## Assignment 6, solutions

#### Problem 1:

A star of mass M and radius R is moving with constant velocity v through a cloud of particles of density ρ.
If all the particles which collide with the star are trapped by it, show that the mass of the star will increase at a rate
dM/dt = πρv(R2 + 2GMR/v2).

Solution:

• Concepts:
Motion in a central potential conservation of energy and angular momentum
• Reasoning:
The particles in the cloud approach the star.  At a large distance from the star they had speed v and varying impact parameters.  We find the impact parameter b for the particles that will just graze the surface of the star.  The particles that are be trapped by the star were, at a large distance, contained in a cylinder of radius b with its axis pointing towards the star and moving towards the star with speed v.
The volume V = πb2vt of this cylinder sweeps over the star in time t. Therefore  dM/dt = ρπb2v.
• Details of the calculation:
Dor a particle that just grazes the surface of the star we have
L = mbv = mRvR, where vR is the speed at R and v = v.
E = ½mv2 = ½mvR2 - GMm/R.  vR2 = v2 + 2GM/R.
b = RvR/v= R(1 + 2GM/(Rv2))½.
dM/dt = πρv(R2 + 2GMR/v2).

#### Problem 2:

A satellite is in a circular orbit of radius r around an airless spherical planet of radius R.  An asteroid of equal mass falls radially towards the planet, starting at zero velocity from a very large distance. The satellite and the asteroid collide inelastically and stick together, moving in a new orbit that just misses the planet's surface.  What was the radius r of the satellite's original circular orbit in terms of R?

Solution:

• Concepts:
Kepler orbits
• Reasoning:
In a central potential the energy E and the angular momentum L are conserved.  The motion is in a plane. For an object with mass m' in a circular orbit about a fixed center with mass M we have
T = ½m'v2 = ½GMm'/r,  U = -GMm'/r,  E = -½GMm'/r,
p = m'v = (GMm'2/r)½,  L = rp = (GMm'2r)½.
• Details of the calculation:
Just before the collision:
For the orbiting mass m, E1 = -½GMm/r,  T1 = ½GMm/r,  p1 = (GMm2/r)½,  L1 = (GMm2r)½.
For the falling mass m, E2 = 0,  T2 = GMm/r,  p2 = (2GMm2/r)½,  L2 = 0.

Just after the collision:
Momentum and angular momentum are conserved in the totally inelastic collision.  The momenta of the two colliding objects are perpendicular to each other.  Therefore the magnitude of the momentum of the combined mass is p = (p12 + p22)½,  p2 = 3GMm2/r.
The magnitude of the angular momentum of the combined object is  L = (GMm2r)½.
Its energy is E = T + U = p2/4m - 2GMm/r = -(5/4)GMm/r.

At the distance of closest approach L = pmaxrmin = pmaxR = (GMm2r)½, pmax2R2 = GMm2r.
E = pmax2/(4m) - 2GMm/R =  -(5/4)GMm/r.
GMm2r/(4mR2) - 2GMm/R =  -(5/4)GMm/r.
r2 - 8Rr + 5R2 = 0,  r = 4R ± √(11)R.
The perihelion of the orbit is at R, so r > R.
r =  4R ± √(11)R.

#### Problem 3:

Consider a particle of mass m moving in the xy plane.  The potential energy function is
U(x,y) = (k/2)/(x2 + y2), with k a positive constant.
(a)  Find the equations of motion.
(b)  Are there circular orbits?  If yes, do they all have the same period?
(c)  Is the total energy conserved?

Solution:

• Concepts:
Central force problem in 2 dimensions
• Reasoning:
We can use the Lagrangian formalism or work in Cartesian coordinates and use Newton's second law.
• Details of the calculation:
Use the Lagrangian formalism.  Choose polar coordinates (r, θ).
(a)  T = ½m[(dr/dt)2 + r2(dθ/dt)2],   U(r) = k/(2r2).
L = T - U
∂L/∂((dr/dt)) = mdr/dt,  d/dt(∂L/∂(dr/dt)) = md2r/dt2,  ∂L/∂r = mr(dθ/dt)2 + k/r3.
md2r/dt2 - mr(dθ/dt)2 - k/r3 = 0.
∂L/∂((dθ/dt)) = mr2(dθ/dt),  ∂L/∂θ = 0.
mr2(dθ/dt) = M = constant.  The angular momentum is conserved.
The equation of motion for θ is dθ/dt = M/(mr2).
md2r/dt2 - M2/(mr3) - k/r3 = 0.
The equation of motion for r is d2r/dt2 - (M2/m2 + k/m)/r3 = 0.
(b)  circular orbit:  d2r/dt2 = 0,  M2/m = -k.  This is not possible.
We have a repulsive potential.  Circular orbits are not possible.
(c)  If the Lagrangian does not explicitly depend on time and the generalized coordinates do not explicitly depend on time, then H = T + U = E and the energy is a constant of motion.  This is the case here.

