Which description of the geoid is correct?

The equipotential surface of the gravity field that best fits mean sea level, perpendicular to the plumb line everywhere

The surface of the best-fitting ellipsoid of revolution, defined only by a semi-major axis and a flattening

The topographic surface of the continents together with the sea floor, smoothed over a few kilometres

The equipotential surface of the gravity field that best fits mean sea level, perpendicular to the plumb line everywhere

The surface on which the value of gravity is the same everywhere, which therefore lies closer to the centre at the poles

Two benchmarks have the same GPS height above the WGS 84 ellipsoid, yet water in a canal between them flows from one to the other. What is the best explanation?

Water follows gravity potential, and the geoid undulation N differs between the two places, so their orthometric heights differ

GPS heights carry a random error of several metres, which hides the true slope of the canal and reverses the apparent flow

Water follows gravity potential, and the geoid undulation N differs between the two places, so their orthometric heights differ

The ellipsoid is flattened at the poles, so every point at one latitude has a different height from every point at another

The gravitational pull of the water itself tilts the plumb line and so shifts the benchmark vertically by a measurable amount

Which of the following fully defines a geodetic datum?

A reference ellipsoid (a and f) together with the position of its centre and the orientation of its axes

Only the semi-major axis and the flattening of the reference ellipsoid, quoted to enough digits

The geoid model that converts ellipsoidal heights into orthometric heights in a given country or region

The map projection used to flatten the surface of the ellipsoid onto a printed or on-screen map sheet

A surveyor plots GPS positions, which are in WGS 84, directly onto an old map drawn on a regional (Everest) datum without any transformation. What is the expected result?

Points shifted by about the same amount in about the same direction, since the ellipsoids differ in size and centre

Random scatter of the plotted points about their true places, because GPS noise is much larger than any datum difference

An error only in the height, because the two datums assign the same horizontal coordinates to every point

Points shifted by about the same amount in about the same direction, since the ellipsoids differ in size and centre

No visible error, because a change of datum moves positions only by a few centimetres on any ordinary map

A GPS receiver needs signals from at least four satellites, although the range to three satellites already fixes a point in space. Why?

The receiver clock has an unknown offset, which is a fourth unknown beside x, y and z

Four satellites are needed so that at least one is always above the horizon of the receiver

The receiver clock has an unknown offset, which is a fourth unknown beside x, y and z

Each satellite broadcasts only one coordinate of its position, so four signals give the full position

The fourth satellite is used to measure the geoid undulation at the receiver location

Dual-frequency GPS receivers remove most of the ionospheric error because

the ionospheric delay varies with frequency as 1/f², so two measurements allow it to be solved for and removed

the troposphere delays both frequencies equally, so the difference between them cancels every atmospheric error at once

the second frequency passes through the ionosphere without any delay and so acts as an exact range reference

the satellite clock error is larger on L2 than on L1 and so subtracts out of the combination of the two

the ionospheric delay varies with frequency as 1/f², so two measurements allow it to be solved for and removed

Measured arcs show that one degree of latitude is about 110.6 km near the equator and about 111.7 km near the poles. What does this show about the Earth?

The Earth is flattened at the poles, because a longer degree means a flatter curve of larger radius

The Earth is elongated at the poles, because a longer degree means a more sharply curved surface there

The Earth is a perfect sphere, because the difference is smaller than the error of the survey measurements

The Earth rotates faster near the poles, which stretches the surface along the meridian and lengthens a degree

Which of the following statements about the geoid and the reference ellipsoid are correct?

The geoid is an equipotential surface of gravity, whereas the reference ellipsoid is a smooth mathematical surface

The geoid is irregular because the Earth’s mass distribution is not perfectly symmetric

The geoid undulation obeys N = h − H, with h the ellipsoidal height and H the orthometric height

The geoid is an equipotential surface of gravity, whereas the reference ellipsoid is a smooth mathematical surface

The geoid is irregular because the Earth’s mass distribution is not perfectly symmetric

The geoid undulation obeys N = h − H, with h the ellipsoidal height and H the orthometric height

The geoid coincides exactly with the best-fitting global ellipsoid at every point, so N is zero

Which of the following statements about GPS positioning are correct?

