Astronomy 111 Recitation #10
10-11 November 2011
Formulas to remember
Tidal acceleration
The difference in tidal acceleration across a satellite with dimensions in the radial and transverse
directions ∆r and ∆ , lying a distance r away from a parent body with mass M:
gtr
=
2GM
M
∆r
r3
GM
gtφ =
− 3 ∆
r
∆
∆r
if it has no angular momentum, and
r
=
gtr
3GM
r3
∆r
if it is in orbit.
Roche limit
The Roche limit is the smallest radius at which a satellite (density ρ ) can orbit a parent body (density
ρplanet , radius Rplanet ) without being torn apart:
aRoche
 ρplanet
≅ 2.46 Rplanet 
 ρ



13
Hill radius
The Hill radius is the range of orbital radii over which a satellite’s (mass m, in orbit with semimajor axis
length a around planet with mass M) tidal forces dominate the motions of smaller bodies; thus
approximately the half-width of the gap it would induce in a ring:
 m 
rH = a 

 3M 
13
.
Perturbations
Transverse acceleration (in orbital plane, perpendicular to radius vector):
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Increases a, small change in ε,
slows down orbit
At periapse:
increases a and ε.
Increases a, small
change in ε, speeds up orbit
At apoapse:
increases a, decreases ε.
Radial acceleration (in orbital plane, parallel to radius vector):
Applied at apoapse: no
change in a and ε;
satellite slows
down in orbit.
Applied here, a and ε
increase.
Applied at
periapse: no
change in a
or ε; satellite
speeds up in
orbit.
Applied here, a and ε
decrease.
Normal acceleration (perpendicular to orbital plane):
Applied at apoapse,
decreases orbital
inclination
Applied here,
precesses orbital plane
counterclockwise.
Applied at periapse,
increases orbital
inclination
Applied here,
precesses orbital plane
clockwise.
Torque and angular momentum
N
L =ω
r × p =I
= rmv sin θ nˆ
dLω d
N =r × F =
=I
dt
dt
= rF
=
sin θ nˆ r⊥ Fnˆ
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θ
F
r
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N is perpendicular to the plane of r and F, in the direction given by the right-hand rule.
Trigonometry reminder
2
2
γ
2
Law of cosines: a = b + c − 2 bc cos α
b

a
β
sin α sin β sin γ
Law of sines: = =
a
b
c
α
Found on the Unit Circle:
sin (π − θ ) =
sin θ
c
cos (π − θ ) =
− cos θ
Workshop problems (do after discussing Homeworks #7 and #8):
Warning! The workshop problems you will do in groups in Recitation are a crucial part of the process of
building up your command of the concepts important in AST 111 and subsequent courses. Do not,
therefore, do your work on scratch paper and discard it. Better for each of you to keep your own account
of each problem, in some sort of bound notebook.
1.
If Earth had an icy ring, where would it be?
2.
Consider the properties of Io and Europa, and without doing any calculations, tell me whether their
tidal heating comes predominantly from Jupiter, or from each other, and why.
3.
Now prove it.
4.
a.
Calculate the difference between the maximum and minimum radial tidal accelerations by Io and
Europa on each other.
b.
Calculate the difference between the maximum and minimum radial tidal accelerations exerted
by Jupiter on Io and Europa.
c.
Revisit problem 1.
“Tidal” torque on a dumbbell. (And a little instruction in how physicists make approximations.)
In this problem you will retrace some of the steps in the tidal-torque calculations from lecture this
week, in a better-defined system in which the “tidal bulge” is a built-in property of the planet. Note
that it is not intended to take long. Refer freely to your lecture notes, and consider dividing the parts
of the problems among your group. Each part will break into at least two sub-parts, for the near and
far distances. There will be a large number of calculations, but all the calculations will be simple.
A moon with mass m executes a circular orbit with radius r about a planet that consists, oddly, of two
spheres with mass M 2 held with their centers a distance 2b  r apart by a rigid, massless bar. The
moon revolves, and the dumbbell rotates about its center of mass, currently with the same angular
frequency, but the near side of the dumbbell is currently ahead of the moon by an angle φ , as shown
below.
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M/2
2b
φ
M/2
r
a.
What are the squares of the distances, rn2 and r f2 , between the moon and the near and far weight
of the dumbbell?
b.
We are told that the moon is very far away from the dumbbell compared to the distance between
the weights (that is, b r  1 ). Use the approximation
( 1 + x )n ≅ 1 + nx
if x  1 ,
and show that the reciprocals of rn2 and r f2 are given by
1
rn2
≅
1 
2b

