SME
Oceanography
Currents and circulation
Use this for “Across the Gulf Stream at 35° N the sea surface rises 1 m over 100 km. How fast is the current?”
Diagram
f
=
2Ω
sin
φ
=
2
×
7.292 ×
10
-5
×
sin
35°
=
8.37 ×
10
-5
s
-1
.
v
=
gΔη
fΔx
=
9.81
×
1
8.37 ×
10
-5
× 1 ×
10
5
=
1.17
m/s.
The current flows into the page: high sea level on its right (north).
This is an example. Type your own values to start a new problem.
f (s⁻¹)
=
2Ω
sin(
φ
°)
v (m/s)
=
g
×
Δη (m)
÷
(
f (s⁻¹)
×
Δx (km)
)
Formulas
f
=
2
×
7.292 ×
10
-5
×
sin(
φ
)
8.365 ×
10
-5
=
2
×
7.292 ×
10
-5
×
sin(35°)
v
=
9.81
×
Δη
f
× Δx × 1,000
1.1727
=
9.81
×
1
8.365 ×
10
-5
× 100 × 1,000
Assumptions
•
Steady flow with no friction, below the wind-driven top layer: the slope’s pressure push and the Coriolis force balance.
•
g = 9.81 m/s² and Earth’s spin rate Ω = 7.292 × 10⁻⁵ rad/s.
•
In the north the water flows along the slope with high sea level on its right.
Step-by-step
WE KNOW
Δη = 1 m, Δx = 100 km, φ = 35°
FIND
Coriolis parameter (f), current speed (v)
Step 1 · Find Coriolis parameter (f)
Use f = 2 × 7.292 × 10⁻⁵ × sin(φ)
The Coriolis parameter is twice Earth’s spin rate times the sine of the latitude.
f
=
2
×
7.292 ×
10
-5
×
sin(35°)
f
=
2
×
(7.292
×
10
-5
)
×
0.57358
f
=
0.00014584
×
0.57358
f
=
8.365 ×
10
-5
s
-1
Step 2 · Find current speed (v)
Use v = 9.81 × Δη ÷ (f × Δx × 1,000)
The current runs just fast enough for its Coriolis force to balance the slope’s push (Δx in meters).
v
=
9.81
×
1
8.365 ×
10
-5
× 100 × 1,000
v
=
9.81
0.008365 × 1,000
v
=
9.81
8.365
v
=
1.1727
m/s
Check
Put the numbers back in. Both sides should match.
8.365 ×
10
-5
=
2
×
7.292 ×
10
-5
×
sin(35°)
✓
1.1727
=
9.81
×
1
8.365 ×
10
-5
× 100 × 1,000
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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