We start by noting Kepler's observations (before Newton) regarding planetary motion.

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1 Phys 325 Lecture 8 Thursday 7 February, 2019 Announcements: HW4 is longer than the others, and is due Tuesday 2/19 Midterm I will be in-class on Thursday 2/21. Material covered will be lectures 1-10 and HWs 1-4 and discussions 1-5 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 focus The 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 predicted by Newton's laws. Circular orbits The general case ( elliptical, parabolic and hyperbolic) orbits will be discussed later. The circular orbits are easy to analyze. We'll start with them. We will for now assume M >>> m, so the central mass 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 v 2 /r) Thus a = v 2 /r = GM/r 2 67

2 v = [ GM/r ] 1/2 The 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 = 2pr/v is P = 2p [ r 3 / GM ] 1/2 While this pertains only to the special case of a circular orbit, it does confirm Kepler's observation that period ~ orbitradius 3/2. Anecdote has it that Newton's first inkling that he had the essential physics right was that this prediction for the period of the moon's orbit corresponded with reality. Q: How did he know G and M? Answer: he didn't, but he did know that g = GM/Rearth 2 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/Rearth 2 = g = 9.8 m/s 2. Thus P is, for a satellite around the earth (where R = radius of earth), P = 2p [ r 3 / R 2 g ] 1/2 If the orbital radius is as low as possible, r ~ R = radius of the planet ( called for the earth, low earth orbit LEO, for which r = R + altitude, with typical altitudes h = 400 km << R = 6400 km) If we ignore h, then P ~ 2p [ R/g ] 1/2 For the earth this is about 85 minutes if h = 0. [Higher orbits have longer periods.] If we substitute into the formula P = 2p [ r 3 / GM ] 1/2 the mass M in terms of the planet's density, M = 4prr 3 /3 ( where r is average mass density) P = 2p [ r 3 / (G4pR 3 r/3) ] 1/2 For low orbit, r ~ R and P = 2p [ 3 / (G4pr ) ] 1/2 This is independent of the size of the planet! Low orbits have a period that just depends on the average mass density r of the planet. Other planets tend to be less dense, so low altitude orbital periods around them are slightly longer than they are around earth. It is noteworthy that the period of a low orbit around any normal solid object is about the same, regardless of its size. The period of a low orbit around a 10 meter asteroid of typical rock density ~3000kg/cubicmeter is not much more than around the earth (average density 5500kg/cubicmeter). Low orbits around the sun -which has an average density about 1/4 that of the earth but has about 100 times the diameter - are about 3 hours

3 The energy e ( per unit satellite mass ) of a circular orbit is ( recall F = U/m and we define t = KineticEnergy/m) e = E/m = t + F = (1/2) v 2 - GM / r = (1/2) v 2 - v 2 = -(1/2)v 2 The potential energy is 2 times the kinetic energy. The total energy of a circular orbit is negative. That total energy is negative ought not be surprising, as we know that F = 0 at infinity, and this orbit is trapped and unable to get to The angular momentum of the satellite in a circular orbit is, per unit mass of the satellite,! = L/ m = r v This circular orbit speed may be compared to escape velocity from the same altitude: Escape velocity from a position a distance r from the center of the earth, must be such that the total energy is not negative. Escape requires e > 0: so e = (1/2) v 2 GM / r 0 vesc = [ 2 GM / r ] 1/2 The minimum speed a projectile must have to escape from a distance r This 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 motion for boosts other than 41.4%. It is perhaps surprising that to get to low earth orbit requires obtaining a speed of about { GM/R } 1/2 ( most of the energy needed to get there is spent on getting speed, not altitude ). To escape to infinity requires only 41.4% more speed. Perhaps surprisingly, it does not matter what direction v is for a trajectory to escape ( as long as the trajectory doesn't intersect the earth ) all that matters is that the kinetic energy is sufficient that the projectile can rise out of the potential energy well, i.e, that the total energy is greater than or equal to zero. Of course if the boost is applied in the direction that the object is already going, we require less fuel. Non circular orbits We continue to assume M >>> m. And we'll continue to see what can be concluded without solving any differential equations. 69

