Weather and the Environment - Answers
Week 1
1. Just note that
5
6
Mg
P0 A 10 × 4π 6.4 × 10
P0 =
∴M =
≈
A
g
10
2
7
2
≈ 5 × 1018 kg
2. (a) The sun emits radiation spherically, therefore the fraction of the sun’s radiation
reaching earth is given by the ratio of the surface area of the sun to that of a sphere
centred on the sun and whose surface touches the earth. Therefore the flux from the
sun at the earth’s surface is given by:
φ=
σT 4πr
2
4πrsun
− earth
4
2
sun
1
4
φr = 1376 149
. × 10 ∴T = σr 5.7 × 10 6.95 × 10 2
sun − earth
2
sun
8
−8
2
5
1
4
= 5770K
(b) Need to find the energy absorbed. This will be the flux multiplied by the cross
sectional area of the earth (since from the sun the earth will appear as a disc)
2
Energy absorbed = φπrearth
and according to the question this is equal to the energy radiated. Now the energy
radiated is given by
2
Energy radiated = σT 4 4πrearth
and therefore equating these we have
φπr
2
earth
= σT 4πr
4
2
earth
φ
∴T = 4σ 1
4
= 278K
(c) For the final part simply replace φ with 0.7 φ and then:
0.7φ T =
4σ 1
4
= 255K
3. (a) Use the hydrostatic equation in the form
p = p0e
−
gh
RT
with h = 10m
∴ ∆p = p0 e
−
gh
RT
− 1 = − p0 × 118
. × 103 ≈ −12
. mbar
(b) Again use the hydrostatic equation with the given estimate of the average
temperature of 260K, then
p = p0e
−
g × 5×104
260 R
≈ p0 × 1.2 × 103 ≈ 12
. mbar
4. As the earth warms so more water evaporates and more clouds form. These clouds
would reflect more sunlight (reducing the earth’s albedo) tending to reduce the
temperature on earth.
Week 2
1. When the parcel and the surrounding air are at the same temperature.
2. When the surrounding air is close to the stability boundary (near DALR) hot gases
from the chimney induce convective updraughts and smoke rises in almost vertical
puffs. If the surrounding air is far from convective instability the smoke is just
advected downstream by the wind.
3. I estimate SVP at 10º C to be 12 mbar and at 20º C to be 22.5 mbar. treat the air as
an ideal gas then the pressure change due to heating from 10º C to 20º C is given by
293
1
∴ RH = 12 ×
×
= 0.55
283 22.5
Week 3
1. Unstable. If the droplet grows then the SVP of the droplet reduces and it is now
surrounded by an atmosphere that is supersaturated. The droplet therefore continues
to grow. If the droplet shrinks the SVP of the droplet will be greater than the vapour
pressure of the surrounding air and hence the droplet will evaporate.
2. Balance gravity and viscous drag:
4
∴ 6πηrv = mg = πr 3 ρg
3
2
2 ρr g
∴v =
9η
3. Use the equation from lectures to calculate the values required
6
2
2
Ew V − v 1 + r 2
dR
R
=
ρ
dt
4
1
6
R × 10-6
mm
30
40
60
80
100
dR
× 10-8ms −1
dt
0.7
1.2
4.2
6.1
10
14.0
17
21.5
26
30.6
dt
dR
138
83
24
16.4
9.9
7.1
5.8
4.7
3.9
3.3
x 10 6sm-1
and then plot a graph
Integrating the area under the
dt vs R
dR graph by counting squares or otherwise
should give an area corresponding to about 1300s.
9
Week 4
1. The centrifugal force is given by
centrifugal force = − mrω 2
the radius of rotation = rearth cos φ
∴ centrifugal force = − m × 6.4 × 10 −6 cos φ
2π = −m × 3.4 × 10
24 × 3600
−2
cosφ N
whilst the Coriolis force is given by
2π 5 sin φ = −m × 7.2 × 10
24 × 3600
Coriolis force = −2mω sin φv = −2m
−4
sin φ N
For most latitudes CorF < CenF < mg
2. There is no Coriolis force at the equator and therefore the wind blows
perpendicular to the isobars.
Week 5
1. Equate the magnitude of the Coriolis force and the pressure gradient force
2mω sin φv g =
m r
∇p
ρ
1 dp
24 × 3600
0.03 × 10 −3 × 105
=
×
2ωρ dx 2 × 2π sin φ × 1.0
103
20.6
∴ vg =
≈ 29 ms-1 at φ = 45
sin φ
∴ vg =
2. At a warm front warm air overlies cold air, i.e. there is an inversion. The cold air
is therefore stable with respect to convection and therefore cumulus clouds are not
formed. The precipitation at a warm front arises from nimbo stratus clouds formed as
the warm air is forced to rise over the cold air mass.
3.
In the southern hemisphere the wind at an altitude greater than one kilometre
circulates in a clockwise direction around a low pressure centre (a cyclone) and in
an anti-clockwise direction around a high (an anticylone), i.e. in the opposite
sense to what goes on in the northern hemisphere. At the earth’s surface the
effect of friction is to cause the winds to flow at an angle across the isobars
towards the low and away from the high.