A pseudorange differs from the geometric range by c times the receiver clock offset plus atmospheric delays and noise

Satellites spread widely across the sky give a low DOP and so a smaller position error for the same range error

The ionospheric delay is larger on the lower-frequency signal, in proportion to 1/f²

A pseudorange differs from the geometric range by c times the receiver clock offset plus atmospheric delays and noise

Satellites spread widely across the sky give a low DOP and so a smaller position error for the same range error

The ionospheric delay is larger on the lower-frequency signal, in proportion to 1/f²

The height reported by GPS is the orthometric height above the geoid, so it needs no further correction for mapping

Which of the following statements about normal gravity γ(φ) are correct?

γ increases from the equator, about 9.780 m/s², to the poles, about 9.832 m/s²

Both the centrifugal acceleration and the flattening of the Earth contribute to the increase with latitude

The north–south gradient of γ, proportional to sin 2φ, is greatest at 45° latitude

γ increases from the equator, about 9.780 m/s², to the poles, about 9.832 m/s²

Both the centrifugal acceleration and the flattening of the Earth contribute to the increase with latitude

The north–south gradient of γ, proportional to sin 2φ, is greatest at 45° latitude

The north–south gradient of γ is greatest at the poles, where the Earth is flattest

Geodesy: Gravity Field, Shape of the Earth, Geoid, Datums and GPS

This chapter answers the heading “Geodesy” of Part B2 of the Geology and Geophysics (GG) paper: “Gravitational Field of the Earth; Clairaut’s theorem, size and shape of Earth; Geoid; Ellipsoid; Geodetic Reference Systems; Datum; GPS.” It teaches the gravity potential and equipotential surfaces, the reference ellipsoid with its flattening, eccentricity and radii of curvature, normal gravity and Clairaut’s theorem, the geoid and its separation from the ellipsoid, datums and reference systems, and how GPS finds a position from pseudoranges. The numericals are eccentricity from the axes, normal gravity at a latitude, Clairaut’s relation, the length of a degree of longitude, geoid undulation from heights, and the pseudorange clock bias.

1. The gravitational field of the Earth: potential, equipotentials and the plumb line

Gravity on the rotating Earth is the sum of Newtonian attraction and the centrifugal acceleration, and both derive from one gravity potential W = V + Φ. For a sphere of mass M, V = GM/r, and the centrifugal part is Φ = ½ ω² r² cos²φ at latitude φ. Gravity is g = −∇W, so it is perpendicular to every equipotential surface (W = constant), and the plumb line follows its direction. Where equipotential surfaces crowd together g is large. Useful constants are GM = 3.986×10¹⁴ m³/s², ω = 7.292×10⁻⁵ rad/s and a mean g of about 9.8 m/s².

Expanded in spherical harmonics, the potential is dominated by GM/r. With the origin at the centre of mass the degree-1 terms vanish, and the largest remaining term is the degree-2 zonal coefficient J₂ ≈ 1.08×10⁻³, a measure of the equatorial bulge. To first order J₂ = (2f − m)/3, with m = ω²a³/GM ≈ 1/289, so a satellite-derived J₂ gives the flattening f ≈ (3J₂ + m)/2 ≈ 3.35×10⁻³. Satellites such as GRACE (time-variable gravity) and GOCE (gravity gradients) and global models such as EGM2008 (to degree and order 2159) map the field to fine detail.

🎯 Why the field is not that of a sphere

Rotation flings material outward, so the Earth settles into an oblate shape in which the surface is a level (equipotential) surface of W. Gravity on that surface is then not constant: it depends on the distance from the centre and on the centrifugal term, both of which change with latitude. That is the origin of the normal-gravity formula of Section 3.

