Equations of Motion, Forces and the Balanced Winds: Geostrophic, Gradient and Cyclostrophic

This chapter belongs to the Engineering Sciences (XE) paper. It answers the first half of the dynamics list of Section XE7.1 Atmospheric Science of XE7 Atmospheric and Oceanic Sciences, one of the nine optional sections, any two of which a candidate chooses in the exam: “Navier-Stokes and Continuity Equations; Compressible and Incompressible Fluids; Pressure Gradient, Centripetal, Centrifugal and Coriolis Forces; Geostrophic, Gradient and Cyclostrophic Balances.” It teaches the continuity and Navier–Stokes equations for air, compressible against incompressible treatment, the real and apparent forces, the equations of motion in a rotating frame and the Rossby number, and the three balanced winds with their directions in each hemisphere. The numericals are the Coriolis parameter, the Rossby number, geostrophic speed from a pressure gradient and the gradient-wind speed. Constants used: Ω = 7.292×10⁻⁵ s⁻¹, Earth radius 6371 km, g = 9.81 m/s², and the air density is stated in each question.

1. Continuity and Navier–Stokes equations; compressible and incompressible fluids

Air is a fluid, so its motion obeys conservation of mass and Newton's second law. In the Eulerian form, mass conservation (continuity) is ∂ρ/∂t + ∇·(ρV) = 0; with the material derivative D/Dt = ∂/∂t + V·∇ it reads Dρ/Dt = −ρ ∇·V. A fluid is incompressible if the density of a moving parcel does not change, Dρ/Dt = 0, and then ∇·V = 0: the flow has no net divergence and whatever flows into a volume flows out. A compressible fluid has ∇·V ≠ 0 where parcels expand or contract.

Air is a compressible gas (its density falls by e every 8 km or so), yet the incompressible form is often used for shallow motions and for flow at speeds well below the speed of sound (Mach number below about 0.3, as for all winds). For deep, large-scale motion meteorologists use ∇·(ρ₀(z)V) = 0 with a reference density profile ρ₀(z) (the anelastic form), or work in pressure coordinates, where continuity takes the exact incompressible-looking form ∂u/∂x + ∂v/∂y + ∂ω/∂p = 0 with ω = Dp/Dt. For horizontal flow, convergence (∇_h·V < 0) must be balanced by vertical motion: air converging at low levels rises.

The Navier–Stokes equation is Newton's second law per unit mass for a viscous fluid. In a frame fixed to the rotating Earth: DV/Dt = −(1/ρ)∇p − 2Ω×V + g + ν∇²V (the last term, molecular friction, with ν = 1.5×10⁻⁵ m²/s for air). Acceleration equals pressure-gradient force, plus the Coriolis term, gravity and friction. Molecular viscosity is negligible in the free atmosphere; friction matters in the boundary layer (lowest 1 km) through turbulent stress, which is parametrised.

2. Pressure-gradient, centripetal, centrifugal and Coriolis forces

  • Pressure-gradient force (PGF), per unit mass, is −(1/ρ)∇p. It is a real force that points from high to low pressure, perpendicular to the isobars, and is larger where isobars are closer. It is the only force that starts a wind from rest.
  • Centripetal acceleration is the inward acceleration V²/R toward the centre of curvature that any body needs to follow a circular path of radius R. It is not a force but a requirement, supplied by the net of real forces.
  • Centrifugal force (per unit mass V²/R, outward) is an apparent force that appears only when we describe the motion in the rotating frame of the curved path or of the Earth. Earth's own centrifugal term, Ω²r, is absorbed into the effective gravity g, which is why g is 9.78 m/s² at the equator and 9.83 m/s² at the poles.
  • Coriolis force (per unit mass −2Ω×V) is an apparent force on a body moving relative to the rotating Earth. Its horizontal part is 2Ω sin φ times the speed, at right angles to the velocity: to the right in the Northern Hemisphere, to the left in the Southern. It does no work (it is perpendicular to V), so it turns the wind without changing its speed, and it acts on moving air only.

