I.—ON THE HEAT GIVEN OUT IN THE CONTRACTION OF THE NEBULA.
§ 1. Fundamental Theorems in the Attraction of Gravitation.
The first theorem to be proved is as follows:—
The attraction of a thin homogeneous spherical shell on any point in its interior vanishes.
Fig. 59.
Take any point P within the sphere. Let this be the vertex of a cone produced both ways, but with a very small vertical angle, so that the small areas S and S´, in which the two parts of the cone cut the sphere, may be regarded as planes. Draw the tangent planes at S and S´. Let the plane of the paper pass through P and be perpendicular to both these tangent planes. Let O P O´ be one of the generators of the cone, and let fall P Q perpendicular to the tangent plane at O, and P Q´ perpendicular to the tangent plane at O´. The volume of the cone with the vertex at P and the base S is ⅓ P Q × S, and the other part of the cone has the volume ⅓ P Q´ × S´.
As the vertical angles of the cones are small, their volumes will, in the limit, be in the ratio of O P3 to O´ P3, and accordingly ⅓ P Q · S ÷ ⅓ P Q´ · S´ = P O3 ÷ O´ P3. But from the figure P Q ÷ P Q´ = P O ÷ P O´, and hence S ÷ O P2 = S´ ÷ O´ P2.
As the shell is uniform, the masses of the parts cut out by the cones are respectively proportional to S and S´. Hence we see that the attractions of S and S´ on P will neutralise. The same must be true for every such cone through P, and accordingly the total attraction of the shell on a particle inside is zero.
The second fundamental theorem is as follows:—
A thin spherical homogeneous shell produces the same attraction at an external point as if its entire mass were concentrated at the centre of the sphere.
This is another famous theorem due to Newton. He gives a beautiful geometrical proof in Section XII. of the first book of the “Principia.” We shall here take it for granted, and we shall consequently assume that—
The attraction by the law of gravitation of a homogeneous sphere on an external point is the same as if the entire mass of the sphere were concentrated at its centre.
§ 2. On the Energy between Two Attracting Masses.
Let m and m´ be two attracting bodies supposed to be small in comparison with their distance x. Let the force between them be ε m m´ ÷ x2 when ε is the force between two unit masses at unit distance. It is required to find the energy necessary to separate them to infinity, it being supposed that they start from an initial distance a. The energy required is obtained by integrating between the limits infinity and a, and is consequently ε m m´ ÷ a.
§ 3. On the Energy Given Out in the Contraction of the Nebula.
We assume that the nebula is contracting symmetrically, so that at any moment it is a homogeneous sphere. We shall consider the shell which lies between the two spheres of radii, r + dr and r respectively.
Let M´ be the mass of the nebula contained within the sphere of radius r, and let dM´ be the mass of the shell just defined. Then it follows from [§ 1] that the condensation of the shell will have been effected by the attraction of the mass M´ solely. The exterior parts of the nebula can have had no effect, for the outer part has always been in symmetrical spherical shells exterior to dM´, and the attraction of these is zero. We see from [§ 2] that the contraction of dM´ from infinity, until it forms a shell with radius r, represents a quantity of energy,
(ε M´dM´)/r ;
for it is obvious that the energy involved in the contraction of the whole shell is the sum of the energies corresponding to its several parts.
If M be the total mass and a the radius of the nebula always supposed homogeneous
M´ = M (r3/a3),
and therefore
dM´ = 3 M (r2/a3) dr.
Hence the work done in the contraction is
(ε/r) M (r3/a3) · 3 M (r2/a3) dr = (3 ε/a6) M2 r4 dr.
Integrating, therefore, the total work of contraction is
⅗ (ε M2/a)
At the present moment a mass of 1 lb. at the surface of the sun would weigh 27 lbs. if tested by a spring balance. Hence
ε M/a2 = 27.
With this substitution we find the expression for the foot-pounds of work corresponding to the contraction of the nebula from infinity to a sphere of radius a to be,
⅗ · 27 a M = 16 a M very nearly.
Hence we have the following fundamental theorem due to Helmholtz, which is the basis of the theory of sun heat.
