67Phys 325 Lecture 8 Thursday 12 February, 2015Announcements: HW4 is longer than the others, and is due Tuesday 2/24 Midterm I will be in-class on Thursday 2/26. Material coveredwill be lectures 1-10 and HWs 1-4 and discussions 1-5 Due to limited space in 151, some of you will be asked to take the exam in 114 Transportation building Office Hours will be adjusted, watch for massmail announcing all this.Orbits in spherically symmetric gravitational fields.We start by noting Kepler's observations (before Newton) regarding planetary motion.Orbits around the sun are planar ellipses, with the sun at one focusThe line between the sun and a planet sweeps out equal areas in equal times( thus the planet moves faster when closer to the sun )Across the solar system, a planet's year is proportional to the 3/2 power of its semi-major axis.We will see that these behaviors are correctly predicted by Newton's laws.Circular orbitsThe general case ( elliptical, parabolic and hyperbolic) orbits will be discussed later. The circularorbits are easy to analyze. We'll start with them.We will for now assume M >>> m, so M does not move and we can treat it as fixed.Later on we will examine the more general case.F=ma implies: (using the familiar centripetal acceleration of a circular path at speed v is v2/r)a = v2/r = GM/r268Thusv = [ GM/r ]1/2The orbital speed of a circular orbit is a diminishing function of distance from the planet.The orbital speed of a planet around the sun scales with the -1/2 power of its distance from the sun.Thus the period of the orbit P = 2πr/v isP = 2π [ r3 / GM ]1/2While this pertains only to the special case of a circular orbit, it does confirm Kepler'sobservation that period ~ orbitradius3/2. Anecdote has it that Newton 's first inklingthat he had the essential physics right was that this prediction for the period of themoon's orbit corresponded with reality. Q: How did he know G and M ? Answer: hedidn't, but he did know that g = GM/Rearth2 so he knew the product GM.We do not need to know G and M to use these formulas for the earth, as we know GM/Rearth2 = g =9.8 m/s2. Thus P is , for a satellite around the earth (where R = radius of earth)P = 2π [ r3 / R2g ]1/2If the orbital radius is as low as possible, r ~ R = radius of the planet ( called for the earth, lowearth orbit LEO, for which r = R + altitude, with typical altitudes = 400 km << R = 6400 km,) thenP ~ 2π [ R/g ]1/2For the earth this is about 87 minutes. [Higher orbits have longer periods.]If we substitute into the formula P = 2π [ r3 / GM ]1/2 the mass M in terms of the planet's density, M= 4πρr3/3 ( where ρ is average mass density)P = 2π [ r3 / (G4πR3ρ/3) ]1/2For low orbit, r ~ R and P = 2π [ 3 / (G4πρ ) ]1/2This is independent of the size of the planet! Low orbits have a period that just depends on theaverage mass density ρ of the planet. Other planets tend to be less dense, so low altitude orbitalperiods around them are slightly longer than they are around earth. It is noteworthy that theperiod of a low orbit around any normal solid object is about the same, regardless of its size. Theperiod of a low orbit around a 10 meter asteroid of typical rock density ~3000kg/cubicmeter is notmuch more than around the earth (average density 5500kg/cubicmeter). Low orbits around thesun -which has an average density about 1/4 that of the earth but has about 100 times the diameter- are about 3 hours.-----------------69The energy e ( per unit satellite mass ) of a circular orbit is ( recall Φ = U/m and we define t =KineticEnergy/m)e = E/m = t + Φ = (1/2) v2 - GM / r = (1/2) v2 - v2 = -(1/2)v2The potential energy is –2 times the kinetic energy.That total energy is negative ought not be surprising, as we know that Φ = 0 at infinity, and thisorbit is trapped and unable to get to ∞.--------------The angular momentum of the satellite is, per unit mass of the satellite,l = L/ m = r v---------------This circular orbit speed may be compared to escape velocity from the same altitude: Escapevelocity from a position a distance r from the center of the earth, must be such that the total energyis not negative. Thus ( Because e = 0 is consistent with zero speed at r = ∞ )e = (1/2) vesc2 – GM / r ≥ 0sovesc = [ 2 GM / r ] 1/2 The minimum speed a projectile must have to escape from a distance rThis is √2 times greater than circular orbital velocity. A rocket in a circular orbit can escape to ∞by suddenly increasing its speed by 41.4%. In the HW you are asked to find the resulting motionfor boosts other than 41.4%.Perhaps surprisingly, it does not matter what direction v is for a trajectory to escape ( as long as theprojectile doesn't hit the earth ) – all that matters is that the kinetic energy is sufficient that theprojectile can rise out of the potential energy well, i.e, that the total energy is greater than or equalto zero . Of course if the boost is applied in the direction that the object is already going, werequire less fuel.Non circular orbitsWe continue to assume M >>> m. And we'll continue to see what can be concluded withoutsolving any differential equations.Planetary orbits are in general not circles. A complete description of them requires that we solvethe differential equations of motion for the planets. We will do this in the next section. It The total energy of acircular orbit is negative.70transpires, however, that we can first get a lot of insight into the orbits without solving thedifferential equations. (always worth doing!) We merely recognize that each orbit is characterizedby its energy and angular momentum, which have the same values everywhere on the orbit.We call the closest approach to the earth the perigee and the furthest distance, the apogee. Thespeeds are different at the two places ( we know this for two reason: conservation of energy andconservation of angular momentum; if r is greater, v must be less ) We may writel = L/ m = rp vp = ra va( By the way, l ≠ rx vx at arbitrary points x on the orbit. Why? )We may also write, for the energy per unit masse = (1/2) va2 - GM / ra = (1/2) vp2 - GM / rp( By the way, this is true for arbitrary points on the orbit e = (1/2) vx 2 - GM / rx )In terms of e and l we may solve for the speeds and distances at the two extreme points. Substitutingv = l /r (valid only at the extreme points where v is