Define the Coriolis parameter f = 2Ω sin φ, with Ω = 7.292×10⁻⁵ s⁻¹ (2π per sidereal day of 86 164 s) and φ the latitude. f is zero at the equator, 7.29×10⁻⁵ s⁻¹ at 30°, 1.03×10⁻⁴ s⁻¹ at 45° and 1.46×10⁻⁴ s⁻¹ at the poles; it is negative in the Southern Hemisphere. Its companion term, with 2Ω cos φ coupling east-west motion to a vertical force, is negligible beside g. Because f is smallest in the tropics, geostrophy fails there.

⚠️ Centrifugal force is not a real force acting on the wind
In an inertial frame there is no outward force: the parcel needs a net inward force, and the centripetal acceleration V²/R is that requirement. Use the centrifugal term only if you also stay in the rotating frame, and never include both a centripetal acceleration and a centrifugal force in the same balance. Similarly the Coriolis force is not a cause of wind: it only deflects moving air.

3. Horizontal equations of motion and the Rossby number

On a mid-latitude plane (the f-plane), with z vertical and friction dropped, the horizontal equations from Navier–Stokes are Du/Dt = −(1/ρ)∂p/∂x + f v and Dv/Dt = −(1/ρ)∂p/∂y − f u. The Coriolis terms are +f v on the east–west equation and −f u on the north–south equation: a wind blowing east (u > 0) is accelerated southward, to its right. Vertically, for large-scale flow the vertical equation reduces to the hydrostatic equation ∂p/∂z = −ρg.

Rossby number. The ratio of the size of the acceleration, U²/L, to that of the Coriolis term, fU, is Ro = U/(fL), for a flow of speed U over length scale L. A synoptic system (U = 10 m/s, L = 1000 km, f = 10⁻⁴ s⁻¹) has Ro = 10 / (10⁻⁴ × 10⁶) = 0.1, so the acceleration is only a tenth of the Coriolis force and the flow is close to geostrophic. A tornado (U = 70 m/s, L = 300 m) has Ro = 70 / (10⁻⁴ × 300) ≈ 2300, so Earth's rotation is irrelevant. Small Ro means rotation dominates.

Rossby number and the balance that applies
RoBalanceExample
≪ 1Geostrophic: PGF = CoriolisStraight jet and large synoptic systems in mid-latitudes
≈ 1Gradient: PGF, Coriolis and centripetalCurved flow around lows and highs, mid-latitude cyclones
≫ 1Cyclostrophic: PGF = centripetalTornado, dust devil, inner core of a tropical cyclone

4. Geostrophic balance and the direction of the wind

For frictionless, unaccelerated, straight flow (Ro ≪ 1), the horizontal equations reduce to f v_g = (1/ρ)∂p/∂x and f u_g = −(1/ρ)∂p/∂y: the pressure-gradient force balances the Coriolis force. In vector form f k × V_g = −(1/ρ)∇p. The speed is V_g = |∇p| / (ρ|f|), directed along the isobars. In pressure coordinates ρ is replaced by height: V_g = (g/f) k × ∇_p z, so winds blow parallel to the height contours of a pressure surface.

Direction. The PGF points toward low pressure. In the Northern Hemisphere (f > 0) the Coriolis force must point away from low pressure, to the right of the wind, so the wind has low pressure on its left (Buys Ballot's rule: stand with your back to the wind and low pressure is on your left). Winds circle anticlockwise around a low and clockwise around a high. In the Southern Hemisphere f < 0, the signs reverse: low pressure is on the right, with clockwise flow round a low. Near the ground, friction weakens the Coriolis force, and the wind crosses isobars toward low pressure at a typical angle of 10 to 45°.

Worked example. Isobars are 4 hPa apart over 400 km at a latitude where f = 1.0×10⁻⁴ s⁻¹, with ρ = 1.2 kg/m³. Pressure gradient = 400 Pa / 4×10⁵ m = 1.0×10⁻³ Pa/m. V_g = 1.0×10⁻³ / (1.2 × 1.0×10⁻⁴) = 8.3 m/s. Halving the isobar spacing doubles V_g; moving to a latitude with half the f doubles it as well. Always convert hPa to Pa and km to m.