If the sun he regarded as a homogeneous sphere of mass M pounds and radius a feet, then the foot-pounds of energy rendered available for sun heat by the contraction of the solar material from, an infinite distance is 16 a M.
§ 4. Evaluation of the Sun Heat Given Out in Contraction.
The number of foot-pounds of work given out in the contraction from infinity is 16 a M. As 772 foot-pounds are equal to one unit of heat, i.e. to the quantity of heat necessary to raise 1 lb. of water 1° Fahrenheit, we see that 772 M is the work required to raise a mass of water equal to the mass of the sun through 1° Fahrenheit. Hence the number of globes of water, each equal to the sun in mass, which would be raised 1° Fahrenheit by the total heat arising from the contraction, is
(16 a)/772,
but a, the radius of the sun in feet, is 2,280,000,000, and hence we have the following theorem:—
The energy liberated in the contraction of the sun from infinity to its present dimensions would, if turned into heat, suffice to raise 47,000,000 globes of water, each having the same mass as the sun, through 1° Fahr.
It is found by experiment that 1 lb. of good coal may develop 14,000 units of heat, and is therefore equivalent to 14,000 × 772 foot-pounds of work. A mass of coal equal to the sun would therefore (granted oxygen enough) be equivalent to 14,000 × 772 × M foot-pounds of work. But we have
(16 a M) 16 × 2,280,000,000
———————————————— = —————————————————— = 3,400.
14,000 × 772 × M 14,000 × 772
Hence we see that
The energy liberated in the contraction of the sun from infinity to its present dimensions, is as great as could be produced by the combustion of 3,400 globes of coal, each as heavy as the sun.
We may speak of 3,400 in this case as the coal equivalent.
§ 5. On the Further Contraction of the Sun and the Heat that may thus be Given Out.
Let us suppose the sun contracts to the radius r, and then, as already proved, [§ 3], the energy it gives out is
⅗ (ε M2)/r,
but we have
ε M/a2 = 27,
whence on contraction to the radius r the total energy given out from the commencement is
16 M (a2/r)
The average density of the sun at present is 1.4. Let us suppose it condenses until it has a density ρ.
r3 ÷ a3 = 1.4 ÷ ρ,
whence the energy becomes
14 a M · ∛ρ;
but the coal equivalent of 16 a M has been found in [§ 4] to be 3,400, and hence the coal equivalent in this case is
3,000 ∛ρ.
If we take ρ to be the density of platinum (21.5), we get a coal equivalent 8,300. This, therefore, seems to represent a major limit to the quantity of heat which can be obtained from the condensation of the nebula from infinity into a sun of the utmost density.
§ 6. On the Present Emission of Sun Heat.
According to Scheiner, “Strahlung und Temperatur der Sonne, Leipzig, 1899,” the value of the solar constant, i.e. the number of cubic centimetres of water which would be raised 1° Centigrade by the quantity of sun heat which, if there were no atmospheric absorption, would fall perpendicularly on a square centimetre, at the earth’s mean distance from the sun, is between 3.5 and 4.0. If we take the mean value, we have (translated into British units), the following statement:—
If at a point in space, distant from the sun by the earth’s mean distance, one square foot was exposed perpendicularly to the solar rays, then the sun heat that would fall upon it in one minute would raise one pound of water 14° Fahr.
This shows that the solar energy emitted daily amounts to
700,000,000,000 × 4 π a2 foot-pounds.
§ 7. On the Daily Contraction of the Sun Necessary to Supply the Present Expenditure of Heat.
We have seen that at the radius r the energy is
16 M (a2/r).
Hence for a change dr it is
–16 M (a2/r2) dr.
At its present size, accordingly, the energy given out by a shrinkage dr is
16 M dr.
One cubic foot of the sun averages 87 pounds, so that
M = 4⁄3 π a3 × 87
16 M dr = 464 × 4 π a3 dr.
We have to equate this to the expression in the last article, and we get
dr = 700,000,000,000/(464 a) = .65.
This is the shrinkage of the sun’s radius expressed in feet. Hence the daily reduction of the sun’s diameter is 16 inches.
One coal equivalent possesses energy represented by M × 14,000 × 772. Hence we can calculate that one coal equivalent would supply the solar radiation at its present rate for about 2,800 years.