Astronomy Principles and Practice Fourth Edition.pdf

172 Celestial mechanics: the two-body problem
Figure 13.4. The Moon’s orbit about the Earth.
In the first second, therefore, a body should fall a distance s M where
s M = 1 2 g M × 1 2 32 × 12
= = 0·0535 inch.
2 × 3600
If gravity did not exist, the Moon would move through space in a straight line (BC in figure 13.4).
But because of the Earth’s attraction, the Moon is continually falling towards the Earth. Its path, BD,
can, therefore, be thought of as being made up of the straight line motion BC and a fall CD. If the
time interval between B and D was 1 second, then CD is the distance fallen by the Moon in 1 second.
According to Newton’s calculation, it should be 0·0535 inch.
Now, CD = CE − DE = CE − BE, if we assume the Moon’s orbit to be circular.
In the right-angled triangle CBE,
or
BE
CE
= cos CEB = cos θ, say
CE = BE/ cos θ.
Hence,
( ) 1
CD = BE
cos θ − 1 .
Now θ is a very small angle—it is the angle swept out by the Moon’s radius vector in one second—
so that we may write
cos θ = 1 − θ 2
2 .
or, by the binomial theorem,
1
cos θ = 1 + θ 2
2 .
Substituting in the expression for CD, we obtain
CD = BE × θ 2
.
2