotations and:

  1. one meridian convergence
  2. one scale factor Correct answer.
  3. one azimuth angle
  4. two scale factors

Explanation

The Helmert (Bursa–Wolf) model uses three translations, three small rotations and a single scale factor to relate Cartesian coordinates in two geocentric datums. The rotations are usually expressed in arcseconds and the scale in parts per million.

Question 32Geodesy and Geodetic Surveying

Grid convergence at a point is the angle between:

  1. the central meridian and the equator
  2. true north and magnetic north
  3. the geoid surface and the ellipsoid
  4. grid north and true north Correct answer.

Explanation

Grid convergence γ is the angle at a point between grid north and true (geodetic) north. It arises because grid lines on a projection are not generally parallel to the meridian.

Question 33Geodesy and Geodetic Surveying

In trigonometric levelling over a distance D expressed in kilometres, the combined curvature and refraction correction to the height difference, in metres, is approximately:

  1. 0.0675 D² metres Correct answer.
  2. 0.0675 D
  3. 0.0675 D² radians
  4. 0.0675 D² seconds

Explanation

Curvature lowers the target by about D²/(2R) and refraction raises the line of sight by about one-seventh of that amount. Combining them gives approximately 0.0675 D² metres when D is in kilometres.

Question 34Geodesy and Geodetic Surveying

Reciprocal trigonometric levelling, observing from both ends of a line, is used chiefly to eliminate:

  1. errors in the vertical circle index
  2. errors in the measured horizontal distance
  3. instrument height measurement errors
  4. refraction and curvature effects Correct answer.

Explanation

Simultaneous reciprocal observations give nearly equal refraction at both ends, and averaging the two height differences cancels the curvature and refraction terms. The method also reduces the influence of vertical-angle index error when the observations are combined properly.

Question 35Geodesy and Geodetic Surveying

The spherical excess of a triangle of area A on a sphere of radius R is:

  1. 2A/R²
  2. A/R
  3. A·R
  4. A/R² Correct answer.

Explanation

Spherical excess is E = A/R², expressed in radians, and it is added to the sum of the observed angles to close the triangle on the sphere. For a small triangle this quantity is very small, which is why it matters mainly in precise geodetic work.

Question 36Geodesy and Geodetic Surveying

Between two points on the same parallel of latitude on an ellipsoid, the shortest surface path is generally:

  1. a geodesic that bends toward the equator
  2. a geodesic that bends toward the pole Correct answer.
  3. a straight line in any conformal projection
  4. along the parallel itself

Explanation

A geodesic on an oblate ellipsoid does not follow a parallel except at the equator. Between two points at the same latitude, the geodesic reaches its maximum latitude between them and lies closer to the pole than the parallel.

Question 37Geodesy and Geodetic Surveying

The ionospheric delay experienced by a GNSS signal is approximately proportional to:

  1. the square of the frequency f²
  2. the signal frequency f
  3. the inverse of the frequency 1/f
  4. inverse square of frequency, 1/f² Correct answer.

Explanation

Ionospheric delay is dispersive and varies as 1/f², so dual-frequency observations can be combined to remove it to first order. This is why the L1/L2 ionosphere-free combination is widely used.

Question 38Geodesy and Geodetic Surveying

The wide-lane combination of the GPS L1 and L2 carrier frequencies has a wavelength of approximately:

  1. 24.4 cm
  2. 19.0 cm
  3. 10.7 cm
  4. 86.2 cm Correct answer.

Explanation

The wide-lane frequency is f1 − f2 ≈ 347.82 MHz, giving a wavelength of about 0.862 m. This long wavelength makes wide-lane ambiguities easier to resolve than the individual L1 or L2 ambiguities.

Question 39Geodesy and Geodetic Surveying

The geopotential number of a benchmark is:

  1. work raising unit mass from the geoid Correct answer.
  2. gravity reduced to sea level at the point
  3. ellipsoidal height minus geoid undulation
  4. orthometric height times mean gravity

Explanation

Geopotential number is the difference in gravity potential between the geoid and the point, equal to the work needed to lift unit mass along the plumb line. It is independent of the path taken and provides the basis for orthometric and dynamic heights.

Question 40Geodesy and Geodetic Surveying

In precise spirit levelling along a long north–south line, an orthometric correction is needed because:

  1. collimation drifts during the run
  2. staff expands with temperature
  3. curvature alters the staff reading
  4. gravity varies with latitude Correct answer.

Explanation

Level surfaces are not parallel: gravity changes with latitude, so equal levelling increments do not correspond to equal potential differences. The orthometric correction accounts for the mean gravity along each plumb line and converts levelled differences to orthometric height differences.

Question 41Geodesy and Geodetic Surveying

At geodetic latitude φ on the ellipsoid, the geocentric latitude ψ satisfies:

  1. tan ψ = tan φ
  2. tan ψ = e² tan φ
  3. tan ψ = (1 + e²) tan φ
  4. tan ψ = (1 − e²) tan φ Correct answer.