⚠️ Geostrophy fails at the equator and for strongly curved flow
Because V_g ∝ 1/f, the formula gives an infinite wind as φ → 0, which is meaningless: flow near the equator is not geostrophic, so it is not used within a few degrees of it. It also overestimates the speed round a low, and underestimates it round a high (the gradient wind corrects this).

5. Gradient and cyclostrophic balances

When the isobars are circles of radius R, the parcel needs a centripetal acceleration V²/R toward the centre. The gradient wind includes it. Write G = (1/ρ)|∂p/∂r| for the pressure-gradient force per unit mass, directed toward the low. For a cyclone (low at the centre) the PGF points inward and Coriolis points outward: V²/R = G − fV, so V² + fRV − GR = 0 and V = −fR/2 + √(f²R²/4 + GR). For an anticyclone the PGF points outward and Coriolis inward: V²/R = fV − G, so V = fR/2 − √(f²R²/4 − GR), which exists only if G ≤ f²R/4.

Compare with geostrophic V_g = G/f. In a cyclone the Coriolis force alone must balance the PGF minus the centripetal need, so V < V_g (subgeostrophic). In an anticyclone V > V_g (supergeostrophic). The condition G ≤ f²R/4 means that pressure gradients near the centre of a high must be weak (a high is flat-topped, a low can be sharp). Example, a cyclone with R = 500 km, f = 10⁻⁴ s⁻¹ and G = 2.0×10⁻³ m/s²: V² + 50V − 1000 = 0, so V = (−50 + √6500)/2 = 15.3 m/s, while V_g = G/f = 20 m/s.

Cyclostrophic balance. Where Ro ≫ 1 (small R, large V) the Coriolis term fV is negligible and V²/R = G, so V = √(R G) = √((R/ρ)|∂p/∂r|). For a tornado with R = 300 m, ∂p/∂r = 20 Pa/m and ρ = 1.2 kg/m³, V = √(300 × 20 / 1.2) = √5000 = 70.7 m/s. The wind can circulate in either sense (it does not depend on f), the flow only exists around a low (the PGF must point inward to supply centripetal acceleration), and so it appears in tornadoes, dust devils, waterspouts and the inner core of a tropical cyclone, but not in highs.

⚠️ Get the sign of the fV term from the geometry
The same formula V²/R + fV = G is used for a cyclone, but for an anticyclone the sign of G flips and the condition becomes V²/R − fV = −G. Memorise the physics, not the algebra: low pressure at the centre means PGF inward, so Coriolis opposes it and V is smaller than geostrophic; high pressure at the centre means the Coriolis force is the larger and V is greater. In the Southern Hemisphere (f < 0) the sense of rotation reverses while these inequalities stay.

Key takeaways

  • Continuity is ∂ρ/∂t + ∇·(ρV) = 0; incompressible flow has ∇·V = 0, a good approximation for winds well below Mach 0.3 and for shallow motions, while deep flow uses the anelastic or pressure-coordinate form. Navier–Stokes adds the pressure-gradient force, Coriolis, gravity and friction.
  • The pressure-gradient force is real; centripetal acceleration is a requirement; centrifugal and Coriolis forces are apparent forces of the rotating frame. Coriolis f = 2Ω sin φ acts at right angles to the motion, does no work and is zero at the equator.
  • The Rossby number Ro = U/(fL) selects the balance: Ro ≪ 1 geostrophic, Ro ~ 1 gradient, Ro ≫ 1 cyclostrophic. A synoptic system has Ro ≈ 0.1 and a tornado has Ro in the thousands.
  • Geostrophic wind V_g = |∇p|/(ρ|f|) blows parallel to the isobars with low pressure on the left in the Northern Hemisphere and on the right in the Southern Hemisphere; it fails near the equator and where friction or curvature matters.
  • Gradient wind: V²/R + fV = G for a cyclone, subgeostrophic; for an anticyclone V is supergeostrophic and G ≤ f²R/4. Cyclostrophic V = √(RG) holds when fV is negligible, only around lows, in either sense of rotation.

