6a The geostrophic flow between two stations A and B is

0.12ms-1. The stations are 150km apart and the water at

station B is 1026.7kgm-3 and is 500m above a reference

point. If the Coriolis parameter is 1.031 x 10-4s-1, what is

the density of the water at station A?

6a • Rearrange the equation to make ρA the subject:

6a • Rearrange the equation to make ρA the subject:

vg =gh

fL

ρA − ρBρA

6a • Rearrange the equation to make ρA the subject:

vg =gh

fL

ρA − ρBρA

fLvg

gh=

ρA − ρBρA

6a • Rearrange the equation to make ρA the subject:

vg =gh

fL

ρA − ρBρA

fLvg

gh=

ρA − ρBρA

ρAfLvg

gh= ρA − ρB

6a • Rearrange the equation to make ρA the subject:

vg =gh

fL

ρA − ρBρA

fLvg

gh=

ρA − ρBρA

ρAfLvg

gh= ρA − ρB

ρAfLvg

gh− ρA = −ρB

6a • Rearrange the equation to make ρA the subject:

vg =gh

fL

ρA − ρBρA

fLvg

gh=

ρA − ρBρA

ρAfLvg

gh= ρA − ρB

ρAfLvg

gh− ρA = −ρB

ρAfLvg

gh− 1 = −ρB

6a • Rearrange the equation to make ρA the subject:

vg =gh

fL

ρA − ρBρA

fLvg

gh=

ρA − ρBρA

ρAfLvg

gh= ρA − ρB

ρAfLvg

gh− ρA = −ρB

ρAfLvg

gh− 1 = −ρB

ρA =−ρB

fLvggh

− 1

6a • Substitute in the values you know:

6a • Substitute in the values you know:

ρA =−1026.7

1.031 x 10−4 x 150,000 x 0.129.81 x 500

− 1= 𝟏𝟎𝟐𝟕. 𝟎𝟗𝐤𝐠𝐦−𝟑

6a • Substitute in the values you know:

ρA =−1026.7

1.031 x 10−4 x 150,000 x 0.129.81 x 500

− 1= 𝟏𝟎𝟐𝟕. 𝟎𝟗𝐤𝐠𝐦−𝟑

• Replace the values for their units, combine like units and cancel

them:

6a • Substitute in the values you know:

ρA =−1026.7

1.031 x 10−4 x 150,000 x 0.129.81 x 500

− 1= 𝟏𝟎𝟐𝟕. 𝟎𝟗𝐤𝐠𝐦−𝟑

• Replace the values for their units, combine like units and cancel

them:

kgm−3

s−1. m.ms−1

ms−2. m− 1

=kgm−3

m2s−2

m2s−2− 1

= 𝐤𝐠𝐦−𝟑

6b At what latitude is the Coriolis parameter equal to 4.97 x

10-5s-1?

6b • Substitute the values you know into the Coriolis equation:

6b • Substitute the values you know into the Coriolis equation:

4.97 x 10−5 = 2 x 7.27 x 10−5 x sin φ

6b • Substitute the values you know into the Coriolis equation:

4.97 x 10−5 = 2 x 7.27 x 10−5 x sin φ

• Rearrange for latitude:

6b • Substitute the values you know into the Coriolis equation:

4.97 x 10−5 = 2 x 7.27 x 10−5 x sin φ

• Rearrange for latitude:

4.97 x 10−5

2 x 7.27 x 10−5= sin φ

6b • Substitute the values you know into the Coriolis equation:

4.97 x 10−5 = 2 x 7.27 x 10−5 x sin φ

• Rearrange for latitude:

4.97 x 10−5

2 x 7.27 x 10−5= sin φ

sin −14.97 x 10−5

2 x 7.27 x 10−5= φ = 𝟏𝟗. 𝟗𝟗°

6c Point A in the Atlantic Ocean is at 42.5°W and has a water

density of 1027.1kgm-3. Point B is at 43.3°W, has a water

density 1026.5kgm-3 and sits 1000m above a reference

isobar.

If the geostrophic flow associated with this slope is 0.9ms-

1, at what latitude do these points sit?

They are both at the same latitude at which 1 degree of

longitude is equivalent to roughly 85km.

6c • Rearrange the geostrophic flow equation for f:

6c • Rearrange the geostrophic flow equation for f:

f =gh

vgL

ρA − ρBρA

6c • Rearrange the geostrophic flow equation for f:

f =gh

vgL

ρA − ρBρA

• Substitute the Coriolis equation for f and rearrange for φ:

6c • Rearrange the geostrophic flow equation for f:

f =gh

vgL

ρA − ρBρA

• Substitute the Coriolis equation for f and rearrange for φ:

2Ω sinφ =gh

vgL

ρA − ρBρA

6c • Rearrange the geostrophic flow equation for f:

f =gh

vgL

ρA − ρBρA

• Substitute the Coriolis equation for f and rearrange for φ:

2Ω sinφ =gh

vgL

ρA − ρBρA

φ = sin−1

ghvgL

ρA − ρBρA

2Ω

6c • Work out the distance between stations by converting between

degrees of longitude and metres:

6c • Work out the distance between stations by converting between

degrees of longitude and metres:

distance m = 85,000 x 43.3 − 42.5 = 68,000m

6c • Work out the distance between stations by converting between

degrees of longitude and metres:

distance m = 85,000 x 43.3 − 42.5 = 68,000m

• Substitute values into the equation for φ:

6c • Work out the distance between stations by converting between

degrees of longitude and metres:

distance m = 85,000 x 43.3 − 42.5 = 68,000m

• Substitute values into the equation for φ:

φ = sin−1

9.81 x 10000.9 x 68,000

1027.1 − 1026.51027.1

2 x 7.27 x 10−5= 𝟒𝟎. 𝟎𝟗°