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The Age of the Earth using Heat Conduction (Topic)

Houstoun in [1] in 1929 estimates the age of the Earth using heat conduction and Earth temperature. Let’s see how close the estimate comes to current day estimate of 4.54 ± 0.05109 years [2].

The general formula for linear flow in a semi-infinite solid is

           [∫  ∞           2      ∫ ∞           2   ]
v = -√-c---      f(ξ)e− (ξ−4κxt)-dξ −     f(ξ)e− (ξ+4xκ)t-dξ
    2  π κt   0                    0
(1)

For this problem let f(ξ) be constant and equal to c, f(ξ) = c.

Then

           [ ∫ ∞    (ξ−x)2      ∫ ∞    (ξ+x)2   ]
v = -√-c---      e− -4κt--dξ −     e− -4κt--dξ
    2  π κt   0                0
(2)

sub in β = ξ x

           [ ∫ ∞     2      ∫ ∞     2   ]         ∫  ∞    2
v = --√c---      e− β4κt d β −    e− β4κt d β = √--c--     e− β4κt dβ,
    2  π κt   −x             +x               π κt  0
(3)

since eβ2
4κt is an even function of β. Write α2 = β2(4κt); then the result takes the simpler form

     2c  ∫ 2x√κt   2
v =  √---      e−α  dα.
       π  0
(4)

Tables of values of this integral have been drawn up, and hence v can be determined as a function of the upper limit.

If we descend into the earth we find that after we pass the points where the diurnal and annual variation cease to be appreciable, the temperature begins to increase. The rate of increase varies from place to place, but may be taken roughly as 1F for every 50 feet of descent for depths up to about 1 mile. This increase of temperature is easily explained on the assumption that the center of the earth is at a high temperature and that heat is flowing outwards.

If we assume that the earth was originally at a uniform temperature c and that its surface has been always at a constant temperature zero, we can use the above result to find how long it has taken to cool. We neglect the convexity of the earth’s surface.

We find from (3) that

∂v-   2c--− x2---1---  --c--- − x2
∂x =  √ πe  4κt 2√ κt-= √ πκt-e 4κt

Kelvin found by the method indicated in paragraph 81 (need to reference new entry) that κ for the material of the earth’s surface has the value 400, the units of length and time being the foot and year. Assume that the earth was initially at the temperature of molten rock, i.e. about 7000F. Insert the value for the gradient at the surface, namely, ddvx = 1F for every 50 feet. Then, writing x = 0 in the exponential, we obtain

√ -
  t    7000
---=  √------.
50      π400

i.e. 

    70002 ×  502
t = ------------ = 108 years.
        400π

If we write x = 100 miles,

  x2      (100×5280)2      3
e−4κt = e− 1600×108  = e− 2.

At that depth the gradient is only 1
4⋅5- of its surface value after 108 years. We see, therefore, from the first result, that according to our assumptions 108 years have elapsed since the earth was at a temperature of 7000F, and we see from the second result that it is permissible to neglect the convexity of the earth’s surface. The assumption throughout all the temperature change of constant conductivity, specific heat and density is, of course, open to questions. Also, the earth may have taken much longer to cool, owing to the liberation of heat due to the radioactive disintegration of some of its material.

So 108 vs 4.54 × 109 is not such a bad estimate.

This article is a derivative work of the public domain in [1].

References

[1] Houstoun, R.A., ”An Introduction to Mathematical Physics” Longmans, Green and CO., London, 1929.
[2] Wikipedia contributors, ”Age of Earth,” Wikipedia, The Free Encyclopedia (accessed February 20, 2025).


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Cross-references: domain, work, conductivity, gradient, heat, function, solid, formula, temperature

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Physics Classification44. (Heat transfer)
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