Astronomy Principles and Practice Fourth Edition.pdf

168 Celestial mechanics: the two-body problem
Figure 13.3. An orbital ellipse, illustrating Kepler’s second law.
where e is the eccentricity of the ellipse, a quantity defined by the relation
e = CS/CA.
The eccentricity gives an idea of how elongated the ellipse is. If the ellipse is a circle, e = 0, since
S and F are coincident with C. The other limit for e is unity, obtained when the ellipse is so narrow
and elongated that the empty focus is removed to infinity.
In fact, the planetary orbits are almost circular so that e is a small fraction.
Let P be the position of the planet in its orbit. Then SP is the planet’s radius vector. The planet
is said to be at perihelion when it is at A. It is then nearest the Sun, since
SA = CA− CS = a − ae = a(1 − e).
When the planet is at A ′ it is said to be at aphelion, for it is then farthest from the Sun. We have,
then,
SA ′ = CA ′ + CS = a + ae = a(1 + e).
The angle ASP is called the true anomaly of the planet.
13.2.3 Kepler’s second law
This law states that the rate of description of area by the planet’s radius vector is a constant.
Let us suppose that in figure 13.3, the planet’s positions at times t 1 , t 2 , t 3 and t 4 are P 1 , P 2 , P 3 and
P 4 . Then between times t 1 and t 2 its radius vector has swept out the area bounded by the radius vectors
SP 1 , SP 2 and the arc P 1 P 2 . Similarly, the area swept out by the radius vector in the time interval
(t 4 − t 3 ) is the area SP 3 P 4 .
Then Kepler’s law states that
area SP 1 P 2
t 2 − t 1
= area SP 3 P 4
t 4 − t 3
= constant. (13.1)
If t 2 − t 1 = t 4 − t 3 , then the area SP 1 P 2 = area SP 3 P 4 .
In particular, if the area is the area of the ellipse itself, the radius vector will be back to its original
position and Kepler’s second law, therefore, implies that the planet’s period of revolution is constant.
Let us suppose the time interval (t 2 − t 1 ) to be very small and equal to interval (t 4 − t 3 ). Position
P 2 will be very close to P 1 ,justasP 4 will be close to P 3 . The area SP 1 P 2 is then approximately the
area of △SP 1 P 2 or
1
2 SP 1 × SP 2 × sin P 1 SP 2 .