SB Efl 2M5
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362
H5
1913
MAIN
GIFT OF
THE
CAUSE
OF
PLANETARY ROTATION
ALSO
A THEORY
AS TO
THE TAIL OF THE
COMET
By I. T. HINTON
SAN FRANCISCO
THE BLAIR-MURDOCK COMPANY
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PLANETARY ROTATION
The following is an attempt to assign the cause of the
moon always presenting the same face to the earth.
Take two small, equal masses, as indicated in the figure, at
equal distances from the line of attraction. B attracts or is
attracted by a force equal tc__ , and B f by an equal force.
The components y and y are equal and opposite, their alge-
braic sum being equal to zero. The combined attraction of the
two small masses in the direction OA is equal to
a
a 2 + 3' 2 ( a 2 + ?)* " " (a 2 -f y*)H
As either a or v increases, the attraction will diminish.
To find the attractive force of a circular disc, a resort
must be had to calculus. Let BB f be the projection of the
disc, and suppose y to be increased by a small quantity dy.
Then the differential of the attractive force of the disc will be
j-rr dy. The integral of this is 2 " >2 . ^ , and tak-
1
282811
ing this between the limits zero and v, we obtain the attraction
of the circular disc, which is
This may be written
<?7rQ 2 + 3' 2 ) /2 a
(#+?)*
Multiplying both terms by (a 2 -\- v 2 )^ + a, we obtain
,?7rj' 2 Area of disc
O 2 + r) + a (a- + /) 54 ^ (a 2 + r 2 ) + j^a (a 2 + )*
This diminishes (v remaining the same) as a; increases; there-
jfore, if we had another disc on the other side of the center of
the sphere and at the same distance from that center, the
attraction of the off disc would be less than that of the near
disc. Now, a sphere may be regarded as made up of an
infinite number of such discs. Hence, the near hemisphere
attracts with a greater force than the off hemisphere.
A more satisfactory proof of this, perhaps, is the following :
When the disc is taken as a section of a homogeneous sphere,
and c is the distance apart of the centers, the distance a be-
comes c + .r and v 2 becomes r 2 .r 2 . Substituting these values
in {A) we have
(2-
V
If this be multiplied by dx, and the integral be taken between
the limits r and -J- r, we shall have the attraction of the
entire sphere. Dropping TT for the present to simplify matters,
we have
, scdx 2x$x
(c* + 2cx + r*)X (c z + sex '+ r*)*
Integrating (using the formula in< = (ndi* -j- (i'dii for the
last term) we have
(c- -f sex -4- r-y/*
2X 2(C~ + 2CX + P*)* 2X -
Making
x = + r
Subtract-
ing
+ 2r
-4^)+
w^-.u
4T4T
- 4 r +
3?
3c'
(c + r) 3 = c 3 + jrV + 3cr- + r 3
( c n.) 3 = c 3 jc 2 r + jcr 2 r 3
That is to say, a homogeneous sphere attracts or is attracted
as if its entire mass were concentrated at its center. (The
geometrical demonstration of this truth given in the books
is faulty.) But it is not a fact that the two hemispheres are
attracted equally. To prove this, let us take the integral first
from r to o, and then from o to + r. The integral (B) is :
2X - 2 ( C- - 2CX -\-r 2 ) 1 / 2
2X
(c 2 +
Making
x = O
.v = r
Subtract-
ing
2Y 2(C r) -f 2Y
(c r)
(V2 l r z\iy 2
^ 8 ^
2
Which, multiplied by TT, is the attraction of the near hemisphere.
3
Making
.1- =
Subtract-
ing
2Y 2(c r)
^ c 2
- 2c + 2 (e-
^ +2^-
2(e-
[u +
(c + r)
2 -f r=)'5*] . (b)
For the attraction of the off hemisphere. Adding (a) and (b),
we obtain
, as a check. To prove algebraically that (a)
is greater than (b) would lead to complexity; hence a resort
will be had to arithmetic, making c = 12 and r = 5. Sub-
stituting these values in (a), the result will be:
12
43
24
26
70
12
-h -750
12 ) 70 ( 5.833
60
IOO
96
40
13
507
169
133 = 2197
7 3 = 343
= 2197
= 343
216 ) 1854 ( 8.583
1728
1260
1080
1800
1728
720
And the attraction of the near hemisphere is .750.
4
For the off hemisphere the result will be :
- 2c + 2(c* + r*)* 2r -+-^-[ (c + r) 8 C^ + r 2 ) 1 *] (b)
12 ) 170 ( I4.i6? 17
12 17
50 119
48 17
20 289
12 17
80 2023
72 289
__, OA
+ 26 i;3 :
I4.I67
+ 12-574
13
13
4913
2197
216 ) 2716 ( 12.574
216
39
13
169
13
556
432
507
169
1240
1080
I3 3 = 2197
1600
1512
-f .407
And the attraction of the off hemisphere is -[-.407.
