Problem 1 ( 15 points) A particle of mass m moves in one-dimension while suffering a drag forceopposed to its motion of magnitude Cv2/3 where C is a positive constant. It starts at the origin x = 0 attime 0 with velocity vo > 0, that then diminishes due to the drag. a) What is its speed after time t? b) At what time does it stop?*c) How far does it go before stopping? mdv / dt = −Cv2 / 3; sodvv2 / 3= −Cdtm; so 3v1 / 3vov= −Ct / mso v1 / 3− vo1 / 3= −Ct / 3m; so v = v0[1− Ct / 3mvo1 / 3]3So tstop= 3mvo1 / 3/ Cx = v0[1− Ct / 3mvo1 / 3]30tstop∫dt =vo4 / 33m4CProblem 2: (15 points) A particle of mass m=2kg moves in 1-d under a force distribution withpotential energy (in Joules) given by the plot (tic marks indicate Joules and meters).Describe the possible motions of the mass qualitatively, andestimate the x-value of any turning points(s), if… a) It has energy E = 4 Joulesanswer: It passes through the region without any turning. b) It has energy E = 1 Jouleanswer: there is a turning point at about x=4.5. If the particle isinitially going rightwards, it turns at x = 4.5 and goes left. Ifinitially going left it just keeps on going, faster and slower inplaces, but without turning. c) It has energy E = -3 Jouleanswer: it is trapped in the well, oscillating periodically between turning points at approximately x =3.5 and x = -3.5 *d) Now assume it starts at position xo = 0 with vo = 2m/sec to the right. What is its kinetic energythere? What is its Potential energy there?KE = (1/2)m vo2 = 4 J; PE = (from reading the plot) –5 J. So total energy is –1 J.What is its speed and position as t goes to ∞?It ends up ultimately going leftwards (after bouncing at about x = 4) to x -> -∞, where PE = -2, so KE= 1J (because total E = -1J ) so speed = 1m/sec (leftwards) from 1 J = (1/2) m v2Problem 3: (20 points) A uniform rod of total mass M and length L has negligible cross sectional area. We are interested inthe gravitational field at a point to the right of the right end. a) Find the gravitational potential Φ at a point a distance 'a' from the end of the rod.Φ = − G dm / r∫; Set dm = (M / L)ds where s! runs! from!!0to L and r = L − s + aΦ = − (MG / L)ds / (s=0s= L∫L − s + a) = (MG / L)ln{L − s + a}s=0L= (MG / L)ln(aa + L) b) Show that your expression for Φ has the correct a-dependence in the limit of large a: a >> L; a < ∞Φ = (MG / L)ln(aa + L) = −(MG / L)ln(1+ L / a) But ln(1+ε) ~ ε so Φ ≈ −(MG / L)(L / a) = −MG / aThis is the correct (1/distance) behavior.Problem 4a: (10 points) Consider a uniform thin spherical shell ofradius R and mass Mshell hanging in otherwise empty space. At thecenter of the shell is a point mass MpointFind the magnitude |rg(r) | of the gravitational acceleration at pointsa distance r from the center and outside the shell r >R.Recall that outside a spherical mass distribution, g is the same as itwould be if all the mass were concentrated at the center. Thus |g| =G(Mshell +mpoint )/r2.Find the magnitude |rg(r) | of the gravitational acceleration at points a distance r from the center andinside the shell r < R.Recall that inside a spherical mass distribution, the mass outside doesn't contribute to g. So|g| = Gmpoint/r2.(*) Choose Φ(∞) = 0 and find the expression(s) for the gravitational potential Φ(r) for all points r insideand outside the shell. Make sure your Φ is continuous at R.We want the derivative of Φ to be the g described above, soOutside R, Φ = Constant – G(Mshell+mpoint)/rWe choose Constant = 0 so that Φ = 0 at ∞,Inside R Φ = Konstant – Gmpoint/rContinuity then demands Konstant = -GMshell/RProblem 4b (10 points) A thin circular hoop of mass Mand radius R hangs in empty space. What is the gravitationalpotential Φ at a point P a distance x above its center ? (Hint:this is very easy.)All points are at the same distance (R2+x2)1/2 from P, henceΦ = −GM/(R2+x2)1/2Problem 5: (30 points) Consider a spaceship of small mass µ in an elliptical orbit around the sun. Theorbit has eccentricity ε < 1 and closest approach(perihelion) a distance Rmin from the sun. Allanswers are to be given in terms of µ, and the sun'smass M, and G, ε, and the given Rmin.a) What is its furthest distance (aphelion) Rmaxfrom the sun? We use r(φ) =α1+εcos(φ);α= l2/ GM;ε= 1+2el2G2M2This implies Rmin = α/(1+ε); hence Rmax = α/(1-ε) = Rmin1+ε1−εb) What is its speed vperihelion at closest approach? We see thatα =Rmin(1+ε); so l =αGM = GMRmin(1+ε) = RminvperhelionSo vperhelion= GM (1+ε) / Rmin .c) What is its speed vaphelion at Rmax?answer: at aphelion the speed must be less than at perhelion by the ratio of the R ( by angularmomentum conservation) so vaphelion =vperhlion(Rmin/Rmax) =GM (1+ε) / Rmin1−ε1+εd) What is the magnitude |L| of its angularmomentum ?|L| = µl =µGMRmin(1+ε)e) What is its energy E?Use ε= 1 +2el2G2M2 to solve for e and set E = eµ.*f) What instantaneous increase of speed Δv isrequired if the spaceship is to boost into an escape trajectory from perihelion r = Rmin ?Escape velocity from a distance R is Vescape= (2GM/R)1/2. The boost needed is therefore Δv = Vescape-vperhlion=(2GM/Rmin)1/2 -GM (1+ε) / RminProblem 6. The gravitational potential on the axis above a thin disk was worked out in class to beΦ = −2πGρh[ R2+ x2− x] where ρ is the volumetric mass density of the disk material and h isits thickness.If two particles are located at different heights, x and x + ε, where ε is small and positive, findthe tidal acceleration between them. Which one accelerates faster and by how much?Answer: the acceleration is −dΦ / dx = −g = 2πGρh[xR2+ x2−1]. (downwards, but slower atgreater heights) The gradient of this is negative: g' = dg / dx = −2πGρhR2(R2+ x2)3/2 which impliesthat a higher object falls less quickly. The differential acceleration is g' ε (where ε is the Δx