Fundamental Astronomy

The energy integral (6.16) is now
h =−µ/2a .
Hence
1
2 v2 − µ
r =−µ
2a ,
which gives
2 1
v = µ −
r a
= 4π2
2 1
−
1.17 1.568
= 6.5044 AU/a ≈ 31 km/s .
Example 6.3 In an otherwise empty universe, two
rocks of 5 kg each orbit each other at a distance of 1 m.
What is the orbital period?
The period is obtained from Kepler’s third law:
P 2 4π
=
2a3 G(m1 + m2)
=
whence
4π 2 1
6.67 × 10 −11 (5 + 5) s2
= 5.9 × 10 10 s 2 ,
P = 243,000 s = 2.8d.
Example 6.4 The period of the Martian moon Phobos
is 0.3189 d and the radius of the orbit 9370 km. What is
the mass of Mars?
First we change to more appropriate units:
P = 0.3189 d = 0.0008731 sidereal years ,
a = 9370 km = 6.2634 × 10 −5 AU .
Equation (6.32) gives (it is safe to assume that mPhobos ≪
mMars)
mMars = a 3 /P 2 = 0.000000322 M⊙
(≈ 0.107 M⊕).
6.12 Examples
Example 6.5 Derive formulas for a planet’s heliocentric
longitude and latitude, given its orbital elements and
true anomaly.
We apply the sine formula to the spherical triangle
of the figure:
sin β
sin i
= sin(ω + f )
sin(π/2)
or
sin β = sin i sin(ω + f ).
The sine-cosine formula gives
cos(π/2) sin β
=−cos i sin(ω + f ) cos(λ − Ω)
+ cos(ω + f ) sin(λ − Ω) ,
whence
tan(λ − Ω) = cos i tan(ω + f ).
Example 6.6 Find the radius vector and heliocentric
longitude and latitude of Jupiter on August 23, 1996.
The orbital elements were computed in Example 6.1:
a = 5.2033 AU ,
e = 0.0484 ,
i = 1.3053 ◦ ,
Ω = 100.5448 ◦ ,
ω = 274.2012 ◦ ,
M = 277.7940 ◦ = 4.8484 rad .
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