Fundamental Astronomy

126
6. Celestial Mechanics
a cloud is much higher than MJ, it will collapse by its
own gravitation.
In (6.52) the pressure can be replaced by the kinetic
temperature Tk of the gas (see Sect. 5.8 for a definition).
According to the kinetic gas theory, the pressure is
P = nkTk , (6.53)
where n is the number density (particles per unit volume)
and k is Boltzmann’s constant. The number density
is obtained by dividing the density of the gas ρ by the
average molecular weight µ:
n = ρ/µ ,
whence
P = ρkTk/µ .
By substituting this into (6.52) we get
3/2 kTk 1
MJ = C √ρ . (6.54)
µG
*Newton’sLaws
1. In the absence of external forces, a particle will
remain at rest or move along a straight line with
constant speed.
2. The rate of change of the momentum of a particle is
equal to the applied force F:
˙p = d
(mv) = F .
dt
3. If particle A exerts a force F on another particle B,
B will exert an equal but opposite force −F on A.
If several forces F1, F2,... are applied on a particle,
the effect is equal to that caused by one force F which
is the vector sum of the individual forces (F = F1 + F2
+ ...).
Law of gravitation: If the masses of particles A and B
are m A and m B and their mutual distance r, the force
exerted on A by B is directed towards B and has the magnitude
GmA m B/r 2 , where G is a constant depending
on the units chosen.
Newton denoted the derivative of a function f by f˙
and the integral function by f ′ . The corresponding notations
used by Leibniz were d f/dt and f dx. Of
Newton’s notations, only the dot is still used, always
signifying the time derivative: f˙ ≡ d f/dt. For example,
the velocity ˙r is the time derivative of r, the acceleration
¨r its second derivative, etc.
6.12 Examples
Example 6.1 Find the orbital elements of Jupiter on
August 23, 1996.
The Julian date is 2,450,319, hence from (6.17), T =
− 0.0336. By substituting this into the expressions of
Table C.12, we get
a = 5.2033 ,
e = 0.0484 ,
i = 1.3053 ◦ ,
Ω = 100.5448 ◦ ,
ϖ = 14.7460 ◦ ,
L =−67.460 ◦ = 292.540 ◦ .
From these we can compute the argument of perihelion
and mean anomaly:
ω = ϖ − Ω =−85.7988 ◦ = 274.201 ◦ ,
M = L − ϖ =−82.2060 ◦ = 277.794 ◦ .
Example 6.2 Orbital Velocity
a) Comet Austin (1982g) moves in a parabolic orbit.
Find its velocity on October 8, 1982, when the distance
from the Sun was 1.10 AU.
The energy integral for a parabola is h = 0. Thus
(6.11) gives the velocity v:
2µ
v =
r =
2GM⊙
r
2 × 4π2 × 1
=
= 8.47722 AU/a
1.10
= 8.47722 × 1.496 × 1011 m
≈ 40 km/s .
365.2564 × 24 × 3600 s
b) The semimajor axis of the minor planet 1982 RA is
1.568 AU and the distance from the Sun on October 8,
1982, was 1.17 AU. Find its velocity.