Physics 325 Discussion 4 Solution2/16/2015Problem 1vc< va– true: Because the potential is ∝ r−1and rc> ra, the potential at c is higher than at a, thenbecause energy is conserved, the kinetic energy, and thus the velocity, must be lower at c than at a.vc< vb– true: Same as above.|Lc| < |Lb| – false: Angular momentum is conserved, and thus |Lc| = |Lb|.Ec> 0 – false: The orbit is bounded, and because the convention is to define a 0 of potential at r = ∞, abounded orbit has E < 0 at all points. Note, E = 0 corresponds to a parabolic orbit, and E > 0 correspondsto a hyperbolic orbit.vara< vbrb– true: Angular momentum, L = r × p = rp sin θ, is conserved, so varasin θa= vbrbsin θb. Ata, sin θa= 1, while at b, sin θb< 1, thus the statement is true. Note that Kepler’s Second Law cannot beused here, because the r˙θ = v only holds when there is no radial velocity (i.e. only at points a and c).Ec= Ea– true: This is the statement of conservation of energy.v2ara=GMr2a– false: This statement is true for uniform circular motion, which can be shown using the radialacceleration in polar coordinates, ar= ¨r − r˙θ2. As in the case of uniform circular motion r˙θ2=v2r, becausethe velocity is only in the azimuthal direction at point a (and also c). However, at point a, ¨r > 0, becausethe radial velocity (i.e. ˙r) is moving from negative to positive. Thus ar> −v2r, where ar= −GMr2is theacceleration due to gravity. Thus −GMr2> −v2r, orGMr2<v2r.v2ara>GMr2a– true: See above.Problem 2For this problem, we need to determine l from the given equation. The equation gives:R =`(GM/`) +p(GM/`)2+ 2e(2.1)R0=`(GM/`) −p(GM/`)2+ 2e(2.2)at perihelion and aphelion, respectively. At this point, we have a system of two equations in terms of two1variables, ` and e. One way to solve this is to rewrite the equations in a simpler form as:R =`2/GM1 +p1 + 2e(`/GM)2=u1 + v(2.3)R0=`2/GM1 −p1 + 2e(`/GM)2=u1 − v(2.4)where we have defined u = `2/GM and v =p1 + 2e(`/GM)2. This system of equation can be solved simplyfor u in terms of R and R0, where:u =2RR0R + R0(2.5)⇒ ` =r2GMRR0R + R0(2.6)We have conservation of angular momentum, and noting L = r × p, at aphelion, because r is perpendicularto v, we have L = R0p sin θ = R0p ⇒ ` = R0v. So then:v =lR0(2.7)=s2GMRR0(R + R0)(2.8)To enter the circular orbit from aphelion, requires that the velocity satisfies v2/R0= GM/R02⇒ v =pGM/R0, thus we can write the required velocity boost as:∆v =rGMR0−s2GMRR0(R + R0)=rGMR0 1 −r2RR +
View Full Document
Unlocking...