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:

2GM M grtr = ∆ r3

GM ∆ gtφ =−∆ r3 ∆r if it has no angular momentum, and r

3GM grtr = ∆ r3 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:

13 ρplanet aRRoche≅ 2.46 planet  ρ

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:

13 m raH =  . 3M

Perturbations

Transverse acceleration (in orbital plane, perpendicular to radius vector):

Increases a, small change in ε, slows down orbit

At periapse: increases a and ε.

At apoapse: Increases a, small increases a, decreases ε. change in ε, speeds up orbit

Radial acceleration (in orbital plane, parallel to radius vector):

Applied at apoapse: no Applied here, a and ε change in a and ε; increase. satellite slows down in orbit. 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, Applied here, decreases orbital precesses orbital plane inclination counterclockwise.

Applied at periapse, Applied here, increases orbital precesses orbital plane inclination clockwise.

Torque and angular momentum

Lrp=×=ω I N = rmvsinθnˆ F ddLω θ N=×= rF =I dt dt r =rFsinθnnˆˆ = r⊥ F

N is perpendicular to the plane of r and F, in the direction given by the right-hand rule. 

Trigonometry reminder γ 2 22 a Law of cosines: a=+− b c2 bc cosα b

sinαβγ sin sin β Law of sines: = = abc α c Found on the Unit Circle:

sin(πθ−=) sin θ cos( πθ−=−) 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.

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.

4. “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 2br 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.

2b M/2

M/2

2 2 a. What are the squares of the distances, rn and rf , 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, br 1 ). Use the approximation

n (1+x) ≅+ 1 nx if x  1 ,

2 2 and show that the reciprocals of rn and rf are given by

11 2bb11  2 ≅+1 cosφφ and ≅−1 cos . 2222  rrnfrrrr 

From here on, use this same first-order approximation: that is, neglect terms proportional to 2 (br) or higher powers of br, in sums with terms proportional just to br, or to terms independent of br.

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 br, the total force on the satellite has magnitude GMm r2 and is directed toward the dumbbell’s center of mass, and thus that the angular frequency of its orbit is ω = GM r3 . 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?

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.

Problem solutions

−3 −3 1. The Roche limit for ice (ρ = 0.94 gm cm ) in orbit around Earth (ρ = 5.5 gm cm ) is 4.43R⊕ , so the outer edge would be 3.43R⊕ = 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:

33Gm∆∆ r Gm r ∆=g − , tr 33 (aEuropa−+ aa Io) ( Europa a Io )

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

∆=×g 2.3 10−4 cm sec -2 on Io from Europa, tr =3.6 × 10−4 cm sec -2 on Europa from Io.

b. But their orbits are not so circular that Jupiter’s tides are the same throughout. The tides are maximum at perijove ( ra=(1 −ε ) ) and minimum at apojove ( ra=(1 + ε ) ):

33GM∆∆ r GM r ∆=g − tr 33 aa(11−+εε) ( )  =4.4 × 10−2 cm sec -2 on Io from Jupiter, =2.4 × 10−2 cm sec -2 on Europa from Jupiter.

c. 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.

4. a. By the law of cosines,

222 rn =+− r b2 rb cosφ

222 22 rf =+− r b2 rb cos(πφ −=++) r b2 rb cosφ

b. Factor out the r2 in each expression:

22bb22 bb  rr22=−+1 cosφφ  ,rr22 =++ 1 cos  . nf22   rrrr  

If br 1 , then br22 is really tiny compared to either 1 or br; let’s approximate by ignoring br22 compared to 1 and br:

2222bb22  rrnf≅−1 cosφφ ,rr ≅+  1 cos . rr 

In the following I will use ≅ to indicate each use of this first-order approximation.

