Astro 162 – Planetary Astrophysics – Solutions to Set 10Problem 1. Dogs and Cats Living TogetherEstimate the size of the meteorite that crashed into the Earth 65 million years ago toform the Chicxulub crater and to wipe out life as the dinosaurs knew it. We saw in classthat the crater was about 180 km in diameter.The crater is larger than the critical size, ∼7 km, above which it takes more energy tolift material out of the crater against gravity than it does to shatter the material. So weare in th e gravity regime. In the gravity regime, the crater diameter D ∼ (2v2/g)1/4d3/4,where v is the impact velocity, g = 103[cgs] is the gravitational acceleration at theEarth’s surface, and d is the projectile diameter. Take D = 180 km and v ≈ 45 km/ s(between 30 km/s = orbital velocity of Earth and 60 km/s = relative velocity betweenEarth and a body revolving about the Sun in a retrograde orbit at 1 AU) and solve ford ∼ 12 km .Note that this rock is so large that it will fail to be slowed by the atmosphere of theEarth, justifying our assumption of neglecting air drag for our estimate of v. T he criticalsize above which solid, competent spheres fail to be slowed is about ∼1 m eter, as weshowed in a p revious problem set. (The Space Shuttle is larger than ∼1 meter and itdoes get slowed down by air, but it is not a solid, competent sphere: it is a plane, whosearea-to-mass ratio is much greater than that of a sphere).Problem 2. J2A body’s J2has simple physical interpretations. Its order of magnitude can be estimatedas eitherJ2∼Requator− RpoleRequator(1)where Requatoris the body’s equatorial radius (measured perpendicular to the spin axis)and Rpoleis the body’s polar radius (measured parallel to its spin axis). Alternatively,we can estimate it asJ2∼Rotational Kinetic EnergyGravitational Potential Energy. (2)(a) Use (2) to derive an analytic expression for J2in terms of the spin freque ncy of theplanet (Ω), the mean density of the planet (ρ), and any natural constants.The nu merator equals IΩ2/2, where I = 2MR2/5 for a uniform density sphere. Thedenominator equals 3GM2/R for a uniform density sphere (derived by assembling the1planet shell-by-shell, as you have done in freshman mechanics). Putting top over bottom,we getJ2∼R3Ω23GM(3)∼Ω24πGρ(4)(b) Estimate J2for the Earth and for Saturn without looking up the answers directly.Then compare your answers to the truth by looking up the answers directly.The mean density of Earth is ρ = 5.5 g cm−3. Its rotation frequency is Ω = 2π/(24 hr).Then for Earth, J2,E∼ 0.0011 . The truth is J2,E= 0.001083 , nearly exactly right.The mean density of Saturn is ρ = 0.7 g cm−3. Its rotation frequency is Ω =2π/(10.8 hr). Then for Saturn, J2,S∼ 0.04 . The truth is J2,S= 0.016298 , a fac-tor of 2 smaller than our estimate (still big enough that you can tell just by looking atSaturn that it is not a s phere). Our error is larger for Saturn th an for Earth because theuniform density approximation is less good f or Saturn (a highly compressible gas/liquid)than for Earth.Problem 3. Weirdness on the Spherea) Long-period comets, unlike their short-period brethren, come from all directions onthe sky. We say that their orbital planes are distributed isotropically; i.e., the orbit polevector (the vector perpendicular to the plane of a comet’s orbit) can point anywhere onthe celestial sphere with equal probability.Define the orbital inclination, i, to be the angle between the orbital plane of a cometand some fixed refere nce plane. The inclination can take any value between i = 0 andi = 180◦. “Prograde” orbits have 0 ≤ i < 90◦, “retrograde” orbits have 90◦< i ≤ 180◦,and “polar” orbits have i = 90◦.Derive the differential inclination distribution, dN/di, for isotropically oriented cometaryorbits, where dN is the differential number of comets having inclinations between i andi + di. Normalize your distribution so thatR180◦0(dN/di)di = 1.First note that co-planar orbits are LESS common than polar orbits. There is only 1orbit having i = 0: the orbit whose pole vector points s tr aight up, perpendicular to thereference plane. By contrast, the pole vectors of polar orbits can point anywhere alongthe circumference of the celestial equator.The differential number of orbits having i between i and i + di is proportional to thedifferential swath of solid angle (on the celestial sphere centered on the Sun) swept out2by pole vectors inclined between i and i + di with respect to the reference plane andhaving arbitary azimuthal orientation. This differential solid angle equals 2π sin i di.Then dN ∝ 2π sin idi, which means dN/di ∝ sin i (where we have dropped the 2π sinceit is a constant). Normalizing this d istribution gives dN/di = (sin i)/2 .(b) Given your answer in (a), are nearly co-planar (i ≈ 0) orbits more common, lesscommon, or just as common to find as nearly polar (i ≈ 90◦) orbits? (Are you weirdedout by your answer?)We already answered this question in (a). Since dN/di ∝ sin i INCREASES with i,co-planar orbits are LESS COMMON than polar orbits. Initially I was weirded out bythis result, bu t now I find it a natural consequence of phase space, spherical geometry,and the breaking of sym metry due to the introd uction of an arbitrary reference plane.(c) Geoff Marcy and his collaborators measure the line-of-sight Doppler shifts of starsto infer the presence of planetary companions around those stars. The extrasolar planettugs gravitationally on its parent star and induces the star to revolve about the commonce nter of mass. A key unknown in the measurement is the angle between the orbitalplane of the planet and the plane of the sky (the plane of the sky is perpendicular tothe observer’s line of sight). Call this angle i. If i = 0, the orbit plane coincides withthe sky plane and there will be no line-of-sight Doppler shift measurable. The maximumline-of-sight Doppler shift arises for i = 90◦. Here, 0 ≤ i ≤ 90◦only.For a g i ven measured Doppler shift, Marcy et al. cannot measure the mass of the planet,m, directly. They can only measure m sin i because of the unknown orientation of theplanet’s orbital plane. For example, if Marcy et al. measure an m sin i = 1 Jupiter mass,they don’t know whether it’s a 1 Jupiter mass planet whose orbit is oriented perpendicularto the sky plane (i = 90◦), or whether it’s a 2 Jupiter mass planet for which i = 30◦.(And in the