2. Size and shape of the Earth: ellipsoid, flattening and radii of curvature

A sphere of radius about 6371 km is a first approximation. A better figure is an oblate ellipsoid of revolution with equatorial semi-axis a and polar semi-axis b < a. The flattening is f = (a − b)/a and the first eccentricity satisfies e² = (a² − b²)/a² = 2f − f². For WGS 84, a = 6 378 137 m and 1/f = 298.257 223 563, so b ≈ 6 356 752 m and a − b ≈ 21.4 km. The flattening is only 1/298, so the Earth is very nearly spherical, and the observed value is very close to, though not exactly, that of hydrostatic equilibrium.

The curvature of the ellipsoid varies with latitude φ. The meridian radius of curvature (north–south) is M = a(1 − e²)/(1 − e² sin²φ)^(3/2) and the prime-vertical radius (east–west) is N = a/(1 − e² sin²φ)^(1/2). One degree of latitude is (π/180) M, 110.57 km at the equator and 111.69 km at the poles, and one degree of longitude is (π/180) N cosφ, 111.32 km at the equator and zero at the poles. Worked example. At φ = 60° with a = 6378.137 km and e² = 0.00669438: 1 − e² sin²φ = 0.99498, N = 6394.21 km, so 1° of longitude = 111.60 × cos 60° = 55.80 km.

Arc measurements. In the 1730s French expeditions to Lapland and to Peru found that a degree of latitude is longer nearer the pole. A longer degree means a flatter curve, hence a larger radius of curvature, so the Earth is flattened at the poles. The Great Trigonometric Survey of India, begun by William Lambton in 1802 and carried on by George Everest, measured a long meridian arc and gave the Everest ellipsoid used for India’s older maps.
Latitudes. The geodetic latitude φ (angle of the ellipsoid normal with the equatorial plane) is what maps and GPS use. The geocentric latitude φ_c is the angle of the radius vector, with tan φ_c = (1 − e²) tan φ. They agree at the equator and the poles and differ most, by about 0.19° (11.5′), near 45°.

3. Normal gravity, Clairaut’s theorem and the latitude effect

Normal gravity γ is the gravity of the reference ellipsoid, itself a level surface of its own potential, and is the reference from which gravity anomalies are measured. It rises from the equator (γe ≈ 9.7803 m/s²) to the pole (γp ≈ 9.8322 m/s²), by about 0.53%, or 5.2 cm/s² (5200 mGal). Two causes act together: the centrifugal acceleration ω²a ≈ 0.034 m/s² at the equator, which vanishes at the poles and accounts for about two thirds of the difference, and the flattening, which puts the poles nearer the centre.

Clairaut’s theorem connects shape to gravity for a rotating body in hydrostatic equilibrium. Write m = ω²a/γe ≈ 0.00347 (about 1/288) for the ratio of centrifugal to gravitational acceleration at the equator and β = (γp − γe)/γe. To first order in f and m, f + β = (5/2) m. Measuring gravity (β) and the rotation rate therefore gives the flattening, with no need to survey the shape. Normal gravity at latitude φ follows the International Gravity Formula (GRS 80): γ(φ) = 9.780327 (1 + 0.0053024 sin²φ − 0.0000058 sin²2φ) m/s². The closed form is Somigliana’s, γ = γe (1 + k sin²φ)/√(1 − e² sin²φ), with k a constant fixed by γe, γp, a and b.

Worked example. At φ = 30°, sin²φ = 0.25 and sin²2φ = 0.75, so γ = 9.780327 × (1 + 0.0013256 − 0.0000044) = 9.7932 m/s². The north–south gradient is dγ/dφ ≈ γe β sin 2φ, about 0.81 sin 2φ mGal per km, largest at 45° and zero at the equator and poles. Gravity falls with height by the free-air gradient of about 0.3086 mGal per metre, so a reading must be reduced to a common level before it is compared with γ.

⚠️ Normal gravity depends on latitude, not on rock

γ(φ) describes a smooth, homogeneous-looking ellipsoid. Local rock density, topography and subsurface structure produce the anomaly, the difference between measured gravity (after reduction) and γ. The sign convention of Clairaut’s theorem also matters: β is positive because gravity is larger at the poles, and both f and β are small positive numbers that add to 5m/2.