1+
cos φ  and
2 
r

r 
1
r f2
≅
1 
2b

1−
cos φ  .
2 
r

r 
From here on, use this same first-order approximation: that is, neglect terms proportional to
( b r )2
or higher powers of b r , in sums with terms proportional just to b r , or to terms
independent of b r .
c.
What is the magnitude of the gravitational force that the moon exerts on each weight?
d. What is the orbital angular frequency? (No, that’s not a trivial question. Show that, to first order
in b r , the total force on the satellite has magnitude GMm r 2 and is directed toward the
dumbbell’s center of mass, and thus that the angular frequency of its orbit is ω = GM r 3 . Hint:
consider using a Cartesian coordinate system with origin at the dumbbell’s center and one axis
through the satellite.)
e.
What is the torque the moon exerts on the dumbbell (magnitude and direction)?
f.
What, therefore, is the total torque the dumbbell exerts on the moon (magnitude and direction)?
g.
What is the magnitude and direction of the moon’s orbital angular momentum? The dumbbell’s
rotational angular momentum?
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h. Does the torque the dumbbell exerts on the moon increase or decrease the moon’s orbital angular
momentum? How will its orbit change as a result?
Learn your way around the sky, lesson 10. (An exclusive feature of AST 111 recitations.) Use the lab’s
celestial globes, TheSky running on the lab computers, and any other resources you would like to use, to
answer these questions about the celestial sphere, the constellations, and the orbits of the planets.
5.
Summer is gone and winter almost upon us, so as the Summer Triangle (Vega, Deneb and Altair; see
Recitation 2) sets early in the evening, the Winter Triangle rises. The Winter Triangle is composed of
three of the brightest stars in the winter sky – Betelgeuse (α Orionis), Sirius (α Canis Majoris), and
Procyon (α Canis Minoris) – and is a useful reference point for finding other constellations.
a.
Calculate the dimensions of the Winter Triangle – that is, the angular width of each side.
b.
Suppose I know where to look, to find the Winter Triangle. Describe to me how I can find the
constellations Orion and Gemini.
c.
The constellation Monoceros (the Unicorn) is notable for the star-formation regions it contains.
Neither these regions, nor any of the stars in the constellation, are very bright to the naked eye:
none of its stars are brighter than fourth magnitude. Nevertheless, few constellations are so easy
to find. Tell me how to find Monoceros.
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Problem solutions
1.
The Roche limit for ice ( ρ = 0.94 gm cm −3 ) in orbit around Earth ( ρ = 5.5 gm cm −3 ) is 4.43R⊕ , so the
outer edge would be 3.43 R⊕ = 21900 km above the surface. The inner edge would be limited by the
atmosphere, since wind resistance would cause the particles to fall, eventually. The top of the
atmosphere, for these purposes, can be taken to be about 400 km. (Two ways to tell: this is the
elevation of the orbit of the Hubble Space Telescope, which NASA did not want to be subject to air
drag. And the Space Shuttle orbit, 280 km up, is not quite high enough; satellites in this orbit,
prominently the International Space Station, need occasional boosting to keep from falling in.)
So: between about 400 and 21900 km elevation, right above the equator.
2.
Io and Europa are not terribly different in size, mass and composition, so if their tidal heating were
dominated by each other, their internal states would be similar. They’re not; Io is much hotter inside
than Europa, as is evident in its extreme volcanic activity. In fact, any small difference with this cause
would be in the wrong direction: since Io is more massive than Europa, one would expect Europa to
be subject to the greater tidal heating.
The other important difference between Io and Europa is distance from Jupiter: Io is a good deal
closer to the planet. Since tidal forces are much larger at smaller distances, this makes it more likely
that Jupiter’s tides heat Io and Europa.
3.
a.
Io and Europa are in very nearly circular orbits; the difference between their perijove and apojove
distances is small compared to their orbital semimajor axes. So their minimum distance is the
difference between their orbital radii, and their maximum distance is the sum of their orbital
radii:
=
∆gtr
(
3Gm∆r
aEuropa − aIo
3Gm∆r
−
) (
3
aEuropa + aIo
)
3
,
where m is the mass of one moon, and ∆r the length of the other along the line between them (i.e.
the diameter of the other). Taking the numbers from the Physical Constants sheet, we get
∆gtr = 2.3 × 10 −4 cm sec-2
= 3.6 × 10
b.
cm sec
on Io from Europa,
-2
on Europa from Io.
But their orbits are not so circular that Jupiter’s tides are the same throughout. The tides are
( r a ( 1 + ε ) ):
maximum at perijove =
( r a ( 1 − ε ) ) and minimum at apojove =
=
∆gtr
c.
−4
3GM ∆r
 a ( 1 − ε ) 
3
−
3GM ∆r
 a ( 1 + ε ) 
3
= 4.4 × 10 −2 cm sec-2
on Io from Jupiter,
= 2.4 × 10 −2 cm sec-2
on Europa from Jupiter.
Thus Jupiter exerts tidal-force differences on these moons about a hundred times larger than
those they exert on each other; Jupiter does essentially all of the tidal heating. This will generally
be true for tidal heating of the satellites of the giant planets.
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4.
a. By the law of cosines,
rn2 = r 2 + b 2 − 2 rb cos φ
r f2 = r 2 + b 2 − 2 rb cos (π − φ ) = r 2 + b 2 + 2 rb cos φ
b.
Factor out the r 2 in each expression:

2b
b2
rn2 =
r2  1 −
cos φ + 2

r
r



2b
b2
2
r2  1 +
cos φ + 2
 , r f =

r
r



 .

If b r  1 , then b 2 r 2 is really tiny compared to either 1 or b r ; let’s approximate by ignoring
b 2 r 2 compared to 1 and b r :
2b
2b




rn2 ≅ r 2  1 −
cos φ  , r f2 ≅ r 2  1 +
cos φ  .
r
r




In the following I will use ≅ to indicate each use of this first-order approximation.
Now let’s take the reciprocals. Note that since b r  1 , then 2 b cos φ r  1 , so with 2 b cos φ r
playing the role of x and -1 playing the role of n, we can use that approximation we were given:
1 
2b

cos φ 
= 2 1−
2
r

rn r 
−1
1 
2b

cos φ 
≅ 2 1+
r

r 
1 
2b

cos φ 
= 2 1+
2
r

rf
r 
−1
1 
2b

cos φ 
≅ 2 1−
r

r 
1
1
c.
The forces are gravitational, and point from the center of each weight toward the center of the
moon:
2b
GMm GMm 

1+
cos φ 
Fn = 2 ≅
2 
r

2 rn
2r 
2b
GMm GMm 

1−
cos φ 
Ff = 2 ≅
2 
r

2r f
2r 
d. Break the forces into vertical and horizontal components according to the following diagram:
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y
M/2
2b
ψn
Fn
φ
rn
m
x
M/2
rf
Ff
ψf
r
where we see that
=
Fn ,x F=
Ff ,x Ff cosψ f
n cosψ n
−Fn sinψ n Ff , y =
Fn , y =
Ff sinψ f
These sines and cosines can be expressed in terms of other dimensions in the diagram by use of
the laws of sines and cosines. Remembering to keep terms through first order in b r , and to leave
out of sums the terms of second order ( (b r )2 ) and higher, we get
sinψ f
sinψ n sin φ
=
=
b
rn
b
b sin φ
sinψ n =
sinψ f =
2b
cos φ
r 1−
r
b sin φ 
b

