Celestial Mechanics
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 TableC.12, we get
a=5.2033, e=0.0484, i=1.3053◦, Ω=100.5448◦, =14.7460◦,
6.12 Examples 137
L= −67.460◦=292.540◦.
From these we can compute the argument of per- ihelion and mean anomaly:
ω=−Ω= −85.7988◦=274.201◦, M=L− = −82.2060◦=277.794◦. Example 6.2 (Orbital Velocity) (a) Comet Aus- tin (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 velocityv:
v= 2μ
r =
2GM r
= 2×4π2×1
1.10 =8.47722 au/a
=8.47722×1.496×1011m
365.2564×24×3600 s ≈40 km/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.
The energy integral (6.16) is now h= −μ/2a.
Hence
1 2v2−μ
r = −μ 2a, which gives
v= μ 2
r −1 a
= 4π2 2
1.17− 1 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 dis- tance of 1 m. What is the orbital period?
The period is obtained from Kepler’s third law:
P2= 4π2a3 G(m1+m2)
= 4π21
6.67×10−11(5+5)s2
=5.9×1010s2, whence
P =243,000 s=2.8 d.
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=6.2634×10−5au.
Equation (6.32) gives (it is safe to assume that mPhobosmMars)
mMars=a3/P2=0.000000322M (≈0.107M⊕).
Example 6.5 Derive formulas for a planet’s he- liocentric longitude and latitude, given its orbital elements and true anomaly.
We apply the sine formula to the spherical tri- angle of the figure:
sinβ
sini =sin(ω+f ) sin(π/2) or
sinβ=sinisin(ω+f ).
The sine-cosine formula gives cos(π/2)sinβ
= −cosisin(ω+f )cos(λ−Ω) +cos(ω+f )sin(λ−Ω), whence
tan(λ−Ω)=cositan(ω+f ).
Example 6.6 Find the radius vector and helio- centric longitude and latitude of Jupiter on Au- gust 23, 1996.
The orbital elements were computed in Exam- ple6.1:
a=5.2033 au, e=0.0484, i=1.3053◦, Ω=100.5448◦, ω=274.2012◦,
M=277.7940◦=4.8484 rad.
Since the mean anomaly was obtained directly, we need not compute the time elapsed since peri- helion.
Now we have to solve Kepler’s equation. It cannot be solved analytically, and we are obliged to take the brute force approach (also called nu- merical analysis) in the form of iteration. For it- eration, we write the equation as
En+1=M+esinEn,
whereEn is the value found in thenth iteration.
The mean anomaly is a reasonable initial guess, E0. (N.B.: Here, all angles must be in radians;
otherwise, nonsense results!) The iteration pro- ceeds as follows:
E0=M=4.8484,
E1=M+esinE0=4.8004, E2=M+esinE1=4.8002,
E3=M+esinE2=4.8002,
after which successive approximations no longer change, which means that the solution, accurate to four decimal places, is
E=4.8002=275.0◦. The radius vector is
r=a(cosE−e)ˆi+a
1−e2sinEjˆ
=0.2045iˆ−5.1772jˆ and the distance from the Sun,
r=a(1−ecosE)=5.1813 au.
The signs of the components of the radius vector show that the planet is in the fourth quadrant. The true anomaly is
f=arctan−5.1772
0.2045 =272.3◦.
Applying the results of the previous example, we find the latitude and longitude:
sinβ=sinisin(ω+f )
=sin 1.3◦sin(274.2◦+272.3◦)
= −0.0026
⇒ β= −0.15◦,
tan(λ−Ω)=cositan(ω+f )
=cos 1.3◦tan(274.2◦+272.3◦)
=0.1139
⇒ λ=Ω+186.5◦
=100.5◦+186.5◦
=287.0◦.
(We must be careful here; the equation for tan(λ−
Ω) allows two solutions. If necessary, a figure can be drawn to decide which is the correct one.) Example 6.7 Find Jupiter’s right ascension and declination on August 23, 1996.
6.12 Examples 139 In Example 6.6, we found the longitude and
latitude, λ=287.0◦, β = −0.15◦. The corre- sponding rectangular (heliocentric) coordinates are:
x=rcosλcosβ=1.5154 au, y=rsinλcosβ= −4.9547 au, z=rsinβ= −0.0133 au.
Jupiter’s ecliptic coordinates must be trans- formed to equatorial ones by rotating them around the x-axis by an angle ε, the obliquity of the ecliptic (see Box2.1):
XJ=x=1.5154 au,
YJ=ycosε−zsinε= −4.5405 au, ZJ=ysinε+zcosε= −1.9831 au.
To find the direction relative to the Earth, we have to find where the Earth is. In principle, we could repeat the previous procedure with the or- bital elements of the Earth. Or, if we are lazy, we could pick up the nearest Astronomical Al- manac, which lists the equatorial coordinates of the Earth:
X⊕=0.8815 au, Y⊕= −0.4543 au, Z⊕= −0.1970 au.
Then the position relative to the Earth is X0=XJ−X⊕=0.6339 au,
Y0=YJ−Y⊕= −4.0862 au, Z0=ZJ−Z⊕= −1.7861 au.
And finally, the right ascension and declination are
α=arctan(Y0/X0)=278.82◦=18 h 35 min, δ=arctan Z0
X20+Y02
= −23.4◦.
If the values given by the Astronomical Almanac are rounded to the same accuracy, the same result
is obtained. We should not expect a very precise position since we have neglected all short-period perturbations in Jupiter’s orbital elements.
Example 6.8 Which is easier, to send a probe to the Sun or away from the Solar system?
The orbital velocity of the Earth is about 30 km/s. Thus the escape velocity from the So- lar system is√
2×30≈42 km/s. A probe that is sent from the Earth already has a velocity equal to the orbital velocity of the Earth. Hence an extra velocity of only 12 km/s is needed. In addition, the probe has to escape from the Earth, which re- quires 11 km/s. Thus the total velocity changes are about 23 km/s.
If the probe has to fall to the Sun it has to get rid of the orbital velocity of the Earth 30 km/s. In this case, too, the probe has first to be lifted from the Earth. Thus the total velocity change needed is 41 km/s. This is nearly impossible with current technology. Therefore a probe to be sent to the Sun is first directed close to some planet, and the gravitational field of the planet is used to acceler- ate the probe towards its final destination.
Example 6.9 An interstellar hydrogen cloud contains 10 atoms per cm3. How big must the cloud be to collapse due to its own gravitation?
The temperature of the cloud is 100 K.
The mass of one hydrogen atom is 1.67× 10−27kg, which gives a density
ρ=nμ=107m−3×1.67×10−27kg
=1.67×10−20kg/m3. The critical mass is
MJ
=
1.38×10−23J/K×100 K 1.67×10−27kg×6.67×10−11N m2kg−2
3/2
× 1
1.67×10−20kg/m3
≈1×1034kg≈5000M. The radius of the cloud is
R=3 3 4π
M
ρ ≈5×1017m≈20 pc.