4. The geoid, the ellipsoid and heights

The geoid is the equipotential surface W = W₀ of the Earth’s gravity field that best fits mean sea level; the plumb line is perpendicular to it, and it is imagined to continue under the continents. It is irregular because the Earth’s mass is not distributed with perfect symmetry. The reference ellipsoid is a smooth mathematical surface chosen to fit the geoid as closely as possible, and their vertical separation is the geoid undulation N, which stays within about ±100 m worldwide: the deepest low, near −100 m, lies south of India in the Indian Ocean, and highs of about +60 to +85 m occur over the North Atlantic and near New Guinea. N is positive where the geoid is above the ellipsoid, over mass excess.

Three heights must be kept apart. The ellipsoidal height h is measured along the ellipsoid normal and is what GPS returns. The orthometric height H is measured along the plumb line above the geoid and is the height of maps and levelling (levelling plus gravity). They differ by the undulation: h = H + N, so N = h − H. Worked example. GPS gives h = 524.8 m and levelling gives H = 507.4 m, so N = +17.4 m and the geoid is 17.4 m above the ellipsoid there. To turn GPS heights into mapping heights one subtracts N taken from a geoid model.

Computing N. With the disturbing potential T (actual minus normal potential), Bruns’ formula gives N = T/γ, and Stokes’ integral over gravity anomalies gives T. Satellite models give the long wavelengths and surface gravity the short ones.
Geoid and sea level. Mean sea level departs from the geoid by up to about 1 to 2 m (sea-surface topography from currents, temperature and salinity), so the two are not identical even over the oceans.

⚠️ Water follows potential, not ellipsoidal height

Two points with the same ellipsoidal height h can lie on different equipotential surfaces, because the geoid is not parallel to the ellipsoid. Water and spirit levels follow equipotentials, so flow direction is decided by orthometric height H (or potential), not by GPS height alone.

5. Geodetic reference systems, frames and datums

A datum ties a reference ellipsoid to the Earth. It fixes the ellipsoid’s size and shape (a and f), the position of its centre and the orientation of its axes. A local (regional) datum, such as the Everest datum with its origin at Kalianpur in India, is chosen so that the ellipsoid fits the geoid best over one country; its centre is displaced from the Earth’s centre of mass and its coordinates are not global. A geocentric (global) datum puts the centre at the centre of mass with the Z axis along the mean rotation axis: WGS 84, the datum of GPS, is of this kind. A vertical datum is the surface from which heights are counted, normally mean sea level at tide gauges.

Local and geocentric datums compared
Feature	Local datum	Geocentric datum
Ellipsoid centre	Offset from the centre of mass	At the centre of mass
Best fit to the geoid	Over one country or region	Over the whole Earth
Typical use	Older national maps and surveys	GPS, satellite geodesy
Examples	Everest datum of India	WGS 84, ITRF

A reference system is the definition (conventions, constants, axes); a reference frame is its realisation by the coordinates of marked points. The International Terrestrial Reference System is realised by ITRF station coordinates, which include velocities because plates move by centimetres a year, so a coordinate needs an epoch. Geodetic coordinates convert to Earth-centred Cartesian ones by X = (N + h) cosφ cosλ, Y = (N + h) cosφ sinλ, Z = [N(1 − e²) + h] sinφ. Moving from one datum to another applies a three-parameter shift (ΔX, ΔY, ΔZ) or a seven-parameter Helmert transformation (three shifts, three rotations and one scale factor). If the transformation is skipped, every position is displaced in much the same direction, a systematic shift, not random scatter.

6. GPS: how a position is computed

GPS has a space segment (nominally 24 or more satellites in six orbital planes inclined at 55°, at an altitude of about 20 200 km, orbital radius about 26 560 km and period about 12 sidereal hours), a control segment of ground stations and a user segment of receivers. Each satellite broadcasts on L1 = 1575.42 MHz and L2 = 1227.60 MHz (154 and 120 times 10.23 MHz), carrying timing codes and its orbit.

