R r Fig. 6.4. The orbit of an object in the gravitational field of another object is a conic section: ellipse, parabola or hyperbola. Vector e points to the direction of the pericentre, where the orbiting object is closest to central body. If the central body is the Sun, this direction is called the perihelion; if some other star, periastron; if the Earth, perigee, etc. The true anomaly f is measured from the pericentre

Fig. 6.4. The orbit of an object in the gravitational field of another object is a conic section: ellipse, parabola or hyperbola. Vector e points to the direction of the pericentre, where the orbiting object is closest to central body. If the central body is the Sun, this direction is called the perihelion; if some other star, periastron; if the Earth, perigee, etc. The true anomaly f is measured from the pericentre and angular momentum depend on the coordinate frame used. Here we have used a heliocentric frame, which in fact is in accelerated motion.

So far, we have found two constant vectors and one constant scalar. It looks as though we already have seven integrals, i. e. one too many. But not all of these constants are independent; specifically, the following two relations hold:

Here v is the speed of the planet relative to the Sun. The constant h is called the energy integral; the total energy of the planet is m 2 h. We must not forget that energy where e and k are the lengths of e and k. The first equation is obvious from the definitions of e and k. To

prove (6.13), we square both sides of (6.10) to get ii or■r ar

¡j2e2 = (k x r) ■ (k x r) + a2 — + 2(k x r) ■ — .

r2 r

Since k is perpendicular to r, the length of k x r is

|k||r| = kv and (k x r) ■ (k x r) = k2v2. Thus, we have a2e2 = k2v2 + a2 + — (kx r■ r).

The last term contains a scalar triple product, where we can exchange the dot and cross to get k ■ r x r. Next we reverse the order of the two last factors. Because the vector product is anticommutative, we have to change the sign of the product:

a2(e2 - 1) = k2v2 - — (k ■ r x r) = k2v2 - — k

But the product r ■ e can also be evaluated using the definition of e:

aa a

Equating the two expressions of r ■ e we get r k /a

This completes the proof of (6.13).

The relations (6.12) and (6.13) reduce the number of independent integrals by two, so we still need one more. The constants we have describe the size, shape and orientation of the orbit completely, but we do not yet know where the planet is! To fix its position in the orbit, we have to determine where the planet is at some given instant of time t = t0, or alternatively, at what time it is in some given direction. We use the latter method by specifying the time of perihelion passage, the time of perihelion t . Telescopes Mastery

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