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one second.* 'This distance turns out to be roughly 1/20 of an inch in a second. |
That fts very well with the inverse square law, because the earth”s radius 1s |
4000 miles, and if something which is 4000 miles from the center of the earth |
* 'That is, how far the circle of the moon”s orbit falls below the straight line tangent to it at |
the point where the moon was one second before. |
--- Trang 144 --- |
falls 16 feet in a second, something 240,000 miles, or 60 times as far away, should |
fall only 1/3600 of 16 feet, which also is roughly 1/20 of an inch. Wishing to |
put this theory of gravitation to a test by similar calculations, Newton made his |
calculations very carefully and found a discrepancy so large that he regarded the |
theory as contradicted by facts, and did not publish his results. Six years later |
a new measurement of the size of the earth showed that the astronomers had |
been using an incorrect distance to the moon. When Newton heard of this, he |
made the calculation again, with the corrected figures, and obtained beautiful |
agrccment. |
This idea that the moon “falls” is somewhat confusing, because, as you see, |
1 does not come any cÍoser. “The idea is suficiently interesting to merit further |
explanation: the moon falls in the sense that # ƒalls auaw jrom the straight line |
that ?t tuould pursue ?ƒ there tuere mo [orces. Let us take an example on the surface |
of the earth. An object released near the earth”s surface will fall 16 feet in the |
first second. An object shot out horizon‡ali will also fall 16 feet; even though it |
is moving horizontally, it stilHl falls the same 16 feet in the same time. Pigure 7-3 |
shows an apparatus which demonstrates this. Ôn the horizontal track ¡is a ball |
which is going to be driven forward a little distance away. At the same height is |
a ball which is going to fall vertically, and there is an electrical switch arranged |
so that at the moment the first ball leaves the track, the second ball is released. |
That they come to the same depth at the same time is witnessed by the fact |
that they collide in midair. An object like a bullet, shot horizontally, might go |
a long way in one second——perhaps 2000 feet——but it will still fall 16 feet 1Í it |
1s aimed horizontally. What happens if we shoot a bullet faster and faster? Do |
not forget that the earth's surface is curved. If we shoot i% fast enough, then |
đZenemee _ Z |
lk⁄ h hị = hạ |
Fig. 7-3. Apparatus for showing the independence of vertical and |
horizontal motions. |
--- Trang 145 --- |
Fig. 7-4. Acceleration toward the center of a circular path. From |
plane geometry, x/S = (2R — S)/x 2R/x, where is the radius of |
the earth, 4000 miles; x Is the distance “travelled horizontally” in one |
second; and S is the distance “fallen” in one second (16 feet). |
when it falls 16 feet it may be at just the same height above the ground as it |
was before. How can that be? It still falls, but the earth curves away, so it falls |
“around” the earth. The question is, how far does it have to go in one second so |
that the earth is 16 feet below the horizon? In Fig. 7-4 we see the earth with |
1ts 4000-mile radius, and the tangential, straight line path that the bullet would |
take If there were no force. Now, IŸ we use one of those wonderful theorems In |
geometry, which says that our tangent is the mean proportional between the two |
parts of the diameter cut by an equal chord, we see that the horizontal distance |
travelled is the mean proportional bebween the 16 feet fallen and the 8000-mile |
điameter of the earth. The square root of (16/5280) x 8000 comes out very close |
to 5 miles. Thus we see that if the bullet moves at 5 miles a second, it then will |
continue to fall toward the earth at the same rate of 16 feet each second, but will |
never get any closer because the earth keeps curving away from it. 'hus it was |
that Mr. Gagarin maintained himself in space while going 25,000 miles around |
the earth at approximately 5 miles per second. (He took a little longer because |
he was a little higher.) |
Any great discovery of a new law is useful only if we can take more out than |
we put in. Now, Newton œseđd the second and third of Kepler?s laws to deduce |
his law of gravitation. What did he predict? Eirst, his analysis of the moon”s |
motion was a prediction because it connected the falling of objects on the earth”s |
surface with that of the moon. Second, the question is, ¡s Éhe orb#t an cllipse? |
W© shall see in a later chapter how it is possible to calculate the motion exactly, |
--- Trang 146 --- |
and indeed one can prove that it should be an ellipse,Š so no extra facE is needed |
to explain Kepler”s #rs law. Thus Newton made his first powerful prediction. |
'The law of gravitation explains many phenomena not previously understood. |
For example, the pull of the moon on the earth causes the tides, hitherto mys- |
terious. 'Phe moon pulls the water up under ¡i§ and makes the tides—people |
had thought of that before, but they were not as clever as Newton, and so they |
thought there ought to be only one tide during the day. 'Phe reasoning was that |
the moon pulls the water up under it, making a high tide and a low tide, and since |
the earth spins underneath, that makes the tide at one station go up and down |
every 24 hours. Actually the tide goes up and down in 12 hours. Another school |
of thought claimed that the high tide should be on the other side of the earth |
because, so they argued, the moon pulls the earth away from the waterl Both of |
these theories are wrong. It actually works like this: the pull of the moon for the |
earth and for the water is “balanced” at the center. But the water which is closer |
to the moon is pulled znmore than the average and the water which is farther away |
from it is pulled /ess than the average. Eurthermore, the water can ow while the |
more rigid earth cannot. The true picture is a combination of these two things. |
What do we mean by “balanced”? What balances? If the moon pulls the |
whole earth toward it, why doesn”t the earth fall right “up” to the moon? Because |
the earth does the same trick as the moon, it goes in a circle around a point |
which is inside the earth but not at its center. 'Phe moon does not just go around |
the earth, the earth and the moon both go around a central position, each falling |
toward this common position, as shown in Eig. 7-5. This motion around the |
_~Z“MOON |
HạO _«< |
X POINT AROUND WHICH |
EARTH & MOON ROTATE |
C3 |
Fig. 7-5. The earth-moon system, with tides. |
* The proof is not given in this course. |
--- Trang 147 --- |
common center is what balances the fall of each. So the earth is not goïng in a |
straight line either; it travels in a circle. The water on the far side is “unbalanced” |
because the moon”s attraction there is weaker than it is at the center of the |
carth, where it just balances the “centrifugal force.” 'Phe result of this imbalance |
1s that the water rises up, away from the center of the earth. Ôn the near side, |
the attraction from the moon is stronger, and the Iimbalance is in the opposite |
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