The receiver times each signal against its own cheap clock, giving the pseudorange ρ = R + c(δt_r − δt_s) + I + T + ε. Here R = √[(x_s − x)² + (y_s − y)² + (z_s − z)²] is the geometric range, δt_r and δt_s are the receiver and satellite clock errors (the latter modelled from the broadcast data), I and T are ionospheric and tropospheric delays and ε is multipath and noise. The unknowns are x, y, z and δt_r, four in all, so at least four satellites are needed, and more are used by least squares. Worked example. If ρ = 22 150 km while R = 22 000 km, then c δt_r = 150 km and δt_r = 1.5×10⁵ / 3.00×10⁸ = 500 µs. A clock error of 1 µs is 300 m of range.

Ionosphere. The delay is dispersive, proportional to 1/f², so two frequencies remove its first-order effect: ρ_IF = (f₁²ρ₁ − f₂²ρ₂)/(f₁² − f₂²). Since (f₁/f₂)² = 1.647, a 5.00 m delay on L1 is 8.23 m on L2. The troposphere is not dispersive at these frequencies and is modelled instead.
Geometry and accuracy. The dilution of precision (DOP) says how satellite geometry magnifies range error: position error ≈ DOP × range error, and satellites spread across the sky give a low DOP. Differential GPS and carrier-phase methods such as RTK use a base station to reach centimetre level.
Relativity. Satellite clocks gain about 38 µs per day net (gravitational and velocity effects combined), so they are set to run slightly slow before launch. Uncorrected, 38 µs would be a range error of about 11.4 km per day.

⚠️ GPS gives h, not H

The height a GPS receiver reports is ellipsoidal. To compare it with a levelled or mapped height one must subtract the geoid undulation, H = h − N. Forgetting this leaves an error of up to tens of metres, which is far larger than the centimetre precision of the positioning itself.

Key takeaways

Gravity is the gradient of a potential W = V + Φ and is perpendicular to equipotentials; the plumb line follows it. The dominant departure from a sphere is J₂ ≈ 1.08×10⁻³, the equatorial bulge, with f ≈ (3J₂ + m)/2.
The reference ellipsoid has f = (a − b)/a and e² = 2f − f² (WGS 84: a = 6 378 137 m, 1/f = 298.257, a − b ≈ 21.4 km). M and N are the meridian and prime-vertical radii; a degree of latitude is longer at the poles, which showed that the Earth is flattened.
Clairaut: f + β = (5/2) m to first order. Normal gravity γ = 9.780327 (1 + 0.0053024 sin²φ − 0.0000058 sin²2φ) m/s² rises about 0.53% from equator to pole, partly through the centrifugal term and partly through flattening; its north–south gradient is about 0.81 sin 2φ mGal/km.
The geoid is the equipotential surface that best fits mean sea level; it differs from the ellipsoid by N within about ±100 m (lowest south of India). Heights obey h = H + N, so N = h − H, and GPS gives h, not H.
A datum fixes an ellipsoid to the Earth (size, centre, orientation). Local datums such as the Everest datum are offset from the centre of mass; geocentric ones such as WGS 84 and ITRF are not. A datum change is a three- or seven-parameter transformation, and skipping it gives a systematic shift.
GPS solves four unknowns (x, y, z and the receiver clock bias) from pseudoranges ρ = R + c(δt_r − δt_s) + errors, so it needs four satellites. Dual frequencies remove the 1/f² ionospheric delay, low DOP means good geometry, and 1 µs of clock error is 300 m.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