Now let’s take the reciprocals. Note that since br 1 , then 2br cosφ  1 , so with 2br cosφ playing the role of x and -1 playing the role of n, we can use that approximation we were given:

− 11 2bb 1 1 2  =−1 cosφφ ≅+ 1 cos 22 2  rrn rr r  − 11 2bb 1 1 2  =+1 cosφφ ≅− 1 cos 22 2  rrf rr r 

c. The forces are gravitational, and point from the center of each weight toward the center of the moon:

GMm GMm2 b F =≅+1 cosφ n 22 22rrn r

GMm GMm2 b F =≅−1 cosφ f 22 22rrf r

d. Break the forces into vertical and horizontal components according to the following diagram:

2b M/2 ψ n Fn r φ n m x

M/2 rf ψ Ff f

where we see that

FFnx,,= ncosψψ n FF f x= fcos f

Fny,,=−= F nsinψψ n FF f y fsin 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 br, and to leave out of sums the terms of second order ( ()br2 ) and higher, we get

sinψ sinφφsinψ f sin (πφ− ) sin n = = = brn b rff r b sinφ φ sinψ = b sin n sinψ f = 2b 2b r 1− cosφ r 1+ cosφ r r bbsinφ  ≅+φ bbsinφ  1 cos ≅−1 cosφ rr rr b sinφ b sinφ ≅ ≅ r r

the sines are equal to first order. Now the cosines:

2 22 2 22 b=+− r rnnn2 rr cosψψb=+− r rf2 rr f cos f rrb222+− 222 ψ = n rrb+−f cos n cosψ = 2rrn f 2rrf r2 112+−(br) cosφ ≅ r2 1++ 1 2(br) cosφ 2 ≅ r 2 1− 2(br) cosφ 2 r 2 1+ 2(br) cosφ bb   ≅−1 cosφφ  1 + cos  bb   rr   ≅+1 cosφφ  1 − cos  rr   ≅ 1 ≅ 1

the cosines are both 1, to first order. (Note that we also could have obtained this from the sines, using sin22ψψ+= cos 1 ). Thus the components of the forces are

FFnx, = n FFfx, = f GMm2 b GMm2 b =1 + cosφ =1 − cosφ 2r2 r 2r2 r bbsinφφsin FF=−=FF ny,, n rrf y f GMm2 b b sinφφGMm 2 b b sin =−+1 cosφφ=− 1 cos 22rr22rr rr GMmb GMmb ≅− sinφφ≅ sin 22rr33

And this gives a rather simple grand total:

GMm22 b  GMm b  GMm FFx=+= nx,, F f x 1 + cosφφ  +1 − cos  = 2rr  22  22r rr GMmb GMmb FFy=+=− ny,, F f y sinφφ + sin = 0 22rr33

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 mω2 r r2

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 F n r Rn n

φ Rf rf Ff α f

Let zˆ be a unit vector which points straight out of the page; then the torque on the far end points along zˆ , and that on the near end points the other way, into the page:

N= RF × Nf= RF ff × n nn =−=bFnnsinααzzˆˆbF fsin f

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:

sinφ sin (πφ− ) sinφ sinαnn= sin ( πα −=) r sinα= sin πα −=rr = r ff( ) n rrff φ sin sinφ = r = − φ r r1 2( br) cos r1+ 2( br) cosφ b ≅+1 cosφφ sin b  ≅−1 cosφφ sin r r

Now we have first-order approximations for the Fs and the αs ; putting everything together gives

Nn=−=bF nnsinαα zˆˆNzffbF sin f GMm2 b  b  =−++b 1 cosφ 1 cos φφ sin zˆ GMm2 b  b  2    =−−b 1 cosφφ  1 cos zˆ 2r rr   2r2 rr   GMmbsinφ  2 b b φ ≅− 1+ cosφφ + cos zˆ GMmbsin 2 b b 2 ≅1 −− cosφ cosφ zˆ 2r rr 2r2 r r GMmbsinφ  3 b φ =−+1 cosφ zˆ GMmbsin 3 b 2  = 1− cosφ zˆ 2r r 2r2 r

And the grand total for the torque exerted by the satellite on the dumbbell is

Ns-d = NNnf + GMmb sinφ  3b 3b = −−1 cosφφ + 1 − cos zˆ 2r2 rr 63GMmb22GMmb =−=sinφφ cos zzˆˆ−sin 2φ 22rr33

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:

3GMmb2 NNd-s=−= s-d sin 2φzˆ 2r3

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 zˆ . The satellite and the ends of the dumbbell all function as point masses here, so their moments of inertia are very simple:

22GM Lzss=Iω ˆ = mr z ˆˆ= GMm r z r3

34 M2 GM GM b Lzdd=Ibω ˆ = 2 z ˆˆ= z 2 rr33

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 Declination In decimal degrees (ICRS J2000) (ICRS J2000)

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=−+arccoscos(αα12) 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.

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...