Explanation

The geocentric latitude is the angle of the radius vector from the ellipsoid centre, and for an oblate ellipsoid it is smaller in magnitude than the geodetic latitude. The exact relation is tan ψ = (1 − e²) tan φ.

Question 42Geodesy and Geodetic Surveying

On a Lambert conformal conic projection, the cone constant n is selected so that:

  1. the central meridian is the only straight meridian
  2. scale is true on both standard parallels Correct answer.
  3. scale is true along the equator
  4. meridians become parallel straight lines

Explanation

A secant cone intersects the ellipsoid along two standard parallels, and n is chosen so that the projection scale equals one on both. Between and outside the standard parallels the scale departs from unity but remains conformal.

Question 43Geodesy and Geodetic Surveying

Converting Earth-centred, Earth-fixed Cartesian coordinates to geodetic latitude and ellipsoidal height normally requires iteration because:

  1. the ellipsoid is not a surface of revolution
  2. the geoid undulation must be known first
  3. latitude is implicit in the curvature radius Correct answer.
  4. longitude cannot be computed from X and Y

Explanation

The conversion involves the prime vertical radius of curvature N, which itself depends on the geodetic latitude, so latitude appears on both sides of the equation. A closed-form solution exists, but the standard approach is iteration until the latitude converges.

Question 44Geodesy and Geodetic Surveying

A station coordinate in ITRF2014 refers to epoch 2010.0. To obtain the coordinate at epoch 2020.0, one applies:

  1. a Helmert transformation between frames
  2. velocity times the 10-year interval Correct answer.
  3. a polar motion correction only
  4. a change of map projection

Explanation

For a fixed reference frame, coordinates change linearly with time because of tectonic plate motion, so the coordinate at a new epoch is the reference coordinate plus velocity times the epoch difference. The station velocity is part of the ITRF solution.

Question 45Geodesy and Geodetic Surveying

According to Clairaut's theorem for the ellipsoid, the gravity flattening β is related to the geometrical flattening f and the ratio m by:

  1. β = (5/2)m − f Correct answer.
  2. β = f − (5/2)m
  3. β = f + m
  4. β = (5/2)m + f

Explanation

Clairaut's equation gives β = (5/2)m − f, where m = ω²a/γ_e measures centrifugal acceleration relative to equatorial gravity. The relation links the ellipsoid's shape, rotation and gravity field.

Question 46Geodesy and Geodetic Surveying

The two-peg test performed with a level is intended to determine:

  1. the sensitivity of the bubble tube
  2. the magnifying power of the telescope
  3. collimation error of the sight line Correct answer.
  4. the error of the horizontal axis

Explanation

The two-peg test compares staff readings taken at equal and unequal sight distances, so curvature and refraction cancel and any residual difference reveals collimation error. Adjusting the crosshair then removes that error.

Question 47Geodesy and Geodetic Surveying

Satellite altimetry determines sea surface height primarily by:

  1. detecting the geoid with a shipborne gravimeter
  2. radar ranging from the satellite to the sea Correct answer.
  3. tracking tide gauges by radio
  4. measuring gravity gradients from orbit

Explanation

An altimeter measures the round-trip travel time of a radar pulse to the sea surface, giving the range from a precisely known orbit. Subtracting this range from the orbital height yields the sea surface height above the reference ellipsoid.

Question 48Geodesy and Geodetic Surveying

Which surface is defined as the equipotential surface of the Earth's actual gravity field that best coincides with mean sea level?

  1. The quasi-geoid of Molodensky
  2. The geoid Correct answer.
  3. The reference ellipsoid adopted for the datum
  4. The sphere of mean Earth radius

Explanation

The geoid is the particular level surface of the real gravity field that coincides with mean sea level over the oceans and is continued mathematically beneath the continents; orthometric heights are reckoned from it. A reference ellipsoid is only a mathematical approximation to that surface and carries no gravity meaning in itself.

Question 49Geodesy and Geodetic Surveying

The UTM projection uses a central-meridian scale factor of 0.9996. A grid distance of 10 000.000 m measured along the central meridian corresponds most nearly to an ellipsoidal distance of:

  1. 9 996.0 m
  2. 10 000.0 m
  3. 10 004.0 m Correct answer.
  4. 10 040.0 m

Explanation

Grid distances are obtained from ellipsoidal distances by multiplying by the point scale factor k, so the ellipsoidal distance is recovered by dividing the grid distance by k. With k = 0.9996 on the central meridian, 10 000.000 / 0.9996 gives about 10 004.0 m.

Question 50Geodesy and Geodetic Surveying

At a station the astronomic latitude is 39°15′20.6″ and the geodetic latitude is 39°15′17.9″. The meridian (north–south) component ξ of the deflection of the vertical is:

  1. +2.7′
  2. +2.7″ Correct answer.
  3. −2.7″
  4. +0.5″

Explanation

The meridian component of the deflection of the vertical is the astronomic latitude minus the geodetic latitude, ξ = Φ − φ, taken positive when the astronomic zenith lies north of the ellipsoid normal. Here Φ exceeds φ by 2.7 seconds of arc, so ξ = +2.7″.