Which description of the geoid is correct?
1. The surface of the best-fitting ellipsoid of revolution, defined only by a semi-major axis and a flattening
2. The topographic surface of the continents together with the sea floor, smoothed over a few kilometres
3. The equipotential surface of the gravity field that best fits mean sea level, perpendicular to the plumb line everywhere
4. The surface on which the value of gravity is the same everywhere, which therefore lies closer to the centre at the poles
Show answer
Answer: C — The equipotential surface of the gravity field that best fits mean sea level, perpendicular to the plumb line everywhere
The geoid is an equipotential surface (W = constant) that best matches mean sea level, so the plumb line is perpendicular to it. The ellipsoid is a different, smoother, purely mathematical surface. A surface of constant gravity is not an equipotential, because gravity varies along the geoid with latitude.
Two benchmarks have the same GPS height above the WGS 84 ellipsoid, yet water in a canal between them flows from one to the other. What is the best explanation?
1. GPS heights carry a random error of several metres, which hides the true slope of the canal and reverses the apparent flow
2. Water follows gravity potential, and the geoid undulation N differs between the two places, so their orthometric heights differ
3. The ellipsoid is flattened at the poles, so every point at one latitude has a different height from every point at another
4. The gravitational pull of the water itself tilts the plumb line and so shifts the benchmark vertically by a measurable amount
Show answer
Answer: B — Water follows gravity potential, and the geoid undulation N differs between the two places, so their orthometric heights differ
Flow follows equipotential surfaces. The geoid is not parallel to the ellipsoid, so equal ellipsoidal heights h = H + N can correspond to different orthometric heights H when N differs. The error of a good GPS height is far smaller than a canal slope would need, and neither the flattening of the ellipsoid nor the pull of the water explains the flow.
Clairaut’s theorem, to first order in the flattening f and in m = ω²a/γe, relates f, m and β = (γp − γe)/γe. Which relation is correct?
1. f β = (5/2) m
2. f − β = (5/2) m
3. f = β + (5/2) m
4. f + β = (5/2) m
Show answer
Answer: D — f + β = (5/2) m
Clairaut’s relation is f + β = (5/2) m: a faster spin (larger m) raises both the flattening and the pole-to-equator gravity increase. Numerically, with m = 0.00347 and β = 0.00530 it gives f ≈ 0.0034, near the true 0.00335. The forms with a difference or a product do not reproduce these sizes.
Which of the following fully defines a geodetic datum?
1. A reference ellipsoid (a and f) together with the position of its centre and the orientation of its axes
2. Only the semi-major axis and the flattening of the reference ellipsoid, quoted to enough digits
3. The geoid model that converts ellipsoidal heights into orthometric heights in a given country or region
4. The map projection used to flatten the surface of the ellipsoid onto a printed or on-screen map sheet
Show answer
Answer: A — A reference ellipsoid (a and f) together with the position of its centre and the orientation of its axes
A datum needs the size and shape of the ellipsoid and also where it sits and how it is turned relative to the Earth. Two datums can share the same a and f yet differ in centre and orientation, so the axes alone are not enough. A geoid model and a projection are separate choices.
A surveyor plots GPS positions, which are in WGS 84, directly onto an old map drawn on a regional (Everest) datum without any transformation. What is the expected result?
1. Random scatter of the plotted points about their true places, because GPS noise is much larger than any datum difference
2. An error only in the height, because the two datums assign the same horizontal coordinates to every point
3. Points shifted by about the same amount in about the same direction, since the ellipsoids differ in size and centre
4. No visible error, because a change of datum moves positions only by a few centimetres on any ordinary map
Show answer
Answer: C — Points shifted by about the same amount in about the same direction, since the ellipsoids differ in size and centre
Different ellipsoids with different centres and orientations assign different coordinates to the same ground point, and over a region that difference is systematic, a shift of similar size and direction. It is not random noise, it affects horizontal position as well as height, and it is far larger than centimetres, which is why a datum transformation is applied.