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.

  1. For an incompressible fluid, the continuity equation reduces to

    1. ∂ρ/∂t = 0 at every point, so the flow must be steady in time and the velocity cannot vary
    2. ∇·V = 0, so the net outflow from any small volume is zero
    3. ∇p = 0, so the pressure is uniform and there is no pressure-gradient force
    4. V = 0 everywhere, so an incompressible fluid must always be at rest relative to the Earth
    Show answer

    Answer: B — ∇·V = 0, so the net outflow from any small volume is zero

    For Dρ/Dt = 0 the continuity equation Dρ/Dt = −ρ∇·V gives ∇·V = 0: the volume of a parcel is conserved, so inflow equals outflow. It does not require steady flow (the flow can still vary in time) or a uniform pressure.

  2. Which statement about the Coriolis force on a moving air parcel is correct?

    1. It speeds the parcel up whenever the parcel moves toward the poles across latitude circles
    2. It acts at right angles to the velocity, so it changes direction but does no work
    3. It acts on air that is at rest relative to the Earth and so it is the force that starts a wind
    4. It is largest at the equator, where the Earth's rotation is fastest, and zero at the poles
    Show answer

    Answer: B — It acts at right angles to the velocity, so it changes direction but does no work

    The Coriolis force −2Ω×V is perpendicular to V, so it does no work and cannot change the speed. It is zero for V = 0 (so it cannot start a wind) and proportional to sin φ, which is zero at the equator and maximum at the poles.

  3. In the Southern Hemisphere, a geostrophic wind blows with the low-pressure region

    1. on its left-hand side, exactly as in the Northern Hemisphere, because the pressure field alone fixes the side
    2. directly ahead of the wind, because a geostrophic wind flows straight toward low pressure
    3. on its right, because f is negative and the Coriolis force is to the left of the motion
    4. behind the wind, since in the Southern Hemisphere the pressure-gradient force is reversed in direction
    Show answer

    Answer: C — on its right, because f is negative and the Coriolis force is to the left of the motion

    In the Southern Hemisphere the Coriolis force deflects the wind to the left, so for balance it must point away from low pressure on the left, leaving low pressure on the right and clockwise flow around a low. The geostrophic wind does not blow toward low pressure; it blows along the isobars.

  4. Two places have the same pressure gradient and air density, one at 30° latitude and the other at 60° in the same hemisphere. The ratio of the geostrophic wind speed at 30° to that at 60° is closest to

    1. 0.58, since the wind is proportional to sin φ
    2. 1.00, since the geostrophic wind does not depend on latitude
    3. 0.50, since the wind is proportional to the latitude
    4. 1.73, since the wind is inversely proportional to sin φ
    Show answer

    Answer: D — 1.73, since the wind is inversely proportional to sin φ

    V_g = |∇p|/(ρ f) ∝ 1/sin φ. The ratio is sin 60° / sin 30° = 0.866/0.5 = 1.73, so the same gradient gives a stronger wind nearer the equator. The 0.58 value is the inverse, from wrongly using proportionality to f.

  5. The geostrophic approximation is unsuitable within a few degrees of the equator mainly because

    1. the pressure gradient vanishes there
    2. gravity changes sign at the equator
    3. f → 0, so the Coriolis force cannot balance the pressure-gradient force
    4. the air there is always saturated and cannot be treated as an ideal gas
    Show answer

    Answer: C — f → 0, so the Coriolis force cannot balance the pressure-gradient force

    With f = 2Ω sin φ → 0, the Coriolis force vanishes and V_g = |∇p|/(ρ f) grows without bound; the balance cannot hold. The pressure gradient does not vanish there and gravity does not change sign.

