At the earth"s distance, which is nearly 200 solar radii, solar gravity will be reduced in the ratio of 1:200 squared.
Hence the force exerted by the sun on the earth is
(25 6 10)/(200) tons weight.
That is to say, it is approximately equal to the weight of 37 107 ordinary tons upon the earth"s surface.
Now steel may readily be found which can stand a load of 37 tons to every square inch of cross-section. The cross-section of a bar of such steel, competent to transmit the sun"s pull to the earth, would therefore have to be
107 square inches,
or say 700 10 square feet.
And this is equivalent to a million million round rods or pillars each 30 feet in diameter.
Hence the statement in the text (page 26) is well within the mark.
_The Pull of the Earth on the Sun._
The pull of the earth on the sun is, of course, equal and opposite to the pull of the sun on the earth, which has just been calculated; but it furnishes another mode of arriving at the result, and may be regarded as involving simpler data--i.e. data more generally known.
All we need say is the following:--
The ma.s.s of the Sun is 316,000 times that of the Earth.
The mean distance of the sun is, say, 23,000 earth radii.
Hence the weight or pull of the sun by the earth is
316000/(23000) 6 10 tons weight.
In other words, it is approximately equal to the ordinary commercial weight of 36 107 tons, as already calculated.
_The Centripetal Force acting on the Earth._
Yet another method of calculating the sun"s pull is to express it in terms of the centrifugal force of the earth; namely, its ma.s.s, multiplied by the square of its angular velocity, multiplied by the radius of its...o...b..t;--that is to say,
F = M (2f/T) r
where T is the length of a year.
The process of evaluating this is instructive, owing to the manipulation of units which it involves:--
F = 6 x 10 tons x (4f x 92 x 106 miles)/(365 days)
which of course is a ma.s.s multiplied by an acceleration. The acceleration is--
(40 x 92 x 106)/133300 x (24) miles per hour per hour
= (3680 x 106 x 5280)/133300 x 576 x (3600) feet per sec. per sec.
= (115 x 5280)/133300 x 576 x 1296 feet per sec. per sec.
= g/1640
Hence the Force of attraction is that which, applied to the earth"s ma.s.s, produces in it an acceleration equal to the 1/1640th part of what ordinary terrestrial gravity can produce in falling bodies; or
F = 6 10 tons g/1640
= 6/1640 x 10 tons weight;
which is the ordinary weight of 37 107 tons, as before.
The slight numerical discrepancy between the above results is of course due to the approximate character of the data selected, which are taken in round numbers as quite sufficient for purposes of ill.u.s.tration.
If we imagine the force applied to the earth by a forest of round rods, one for every square foot of the earth"s surface--i.e. of the projected earth"s hemisphere or area of equatorial plane,--the force transmitted by each would have to be 2700 tons; and therefore, if of 30-ton steel, they would each have to be eleven inches in diameter, or nearly in contact, all over the earth.
_Pull of a Planet on the Earth._
While we are on the subject, it seems interesting to record the fact that the pull of any planet on the earth, even Neptune, distant though it is, is still a gigantic force. The pull of Neptune is 1/20,000th of the sun"s pull: i.e. it is 18 billion tons weight.
_Pull of a Star on the Earth._
On the other hand, the pull of a fixed star, like Sirius--say a star, for example, which is 20 times the ma.s.s of the sun and 24 light years distant--is comparatively very small.
It is easily found by dividing 20 times the sun"s pull by the squared ratio of 24 years to 8 minutes; and it comes out as 30 million tons weight.
Such a force is able to produce no perceptible effect. The acceleration it causes in the earth and the whole solar system, at its present speed through s.p.a.ce, is only able to curve the path with a radius of curvature of length thirty thousand times the distance of the star.
_Force required to hold together the Components of some Double Stars._
But it is not to be supposed that the transmission of any of these forces gives the ether the slightest trouble, or strains it to anywhere near the limits of its capacity. Such forces must be transmitted with perfect ease, for there are plenty of cases where the force of gravitation is vastly greater than that. In the case of double stars, for instance, two suns are whirling round each other; and some of them are whirling remarkably fast. In such cases the force holding the components together must be enormous.
Perhaps the most striking case, for which we have substantially accurate data, is the star Aurigae; which, during the general spectroscopic survey of the heavens undertaken by Professor Pickering of Harvard, in connexion with the Draper Memorial, was discovered to show a spectrum with the lines some days double and alternate days single. Clearly it must consist of a pair of luminous objects revolving in a plane approximately containing the line of vision; the revolution being completed every four days. For the lines will then be optically displaced by the motion, during part of the orbit--those of the advancing body to the right, those of the receding body to the left,--while in that part of the orbit which lies athwart the direction of vision, the spectrum lines will return to their proper places,--opening out again to a maximum, in the opposite direction, at the next quadrant.
The amount of displacement can be roughly estimated, enabling us to calculate the speed with which the sources of light were moving.
Professor Pickering, in a brief statement in _Nature_, Vol. XLI, page 403, 1889, says that the velocity amounts to about 150 miles per second, and that it is roughly the same for both components.
Taking these data:--
Equality and uniformity of speeds, 150 miles per second each, Period 4 days,
we have all the data necessary to determine the ma.s.ses; and likewise the gravitative pull between them. For the star must consist of two equal bodies, revolving about a common centre of gravity midway between them, in nearly circular orbits.