A GPS receiver needs signals from at least four satellites, although the range to three satellites already fixes a point in space. Why?
1. Four satellites are needed so that at least one is always above the horizon of the receiver
2. The receiver clock has an unknown offset, which is a fourth unknown beside x, y and z
3. Each satellite broadcasts only one coordinate of its position, so four signals give the full position
4. The fourth satellite is used to measure the geoid undulation at the receiver location
Show answer
Answer: B — The receiver clock has an unknown offset, which is a fourth unknown beside x, y and z
Pseudoranges contain c·δt_r because the receiver’s inexpensive clock is not synchronised with satellite time. With x, y, z and δt_r unknown, four equations are needed. A satellite broadcasts its full orbit, not one coordinate, and the geoid undulation comes from a model, not from a satellite signal.
Dual-frequency GPS receivers remove most of the ionospheric error because
1. the troposphere delays both frequencies equally, so the difference between them cancels every atmospheric error at once
2. the second frequency passes through the ionosphere without any delay and so acts as an exact range reference
3. the satellite clock error is larger on L2 than on L1 and so subtracts out of the combination of the two
4. the ionospheric delay varies with frequency as 1/f², so two measurements allow it to be solved for and removed
Show answer
Answer: D — the ionospheric delay varies with frequency as 1/f², so two measurements allow it to be solved for and removed
Because the ionosphere is dispersive with delay proportional to 1/f², measuring the same range on two frequencies gives two equations that eliminate the first-order delay. The troposphere is non-dispersive at these frequencies and is not removed this way, and no frequency passes the ionosphere without delay.
Measured arcs show that one degree of latitude is about 110.6 km near the equator and about 111.7 km near the poles. What does this show about the Earth?
1. The Earth is flattened at the poles, because a longer degree means a flatter curve of larger radius
2. The Earth is elongated at the poles, because a longer degree means a more sharply curved surface there
3. The Earth is a perfect sphere, because the difference is smaller than the error of the survey measurements
4. The Earth rotates faster near the poles, which stretches the surface along the meridian and lengthens a degree
Show answer
Answer: A — The Earth is flattened at the poles, because a longer degree means a flatter curve of larger radius
Arc length per degree is (π/180) times the radius of curvature, so a longer degree near the poles means a larger radius there, that is, a flatter, less curved surface: the Earth is oblate (M is 6335 km at the equator and 6400 km at the pole). A longer degree at the poles is the reverse of what an elongated figure would give.
Which of the following statements about the geoid and the reference ellipsoid are correct?
1. The geoid is an equipotential surface of gravity, whereas the reference ellipsoid is a smooth mathematical surface
2. The geoid is irregular because the Earth’s mass distribution is not perfectly symmetric
3. The geoid undulation obeys N = h − H, with h the ellipsoidal height and H the orthometric height
4. The geoid coincides exactly with the best-fitting global ellipsoid at every point, so N is zero
Show answer
Answer: A — The geoid is an equipotential surface of gravity, whereas the reference ellipsoid is a smooth mathematical surface; B — The geoid is irregular because the Earth’s mass distribution is not perfectly symmetric; C — The geoid undulation obeys N = h − H, with h the ellipsoidal height and H the orthometric height
That the geoid is an equipotential surface, that it is irregular because of the uneven mass distribution, and that N = h − H are standard. The claim that the geoid coincides exactly with the ellipsoid is false: the geoid departs from the best-fitting ellipsoid by up to about 100 m, which is the geoid undulation N, so N is not zero in general.
Which of the following statements about GPS positioning are correct?
1. A pseudorange differs from the geometric range by c times the receiver clock offset plus atmospheric delays and noise
2. Satellites spread widely across the sky give a low DOP and so a smaller position error for the same range error
3. The ionospheric delay is larger on the lower-frequency signal, in proportion to 1/f²
4. The height reported by GPS is the orthometric height above the geoid, so it needs no further correction for mapping