  6. Compared with the geostrophic wind computed from the same pressure gradient, the gradient wind around a high-pressure centre in the Northern Hemisphere is

    1. subgeostrophic, because the centripetal acceleration opposes the Coriolis force and slows the flow
    2. supergeostrophic, because the Coriolis force must exceed the pressure-gradient force to supply the inward acceleration
    3. exactly equal to it, because the isobars are closed curves and so the curvature has no effect
    4. directed across the isobars toward the centre of the high, since the pressure-gradient force is inward
    Show answer

    Answer: B — supergeostrophic, because the Coriolis force must exceed the pressure-gradient force to supply the inward acceleration

    Around a high the PGF points outward and Coriolis inward, and the inward net force must supply V²/R: fV − G = V²/R, so fV = G + V²/R > G and V > V_g. The subgeostrophic result belongs to a low, where the same logic reverses.

  7. Cyclostrophic balance, V²/R = (1/ρ)|∂p/∂r|, is a good approximation in which situation?

    1. A broad mid-latitude high with weak winds, a large radius and Rossby number below one
    2. A long straight upper-level jet stream with a Rossby number of about 0.1
    3. The slow drift of air in a large, deep anticyclone situated near the pole
    4. A tornado or dust devil, with small R and a very large Rossby number
    Show answer

    Answer: D — A tornado or dust devil, with small R and a very large Rossby number

    The Coriolis term fV is negligible only when Ro = V/(fR) ≫ 1, as in tornadoes and dust devils. The other situations have Ro ≤ 1, where Earth's rotation matters, and cyclostrophic flow cannot occur around a high at all because the inward PGF is absent.

  8. A vortex of radius R = 100 km has wind speed V = 10 m/s at latitude 30° (f = 7.3×10⁻⁵ s⁻¹). Which balance is the best first approximation?

    1. Geostrophic, since the Rossby number is about 0.1 and rotation dominates
    2. Gradient wind, since Ro is about 1.4 and the three terms are comparable
    3. Cyclostrophic, since the Rossby number is about 100 and Coriolis is negligible
    4. Hydrostatic, since the vortex has a small vertical extent compared with its width
    Show answer

    Answer: B — Gradient wind, since Ro is about 1.4 and the three terms are comparable

    Ro = V/(fR) = 10 / (7.3×10⁻⁵ × 10⁵) = 1.4. Both the Coriolis term fV = 7.3×10⁻⁴ m/s² and the centripetal term V²/R = 1.0×10⁻³ m/s² are of similar size, so the full gradient-wind balance is needed. Hydrostatic balance concerns the vertical direction and is not an alternative to these horizontal balances.

  9. A parcel moves in a horizontal circle of radius R at speed V. Viewed from an inertial frame, the inward net force per unit mass is

    1. V²/R, which is the centripetal acceleration times the unit mass
    2. zero, because the centrifugal force cancels the inward force
    3. V²/R outward, since the parcel tends to move away from the centre
    4. 2ΩV, which is the Coriolis force alone
    Show answer

    Answer: A — V²/R, which is the centripetal acceleration times the unit mass

    Circular motion requires a net inward force per unit mass of V²/R. The centrifugal force is not present in an inertial frame; it appears only in the frame of the circling parcel and would be balanced against that net force there. The net force is the resultant of the real forces, such as the pressure-gradient force, not the Coriolis force alone.

  10. Select ALL correct statements about forces in the atmosphere.

    1. The pressure-gradient force is perpendicular to the isobars and points from high to low pressure
    2. The Coriolis parameter f is zero at the equator and changes sign across it
    3. The centrifugal force is a real force that the isobars exert on moving air
    4. The Coriolis force can increase the kinetic energy of a parcel moving across latitudes
    Show answer

    Answer: A — The pressure-gradient force is perpendicular to the isobars and points from high to low pressure; B — The Coriolis parameter f is zero at the equator and changes sign across it

    The PGF is −(1/ρ)∇p and f = 2Ω sin φ is odd about the equator. Centrifugal force is an apparent force, not exerted by anything. The Coriolis force is perpendicular to the motion, so it cannot change kinetic energy even while the parcel moves across latitudes.