1 + cos φ 
≅
≅
r 
r

b sin φ
≅
≅
r
sin (π − φ ) sin φ
=
rf
rf
b sin φ
2b
cos φ
r
b sin φ 
b

1 − cos φ 

r 
r

b sin φ
r
r 1+
the sines are equal to first order. Now the cosines:
b 2 = r 2 + rn2 − 2 rrn cosψ n
cosψ n =
≅
b 2 = r 2 + r f2 − 2 rr f cosψ f
r 2 + rn2 − b 2
2rrn
cosψ f =
r 1 + 1 − 2 ( b r ) cos φ
2
≅
2
r 2 1 − 2 ( b r ) cos φ
b
b



≅  1 − cos φ  1 + cos φ 
r
r



≅1
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r 2 + r f2 − b 2
2rr f
r 2 1 + 1 + 2 ( b r ) cos φ
r 2 2 1 + 2 ( b r ) cos φ
b
b



≅  1 + cos φ  1 − cos φ 
r
r



≅1
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Astronomy 111, Fall 2011
the cosines are both 1, to first order. (Note that we also could have obtained this from the sines,
using sin 2 ψ + cos 2 ψ =
1 ). Thus the components of the forces are
Ff ,x = Ff
Fn ,x = Fn
GMm 
2b

GMm 
2b

1+
cos φ 
=
1−
cos φ 

2 
2
r
r

2r 

2r 
b sin φ
b sin φ
=
−Fn
Ff , y =
Ff
r
r
GMm 
GMm 
2b
2b
 b sin φ
 b sin φ
1+
cos φ 
1−
cos φ 
=
−
=

2 
2
r
r
 r
 r
2r 
2r 
GMmb
GMmb
≅−
≅
sin φ
sin φ
2r 3
2r 3
=
Fn , y
And this gives a rather simple grand total:
2b
2b
GMm 
 GMm 
 GMm
1+
cos φ  +
1−
cos φ  =

2 
2
r
r
 2r 

r2
2r 
GMmb
GMmb
−
Fy =
Fn , y + Ff , y =
sin φ +
sin φ =
0
3
2r
2r 3
Fx = Fn ,x + Ff ,x =
So to first order the force on the satellite from the dumbbell is the same as if the dumbbell were
concentrated at its center of mass, and the angular speed works out as usual:
GMm
F=
ma =
mω 2 r
−
=
2
r
GM
ω=
r3
e.
Here we illustrate first the geometry of the two cross products that give us the torques due to the
forces on the ends of the dumbbell:
αn
Rn
Rf
αf
Fn
rn
φ
rf
Ff
r
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Let ẑ be a unit vector which points straight out of the page; then the torque on the far end points
along ẑ , and that on the near end points the other way, into the page:
N=
R f × Ff
f
N=
n Rn × Fn
=
−bFn sin α n zˆ
=
bFf sin α f zˆ
So we need the sines of the angles α to first order. Noting that each of them is a 180°
complement to an angle within one of the triangles on the diagram, we can use the law of sines
again:
sin α n= sin (π − α n )= r
=r
sin φ
sin φ
rn
r 1 − 2 ( b r ) cos φ
b


≅  1 + cos φ  sin φ
r


(
)
sin α f = sin π − α f = r
=r
sin (π − φ )
rf
= r
sin φ
rf
sin φ
r 1 + 2 ( b r ) cos φ
b


≅  1 − cos φ  sin φ
r


Now we have first-order approximations for the Fs and the α s ; putting everything together
gives
Nn =
Nf =
−bFn sin α n zˆ
bFf sin α f zˆ
GMm 
b
2b