4 Planetary orbits are in general not circles. A complete description of them requires that we solve the differential equations of motion for the planets. We will do this in the next section. It transpires, however, that we can first get a lot of insight into the orbits without solving the differential equations. (always worth doing!) We merely recognize that each orbit is characterized by its energy and angular momentum, which have the same values everywhere on the orbit. The picture shows an ellipse, though we haven t yet derived that shape. We call the closest approach to the earth the perigee and the furthest distance, the apogee. The speeds are different at the two places ( we know this for two reason: conservation of energy and conservation of angular momentum; if r is greater, v must be less ) We may write! = L/ m = rp vp = ra va ( By the way,! rx vx at arbitrary points x on the orbit. Why? ) We may also write, for the energy per unit mass e = (1/2) va 2 - GM / ra = (1/2) vp 2 - 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! we may solve for the speeds and distances at the two extreme points. Substituting v =! /r (valid only at the extreme points where v is perpendicular to r) into the expression for e: e = (1/2) v 2 - GM / r = (1/2) v 2 - GMv /! We get a quadratic eqn for v at either of these two special positions v = (GM / l) ± (GM / l) 2 + 2e 70

5 If you know e and!, this gives you the apogee and perigee values of v The corresponding distances are also determinable; recall for such positions, r =! /v and conclude r = (GM / l) ± l (GM / l) 2 + 2e = rmin and rmax The two roots are the two extreme points. The larger v and smaller r ( the + sign on the ) correspond to the perigee, the other pair corresponds to the sign on the and to the apogee. These two values rmin and rmax are the turning points in the figure of U eff in lecture 6. Several special cases: a) e = 0 (just barely sufficient to be unbound) In this case the smaller root( the perigee) r is l r = where v = 2GM/! (2GM /l) = l2 2GM The other root is r =, v = 0. This is its state after it escapes. The orbit (as we will see later) is a parabola. The particle escapes (just barely) to. ====================== b) e > 0 ( an unbound orbit) As we will see later, this orbit is a hyperbola. There are two roots. The one with r and v both positive corresponds to the closest approach, the perigee. The other has r and v both negative. This root is unphysical (corresponding to the other branch of the hyperbola.) 2b is the angle through which a particle coming from is deflected. It is not clear (yet) how b is related to e and!. ======================== c) e = (GM / l ) 2 There is a double root, at r = l2. GM ; v = l / r = GM l This is the circular orbit discussed above. We readily see that v 2 = GM/r. just like we found earlier. 71

6 ======================== d) 0 > e > (GM / l ) 2 There are two distinct roots for r, rmax and rmin, (at which of course dr/dt = 0.) These are the turning points in plots of U effective. As we will see later, the orbit is elliptical. Energy e is negative and so is insufficient for escape. But for the specified angular momentum, e is greater than the energy for a circular orbit. =========== Each of these four cases can be identified in our plot of effective U, for different values of e. Recall how we found effective U: Write e = (1/2) v 2 GM/r = (1/2)( dr/dt) 2 + (1/2)r 2 (df/dt) 2 - GM/r Then substitute df/dt =!/ r 2, to obtain e = (1/2)( dr/dt) 2 +! 2 / 2r 2 - GM/r We define U effective =! 2 / 2r 2 - GM/r, which for any! 0, always looks qualitatively like the figure here. It has a centrifugal barrier at small r and an attractive well at moderate r. 72

7 The possible motions and allowed ranges for r(t) are readily identified in the figure for effective U. Hints on how to do HW 4A problem 1 ( not presented in lecture ) Problem statement: A satellite is in a circular orbit around the earth in an orbit of radius r. (so you know its speed) It applies its thrusters and suddenly increases its speed (maintaining the direction of its motion) by a factor c > 1. What must c be if the final orbit is to be parabolic? If c is less than this, so that the final orbit is elliptical, what is the apogee of that orbit? What is the perigee? Hint: Its initial radius is r, its initial speed is [GM/r] 1/2 After the burn it is still at r, but now its speed is c [GM/r] 1/2 in the same direction. Calculate the new orbit's! and e From these, use the formulas developed above to calculate the extreme values of r. These are the distances of perigee and apogee. A parabolic orbit means r max =. An elliptic orbit means r min < r max < ============ The above analysis has told us lots, without our having to solve the differential equations. However it has not told us details of the shapes of the orbits or how speed and direction varies along the orbits, or even that the orbits are, as claimed, ellipses and hyperbolas and parabolas, or even that the (negative energy) orbits are closed. Nor has it allowed the possibility that M and m might be comparable. Thus we are now motivated to take a deeper look. More formal derivation of the orbits, in which it isn't necessarily the case that M >>> m. Also determination of the orbit's shape (not just its extreme points) and period, by solving the ODE Consider the two body problem of two point masses m1 and m2. They have position vectors r1 and r2 relative to some origin. The equations of motion are m 1 d 2 dt 2! r1 = Gm 1 m 2! r 1! r 2 3 (! r 1! r 2 ); m 2 d 2 dt 2! r2 = Gm 1 m 2! r 1! r 2 3 (! r 2! r 1 ) 73