  11. Select ALL correct statements about balanced winds.

    1. Around a low in the Northern Hemisphere the gradient wind is slower than the geostrophic wind for the same pressure gradient
    2. For an anticyclone, the gradient-wind solution exists only if (1/ρ)|∂p/∂r| ≤ f²R/4
    3. Cyclostrophic wind can occur around a high-pressure centre as readily as around a low
    4. Cyclostrophic wind speed depends on the Coriolis parameter
    Show answer

    Answer: A — Around a low in the Northern Hemisphere the gradient wind is slower than the geostrophic wind for the same pressure gradient; B — For an anticyclone, the gradient-wind solution exists only if (1/ρ)|∂p/∂r| ≤ f²R/4

    A cyclone's centripetal need is met by part of the PGF, so V < V_g, and the anticyclone discriminant must stay non-negative. Cyclostrophic flow needs an inward PGF, so it exists around lows only, and V = √(RG) has no f in it.

  12. A Northern Hemisphere geostrophic wind blows from west to east. Select ALL statements that must be true.

    1. Pressure is lower to the north of the wind than to the south
    2. The pressure-gradient force on the air points toward the north
    3. The Coriolis force on the air points toward the south
    4. The wind has low pressure on its right-hand side
    Show answer

    Answer: A — Pressure is lower to the north of the wind than to the south; B — The pressure-gradient force on the air points toward the north; C — The Coriolis force on the air points toward the south

    The Coriolis force deflects eastward motion to its right, toward the south. In geostrophic balance the PGF must be equal and opposite, pointing north, and the PGF points toward low pressure, so pressure is lower in the north, which is on the left of the wind. Low pressure on the right is the Southern Hemisphere rule.

  13. Using Ω = 7.292×10⁻⁵ s⁻¹, find the Coriolis parameter f = 2Ω sin φ at latitude 30°N, expressed in units of 10⁻⁵ s⁻¹, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 7.29

    f = 2 × 7.292×10⁻⁵ × sin 30° = 2 × 7.292×10⁻⁵ × 0.5 = 7.292×10⁻⁵ s⁻¹, i.e. 7.29 in units of 10⁻⁵ s⁻¹.

  14. A weather system has a typical wind speed U = 12 m/s and length scale L = 800 km at a latitude where f = 1.0×10⁻⁴ s⁻¹. Find its Rossby number Ro = U/(fL), to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 0.15

    Ro = 12 / (1.0×10⁻⁴ × 8×10⁵) = 12/80 = 0.15, so the flow is nearly geostrophic.

  15. The pressure falls by 3 hPa over 300 km, perpendicular to straight isobars. With ρ = 1.1 kg/m³ and f = 1.0×10⁻⁴ s⁻¹, find the geostrophic wind speed in m/s, to two decimal places.

    Numerical answer — type the value.

    Show answer

    Answer: 9.09

    Gradient = 300 Pa / 3×10⁵ m = 1.0×10⁻³ Pa/m. V_g = 1.0×10⁻³ / (1.1 × 1.0×10⁻⁴) = 9.09 m/s.

  16. A cyclone has circular isobars of radius R = 400 km, f = 1.0×10⁻⁴ s⁻¹ and a pressure-gradient force per unit mass G = 1.5×10⁻³ m/s² toward the centre. Solving V²/R + fV = G for the positive root, find the gradient wind speed in m/s, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 11.6

    fR = 40 m/s, so V² + 40V − GR = 0 with GR = 600. V = (−40 + √(1600 + 2400))/2 = (−40 + 63.25)/2 = 11.6 m/s (the geostrophic value G/f = 15 m/s is larger).

  17. In a dust devil the cyclostrophic balance holds. Take R = 150 m, a radial pressure gradient ∂p/∂r = 24 Pa/m toward lower pressure at the centre and ρ = 1.2 kg/m³. Find the wind speed V = √((R/ρ)∂p/∂r) in m/s, to one decimal place.

    Numerical answer — type the value.

    Show answer

    Answer: 54.8

    V² = (150/1.2) × 24 = 125 × 24 = 3000, so V = 54.77 ≈ 54.8 m/s.