4r 3 ___ 500
jr 2 " ~ 432 Attraction of whole sphere.
432 ) 500 ( 1.157 Near hemisphere .............. 750
432 Off hemisphere ............... 407
680
432 Whole sphere ............ 1.157
2480
2160
3200
3024
Now, suppose a body to be at the point C, and moving at
right angles to OC, so that it will travel to C l in a certain in-
terval of time, as indicated in the figure, C being the center
of the sphere, and A and B being the centers of the mass of
the respective hemispheres. If there were no force other than
this tangential force, the center C would move to C\. Suppose,
again, that afterwards an attractive force acts for the same
length of time in the direction OC l . If we regard the forces
as acting on the whole sphere, the center will take some such
position as C., ; and if we regard the forces as acting upon
the two hemispheres separately, the points A and B will assume
the positions A,, and B 2 . The component distances AA 2 ,
CjC 2 , B B 2 , and A 2 A 3 , C 2 C 3 , B 2 B 3 are not in the above ratio
75 : /^ (1.157) ' -47> though nearly so. However, the smal-
ler the intervals of time and space are taken, the more nearly
Will the ratios approach equality, until, when the calculus limit
is reached and the orbit becomes a curve, instead of a polygon,
the lines OA, OC, OB will coincide within an infinitely small
angle, and the lines A 3 A 2 , C 3 C 2 , B 3 B 2 will attain the ratio
.750:^2(1.157) 1.407. In other words, the original points
A, C, B will assume the position OA 2 C 2 B 2 , all on the same
straight line. This clearly indicates a rotation going part />a^w
with the revolution. Hence, when gravity acts as a central
acceleration, the secondary will always present the same face
to the primary. Q. E. D.
The earth does not rotate according to this law, but here
other forces have been or are now in action. The following is
a highly improbable, but may be possible explanation: If the
earth had as regards the sun no rotation at all, or a rotation
(due to the moon's influence) once in about twenty-eight days,
the face towards the sun would soon become very hot, and
the water on that side would evaporate rapidly and condense
as ice on the surface of the earth's off hemisphere. After a
while the condition would be somewhat as in the figure, the
shaded portion representing ice.
Such a body would (I think) rotate with a continuous ac-
celeration, the ratio of acceleration diminishing as the ice
melted faster through the more frequent turning of the earth's
face to the sun, until the acceleration became nil. The body
would then continue to rotate at the speed attained when the
acceleration became zero.
WEIGHING THE EARTH.
If a known weight were in a well at the point A in the figure
on page I, its attraction towards the center would be equal
to the attraction of the mass BEB' less that of the mass BDB' ' ,
which can be found by integration, and the attraction or
weight of the earth could be found if we knew the weight
of the spherical segment BDB' . This would be exceedingly
difficult to obtain with any degree of accuracy on the land, but
at sea we can go down five miles, and the weight of a spherical
segment of water whose height is five miles can be calculated
to a nicety.
THE TAIL OF THE COMET.
In endeavoring to explain this phenomenon by the laws of
mechanics, the writer chanced upon the preceding demon-
stration, without making any progress on the main problem.
However, he ventures the following explanation of the comet's
tail from physical laws:
The nucleus of the comet, it appears to be admitted, is a
mass of very hot vapor or vapors, and if this is true it must
be surrounded by a vast mass of less hot and less condensed
vapors, this large mass, or a portion thereof vastly larger
than the nucleus, being somtimes faintly visible by starlight.
The comet is therefore an illuminated nucleus surrounded by
an immense envelope of vapor, hot, but not as hot as the
nucleus. If a cannon-ball were lying on the surface of the
earth, it would weigh more at 12 midnight than at noon, since
in the one case the sun's attraction would be added to gravity
and in the other it would be subtracted from it. The moon
at conjunction would have a greater effect, perhaps enough
to be noticed on a fine spring balance. So, the comet would
be flattened on the side away from the sun, and elongated on
the near side, thus assuming an egg-like form. However,
when the distance is great, the form is probably nearly a
sphere. Now, the nucleus of a comet is dense enough to cast
a shadow, and against the dark background of that shadow the
heated particles immediately surrounding that shadow be-
come visible, while all the rest of the vast volume of the comet
except the nucleus fail to give out sufficient light to be visible.
Their light is obscured just as that of the moon is obscured in
the daytime, while the particles around the shadow of the
nucleus are more bright from the same reason that the corona
of the sun appears brighter during an eclipse. The length of
the tail depends on the heat of the comet, the distance from
the sun and the angle of view from the earth. The curvature
is due to the difference in time taken by the light coming from
the nucleus and from the tail. It is most noticeable when the
comet is near the sun, since then the velocity of the nucleus is
immense, and that of the end of its shadow is very much
more so.
8
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