85Phys 325 Lecture 10 Thursday 19 February, 2015NB Exam next week!Modified Gravitational lawsIt is amusing, and instructive as an exercise, to consider modified Gravitational laws, with adifferent power of distance. For example. F = GMµ/ r3 . Of course the dimensions of theconstant G are now different.If the power is three, as many would argue would obtain in a universe with 4 spatial dimensions,we would then have for the potential energy per massΦ = - GM/2r2 ( as opposed to Newtonian case - GM/r )The same arguments apply as before, l = r2dφ/dt is a constant of the motion. So is e, but nowthe expression for e ise = (1/2) ( (dr/dt)2 + r2 (dφ/dt)2) - G M / 2r2ore = (1/2) ( (dr/dt)2 + l 2/r2 ) - G M / 2r 2 = (1/2) (dr/dt)2 + Ueff(r)where Ueff(r), the effective potential (forthe motion r(t) ) isUeff = l2/2r2 - G M / 2r2This implies entirely different behaviorfrom the usual case.For l2 > GM, e must be positive, there isa minimum r, and there is no maximum r,i.e, there are no trapped orbits.For small angular momentum, l2 < GM,e could be positive or negative.If e is positive while l2 < GM thefate of the particle depends on its initialdirection; if going ourwards , its escapesto ∞; if going inwards, it spirals into the origin.If e is negative while l2 < GM then there is a maximum r, but no minimum. Alltrajectories that start with dr/dt < 0 spiral directly into the sun. Trajectories that start with dr/dt >0 spiral out for a while to a maximum r, and then turn and spiral in.In physical terms, we could say that angular momentum tends to repel r from the origin, butgravity increases fast enough at small r to compensate.86But wait! Can't you still look for circular orbits like we did earlier? In that case we'd set v2/r =acceleration = GM/r3. This implies v= [ GM / r2 ]1/2. and KE per mass = v2/2 and PE per mass= -GM/2r2 , so total energy per mass e = 0. Also l=vr = [GM]1/2. Yes, there are circular orbits.They have angular momentum such that Ueff = 0. But now imagine perturbing a circular orbitslightly, changing its angular momentum such that l ≠ [GM]1/2. Then (according to the figure)the planet either spirals into the sun or spirals out to ∞. Circular orbits are unstable in thisuniverse.In our universe, however, circular orbits, when perturbed just become elliptical. Muchsafer.-----If the power of r in the formula for gravitational force were 1: F = -GM/r ( as could be argued itwould be in a universe with 2 spatial dimensions ) thenΦ = GM loge(r) ( as opposed to Newtonian case - GM/r )The same arguments apply as before, l = r2dφ/dt is a constant of the motion. So is e (because allforces are conservative), but now the expression for e ise = (1/2) [ (dr/dt)2 + r2 (dφ/dt)2 ] + G M loge(r)ore = (1/2) ( (dr/dt)2 + l2/r2 ) + G M log(r) = (1/2) (dr/dt)2 + Ueff(r)where Ueff(r), the effective potential isU = l2/2r2 + G M loge(r)Now, regardless of e, all orbits are trapped.None escape to ∞, Nor can any get to r = 0( unless l is 0 )What shape are these orbits?The equation for u(φ) now (after a processsimilar to that used to get the eqn for u(φ)earlier) becomesd2u/dφ2 + u = = -r2F(r)/m = (GM/l2)/uThis differential equation is nonlinear, which makes it much harder to solve than what we hadpreviously: d2u/dφ2 + u = (GM/l2).Often for nonlineaer ODEs we get insight by considering perturbations on simple solutions.87One simple solution is u = constant, i.e a circular orbit.u0 = ( GM/l2 )1/2 ( i.e. orbit radius 1/uo = l/ √GM = constant)which has period 2π / (dφ/dt) = 2π / l uo2 = 2π l /GMLet us consider an orbit of arbitrary angular momentum l, that is nearly circular. Let it have ashape u(φ) = u0 + η(φ), where η is small and represents deviations from the circular orbit u = u0 = (GM/l2 )1/2 . Then the eqn for η isd2η/dφ2 + uo + η = GM/( uo + η)l 2Now, by expanding the right side to leading order in the small quantity η, (using [ 1 + ε ]-1 ~ 1 - ε )we getd2η/dφ2 + uo + η = GM/ uo l 2 – GM η / (uo 2 l2)Note that the terms independent of η (the second term on the left and the first term on the right)cancel (see the formula for uo) They must cancel because we know η=0 has to be a solution as itcorresponds to the simple solution identified above.Thus the small perturbations satisfyd2η/dφ2 + η (1+ GM / (uo 2 l2 ) )= 0ord2η/dφ2 + 2 η = 0This equation for the smallperturbations η is linear; we cansolve it. Typical solutions areη(φ) = Α cos(√2φ)The perturbations ( if smallenough to justify ourapproximations) to the circularorbit r = constant have period2π/√2 in φ; these are notcommensurate with 2π. Hencethe perturbed orbit is NOT anellipse. It isn't even closed.88An example is illustrated here. r = [ 1 + 0.1 cos√2φ) ]−1This modified force law implies that orbits retard their perihelions. Each year's closest approachis at a value of φ greater by 360° / √2 ~ 256° , equivalent to -105° . Interestingly, somethingsimilar happens in our solar system. Mercury, and to a lesser extent other planets, advance theirperihelions, albeit very slowly (seconds of arc per century). This is partly due to perturbationsfrom the other planets, but not completely. For many years this was a mystery; but GeneralRelativity showed that it is a relativistic effect.Orbits in repulsive 1/r2 central forces ( e.g. alpha particle going by a nucleus )The mathematics we have been looking at for particle trajectories in the solar system is very likethat we'd need for the Coulomb interaction. If the charges are opposite, the force is still attractive,and the only difference is that we'd have some spare factors of q/m charge to mass ratio (that wereunity for gravitation). If , however the charges are the same, like they were for Rutherford'sexperiment, then the force is repulsive, some of the signs are different, and bound orbits are notpossible.Consider a fixed charge Q with a particle of mass µ and charge q in its vicinity. In this case theforce is qQ/r2 and the potential energy isU(r) = qQ / rtotal energy is E = qQ / r +µ2(&r2+ r2&φ2)Angular momentum is L =µr2&φin the direction perpendicular to the planar trajectory.Thus E =µ2&r2+ qQ / r +L22µr2from which we recognizeUeff(r) =qQr+L22µr289If the charges are opposite, plots of Ueffective look much like they do for gravitation, with a potentialwell in which we can have bound orbits.If