8 This is six coupled nonlinear differential equations for the three components of each of the r's. We can greatly simplify it by defining a change of variables to C and r instead of r1 and r2 C = (m1 r1 + m2 r2 ) / ( m1 + m2 ) r = r1- r2 C is the vector position of the center of mass. r is the vector between the two masses pointing from #2 towards #1. By adding the above two differential equations, we get an eqn for C: (m 1 + m 2 ) d 2 dt 2! C = 0 showing that the center of mass of the two particle system has no acceleration. The center of mass! C moves at constant velocity C = V! C t + C! o. By dividing the first of the above equations by m 1 and the second by m 2 d 2! r1 = Gm 2 dt 2 r!! d 2! r; r2 = Gm 1 3 dt 2 r!! r 3 And subtracting them, we find: d 2! G(m r = 1 + m 2 ) dt 2 r!! GM r = 3 r!! r; 3 The relative position vector r, is governed by the same differential equation that would govern a test mass in orbit around a fixed central mass of the sum of the individual masses: M = m1 + m2. [ For example two equal mass bodies M have their relative position ( the vector between them ) evolve just like the position of a tiny test mass around a central fixed mass 2M ] The problem has thus been reduced to three degrees of freedom, the components of the relative position vector r = r 1-r 2 It is conventional to multiply the above eqn by the "reduced mass" µ = m 1 m 2 / M ( in practice slightly less than the smaller of the two masses m1 and m2). Then µ d 2 dt 2! r = GMµ! r 3! r; (**) 74

9 So now it looks like a mass µ in orbit around a fixed mass M. system energy The virtue of µ is that the total E = 1 2 m 1 (d! r 1 / dt) m 2 (d! r 2 / dt) 2 Gm 1 m 2 /! r 1! r 2 is ( do the algebra if you don't believe it ) E = 1 2 µ(d r! / dt) 2 GMµ/ r! M (d C! / dt) 2 the sum of the energies of each part, C and r. Solution of the ODE for r The presence of the factor µ of course makes no difference for the solution of eqn (**) We divide it out. Using spherical coordinates, r = r, q = latitude below the north pole of our coordinate system f = east longitude, (see figure) the ODE a = F/µ becomes = ˆr GM/r 2 We will solve this next time. 75

10 Miniquiz 8 Thursday Feb 7, 2019 name A particle of mass m moves in orbit in a two-dimensional attractive central force field F = - k r (Thus the potential energy is U = kr 2 /2, like a spring) ˆr Find the relationship between the radius of a circular orbit r and its speed v by equating the centripetal acceleration of the circular orbit to F/m. Answer: F=ma with a = v 2 /r as it is for a circular orbit at speed v means: kr = mv 2 /r, So v = r k/m is proportional to r ( unlike Kepler case that has v proportional to 1/ r) What is the period of a circular orbit of radius r? Answer: The period is P = 2p r/v = 2p m/k, independent of orbit radius r. For specified angular momentum L = m r 2 df/dt, find and sketch the effective potential U eff that appears in E = m 2!r 2 +U eff (r) Answer: E = (m/2)[ (dr/dt) 2 + (rdf/dt) 2 ] + (k/2)r 2 is energy. Substituting for df/dt using L = m r 2 df/dt, we find E = (m/2) (dr/dt) 2 + L 2 /2mr 2 + (k/2)r 2. U eff is the last two terms: U eff = L 2 /2mr 2 + (k/2)r 2 The centrifugal barrier L 2 /2mr 2 dominates at small r; the true potential (k/2)r 2 dominates at large r. There is a well in-between. All orbits are trapped in the well because for finite energy E, regardless of L, there is always a maximum possible r. 76