Show answer
Answer: A — A pseudorange differs from the geometric range by c times the receiver clock offset plus atmospheric delays and noise; B — Satellites spread widely across the sky give a low DOP and so a smaller position error for the same range error; C — The ionospheric delay is larger on the lower-frequency signal, in proportion to 1/f²
The pseudorange equation, the DOP geometry factor and the 1/f² dispersion are all correct. GPS heights are ellipsoidal, so the geoid undulation must be subtracted (H = h − N) before they are used as mapping heights.
Which of the following statements about normal gravity γ(φ) are correct?
1. γ increases from the equator, about 9.780 m/s², to the poles, about 9.832 m/s²
2. Both the centrifugal acceleration and the flattening of the Earth contribute to the increase with latitude
3. The north–south gradient of γ, proportional to sin 2φ, is greatest at 45° latitude
4. The north–south gradient of γ is greatest at the poles, where the Earth is flattest
Show answer
Answer: A — γ increases from the equator, about 9.780 m/s², to the poles, about 9.832 m/s²; B — Both the centrifugal acceleration and the flattening of the Earth contribute to the increase with latitude; C — The north–south gradient of γ, proportional to sin 2φ, is greatest at 45° latitude
The rise from 9.780 to 9.832 m/s² has a centrifugal part and a shape part, and dγ/dφ ≈ γe β sin 2φ peaks at 45°. At the poles sin 2φ = 0 and γ is at its maximum value but flat, so its gradient there is zero, not largest.
For a reference ellipsoid the semi-major axis is a = 6378.137 km and the semi-minor axis is b = 6356.752 km. The first eccentricity squared is e² = (a² − b²)/a². Find e² × 10³ to two decimal places.
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Answer: 6.69
b/a = 6356.752 / 6378.137 = 0.996647, so (b/a)² = 0.993306 and e² = 1 − 0.993306 = 0.0066945, which is 6.69 × 10⁻³. As a check, f = (a − b)/a = 0.0033529 and 2f − f² = 0.0066945.
Use the International Gravity Formula γ(φ) = 9.780327 (1 + 0.0053024 sin²φ − 0.0000058 sin²2φ) m/s². Find normal gravity at latitude 30°, in m/s² to three decimal places.
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Answer: 9.793
At 30°, sin²φ = 0.25 and sin²2φ = sin²60° = 0.75. The bracket is 1 + 0.0053024 × 0.25 − 0.0000058 × 0.75 = 1 + 0.0013256 − 0.00000435 = 1.0013213. γ = 9.780327 × 1.0013213 = 9.79325, which is 9.793 m/s² to three decimals. sin²2φ = 0.75, not 2 sin²φ = 0.5; the sin²2φ term changes γ by only about 4×10⁻⁵ m/s², so it does not affect the third decimal, and the sin²φ term does almost all the work.
At a station a GPS receiver gives an ellipsoidal height h = 524.8 m. A geoid model gives the geoid undulation N = +17.4 m there. Find the orthometric height H in metres, to one decimal place.
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Answer: 507.4
Heights satisfy h = H + N, so H = h − N = 524.8 − 17.4 = 507.4 m. Adding N instead gives 542.2 m, the usual sign slip; a positive N means the geoid lies above the ellipsoid, so the ground is higher above the ellipsoid than above the geoid by that amount.
Take a = 6378.137 km and e² = 0.00669438. The length of one degree of longitude at latitude φ is (π/180) N cosφ, with N = a/√(1 − e² sin²φ). Find it at latitude 60° in km, to one decimal place.
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Answer: 55.8
At 60°, sin²φ = 0.75, so 1 − e² sin²φ = 1 − 0.00502 = 0.99498 and its square root is 0.99749. N = 6378.137 / 0.99749 = 6394.2 km. Then (π/180) × 6394.2 = 111.60 km and multiplying by cos 60° = 0.5 gives 55.80 km, which is 55.8 km. Using the meridian radius M instead of N is the common mistake.
A receiver measures a pseudorange of 22 150.00 km to a satellite whose true geometric range is 22 000.00 km. Ignoring every error except the receiver clock offset, find that offset in microseconds. Use c = 3.00 × 10⁸ m/s.
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Answer: 500
The pseudorange excess is 22 150 − 22 000 = 150 km = 1.5 × 10⁵ m, and this equals c δt_r. So δt_r = 1.5 × 10⁵ / 3.00 × 10⁸ = 5.0 × 10⁻⁴ s = 500 µs. Giving the answer in milliseconds (0.5) or forgetting to convert km to m are the usual slips.
To first order Clairaut’s theorem states f + β = (5/2) m. For a rotating Earth with m = 0.00346 and gravity increase β = 0.00530 from equator to pole, find the flattening f, in units of 10⁻³, to two decimal places.
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Answer: 3.35
f = (5/2) m − β = 2.5 × 0.00346 − 0.00530 = 0.00865 − 0.00530 = 0.00335 = 3.35 × 10⁻³ (1/f ≈ 298.5). Using m instead of (5/2) m gives a negative flattening, which signals the slip.