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By Charles Rhodes, Xylene Power Ltd.

**TEMPERATURE OF AN IDEAL ROTATING SPHERICAL BODY:**

It is helpful to first investigate the temperature of an ideal rotating nonconducting spherical body with a partially reflecting surface. Assume that the reflectance and emissivity of the surface are frequency independent. Then from the Radiation Physics section the theoretical surface temperature Te is given by:

**Te = (Ho dAc / Cb dAs)^.25**

where:

dAc = an element of cross sectional area absorbing energy from the sun

and

dAs = the corresponding element of radiating surface area.

**EARTH MODEL SHOWING DEPENDENCE OF Te ON LATITUDE AT AN EQUINOX:**

At an equinox the geometry of the spherical Earth is sufficiently simple to permit easy calculation of the dependence of Te on latitude.

Let R = the radius of the earth

Let L = the angle of latitude (L= 0 at the equator, L = 90 degrees = (Pi / 2) radians at the north pole)

Consider a narrow latitude band around the earth's axis that is tangent to the earth's surface.

The width of the band is (R dL)

The radius of the band is R cos(L)

The circumference of the band is 2 Pi R cos(L)

The surface area of the band is given by:

dAs = 2 Pi R cos(L) R dL

At an equinox the cross-sectional area of the latitude band exposed to the sun at any instant in time is given by:

dAc = 2 R cos(L) R dL cos(L)

Hence:

dAc / dAs = [2 R cos(L) R dL cos(L)] / [2 Pi R cos(L) R dL]

or

**dAc / dAs = [cos(L) / Pi]**

At an equinox numerical evaluation of (dAc / dAs) for a spherical Earth gives:

dAc / dAs = 1 / Pi = .31831 at the equator (latitude L = 0 degrees = 0 radians)

dAc / dAs = 3^0.5 / 2 Pi = .27566468 at latitude L = 30 degrees = (Pi / 6) radians

dAc / dAs = 1 / (2^0.5 Pi) = .225079 at latitude L = 45 degrees = (Pi / 4) radians

dAc / dAs = 1 / 2 Pi = .159155 at latitude L = 60 degrees = (2 Pi / 6) radians

dAc / dAs = (.7233702 / Pi) = .230256 at latitude L = 43 40' degrees

dAc / dAs = (.7087508 / Pi) = .225603 at latitude L = 44 52' degrees

Recall that:

Te = [(Ho dAc) / (Cb dAs)]^.25

= [(1367 / 5.6697)(dAc / dAs)]^.25 X 100 K

= 394.0507 X (dAc / dAs)^.25 K

Hence, at an equinox numerical evaluation of Te gives:

Te = 295.98 K at latitude = 0 degrees = 0 radians (the equator)

Te = 285.53 K at latitude = 30 degrees = (Pi / 6) radians

Te = 271.42 K at latitude = 45 degrees = (Pi / 4) radians

Te = 248.89 K at latitude = 60 degrees = (2 Pi / 6) radians

Te = 272.96 K at 43 40' N latitude (Toronto Airport)

Te = 271.57 K at 44 52'N latitude (Halifax Airport)

**EFFECTIVE AVERAGE SURFACE TEMPERATURE:**

At an equinox the effective average surface temperature Tea for this ideal rotating spherical body is given by:

Tea = [Ho Ac / Cb As]^.25

= (Ho Pi R^2 / Cb 4 Pi R^2)^.25

= [Ho / (4 Cb)]^.25

= [(1367 W / m^2) / (4 X 5.6697 X 10^-8 W / m^2-K^4)]^.25

= **278.6359256 degrees K**

This web page last updated October 8, 2008

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