=
−b
1+
cos φ  1 + cos φ  sin φ zˆ
2 
r
r


2r 
GMmb sin φ 
b
2b

≅−
 1 + r cos φ + r cos φ  zˆ
2


2r
GMmb sin φ 
3b

=
−
 1 + r cos φ  zˆ
2


2r
GMm 
b
2b


1−
cos φ  1 − cos φ  zˆ
2 
r
r


2r 
GMmb sin φ 
b
2b

≅
 1 − r cos φ − r cos φ  zˆ
2


2r
=
=b
3b
GMmb sin φ 

 1 − r cos φ  zˆ
2


2r
And the grand total for the torque exerted by the satellite on the dumbbell is
N s-d
= Nn + N f
=
3b
3b
GMmb sin φ 

 −1 − r cos φ + 1 − r cos φ  zˆ
2


2r
6GMmb 2
3GMmb 2
=
−
−
sin φ cos φ zˆ =
sin 2φ zˆ
2r 3
2r 3
The minus sign indicates that this torque points into the page, for values of φ between zero and
90°. (All the other quantities in the equation are positive definite.)
f.
By Newton’s Third Law, or equivalently by conservation of angular momentum, the torque the
dumbbell exerts on the satellite is equal and opposite:
3GMmb 2
Nd-s =
− N s-d =
sin 2φ zˆ
2r 3
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which points out of the page, for values of φ between zero and 90°.
g.
Both the dumbbell’s rotation and the satellite’s revolution are counterclockwise, so according to
the right-hand rule their angular momentum vectors point out of the page, along the unit vector
ẑ . The satellite and the ends of the dumbbell all function as point masses here, so their moments
of inertia are very simple:
2
=
Ls I=
sω zˆ mr
GM
=
zˆ
r3
GMm2 r zˆ
 M 2  GM ˆ
b  =
=
Ld I=
z
dω zˆ 2 
 2
 r3
GM 3 b 4
r3
zˆ
h. If φ lies between zero and 90°, as drawn, then the torque is opposite L for the dumbbell and
along L for the satellite. Thus the dumbbell will spin down, and the orbital angular momentum of
the satellite will increase. As this increase can’t be reflected in an increase in G, M or m, it shows
up as an increase in r: the satellite drifts away from the dumbbell over time.
5.
a.
I’ll do it in full numerical precision, even though this isn’t really called for. The coordinates, from
SIMBAD, are accurate to a few thousandths of an arcsecond, so when expressed in decimal
degrees one is actually justified in seven decimal-place precision:
Name
Right ascension
(ICRS J2000)
Declination
(ICRS J2000)
In decimal degrees
h
m
s
d
m
s
RA
Dec
Betelgeuse
05
55
10.3053
+07
24
25.426
88.7929375
7.4070628
Sirius
06
45
08.9173
-16
42
58.017
101.2871554
-16.7161158
Procyon
07
39
18.1183
+05
13
29.975
114.8254929
5.2249931
Using the accurate formula for angular distances,
ψ =arccos  cos (α 1 − α 2 ) cos δ 1 cos δ 2 + sin δ 1 sin δ 2  ,
because these don’t look like they’ll be small angles, we get:
Betelgeuse-Sirius
27.1045282°
Sirius-Procyon
25.7013643°
Procyon-Betelgeuse
25.9620289°
It’s pretty close to being an equilateral triangle.
b.
Betelgeuse is α Ori, the upper-left-hand star in Orion, so Orion hangs from the triangle’s
northwestern vertex. North and east of the triangle the next two bright stars one encounters are
the similarly-bright Castor and Pollux, otherwise known as α and β Geminorum. The rest of
Gemini stretches southward, toward Orion, from these two stars.
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c.
Look right in the center of the Winter Triangle. Looks as if Orion and his hunting dogs are after
the unicorn, and have their prey surrounded. Perhaps a celestial-allegorical explanation for why
there are no unicorns...
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