#### Problem 4:

(a)  Derive the relationship between the impact parameter b and the scattering angle θ for Rutherford scattering of a projectile of mass m by a fixed target particle of mass M.
(b)  If 4 MeV alpha particles are incident on a gold foil (Z = 79), calculate the impact parameter that would give a deflection of 10 degrees.
(c)  Explain what modifications of this calculation must be made if the gold atom is allowed to recoil during the collision.

Solution:

• Concepts:
Motion in a central potential, elastic scattering
• Reasoning:
For the case of the repulsive Coulomb potential U(r) = α/r, we are asked to find the relationship between the impact parameter and the scattering angle.
• Details of the calculation:
In a central potential the motion is in a plane and M and E are constant.
E = ½m(dr/dt)2 + b2E/r2 + U(r),  dr/dt = ±[(2/m)(E(1 - b2/r2) - U(r))]½.
From
dr/dt = [(2/m)(E(1 - b2/r2) - U(r))]½,  d/dt = [b(2mE)½/(mr2)] d/dφ,
we find the equation for the orbit.
dφ = [b(2mE)½/(mr2)]dr [(2/m)(E(1 - b2/r2) - U(r))].
φ(u) = b∫0udu'/[1 + b2u'2 - U(u')/E]½,
with u = 1/r and φ(z = -∞) = 0.

Let θ be the angle between the incident and the scattered direction and φ0 be the angle between r(z = -∞) and rmin.  Then
φ0 = b∫0umaxdu'/[1 + b2u'2 - U(u')/E]½,
and
E = M2/(2mr2min) + U(rmin) = b2E/r2min + U(rmin) =  b2Eu2max + U(umax)
determines umax.  We have
θ = π - 2φ0 for a repulsive potential,
θ = 2φ0 - π for an attractive potential, (or θ = π - 2φ0, θ < 0).

If U(r) = α/r,  U(u) = αu,  U(umax) = αumax, then
umax = -α/(2b2E) + [α2/(4b4E2) + 1/b2]½ and φ0 = cot-1(α/(2bE)).
cotφ0 = α/(2bE),  cot(θ/2) = cot(π/2 - φ0) = tanφ0 = 1/cotφ0.
b = [α/(2E)] cot(θ/2)
is the he relationship between the impact parameter b and the scattering angle θ for Rutherford scattering.

(b)  E = 4 MeV = 4*10eV(1.6*10-19 J/eV) = 6.4*10-13 J.
α = 1/(4πε0)(2)(79)(1.6*10-19 C)2
=  (9.0*109 Nm2/C2)(158)(2.56*10-38 C2)
=  3640*10-29 Nm2 = 3.6*10-26 Jm.
θ = π - 2Φ0 = 10o.
cot 5o  = 11.43
b = (3.6*10-26 Jm)(11.43)/2(6.4*10-13 J)
= 3.2*10-13  m.

(c)  If the target is allowed to recoil, then we solve for the motion of the CM and the motion about the CM, i.e. the relative motion.  The CM moves with constant velocity V = m v1/(m + M).  The problem of two interacting particles in their CM frame is equivalent to the problem of a fictitious particle of reduced mass μ moving in a central potential U(r).  The reduced mass of the alpha particle-gold nucleus system is μ = mM/(m + M).  If the energy of the alpha particle in the lab frame is E, then the energy of the fictitious particle is E' = ME/(M + m).  Given a scattering angle θ in the lab frame we need to find the corresponding scattering angle θ' in the CM frame.

v'sinθ' = vsinθ,  v2 = v'2 + V2 + 2v'Vcosθ'.
Then we use
b = (α/2E')cot(θ'/2)
to find the impact parameter b for θ'.

#### Problem 5:

Three stars with masses m1, m2, and m3 are forming a peculiar triple-star system, where each of the stars is situated in the corners of an equilateral triangle with a side length d.  The stars are attracting each other with gravitational forces.  Determine the direction and magnitude of the rotational velocity ω which will leave the relative position of the three stars unchanged.

Solution:

• Concepts:
Orbiting, uniform circular motion, the center of mass (CM)
• Reasoning:
For the relative positions of the stars to remain unchanged all stars must have circular orbits lying in the plane of the triangle about a common point.  The force on each star must be directed towards this point, the center of the circular orbits.   We show that this point is the CM.
• Details of the calculation:

Place the triangle into the xy plane.  Place the CM of the system at the origin.
M = m1 + m2 + m3,  MR = m1r1 + m2r2 + m3r3 = 0.
Star 1:  F1 = F12 + F13 = Gm1m2(r2 - r1)/d3 + Gm1m3(r3 - r1)/d3.
F1 = (Gm1/d3)[m2r2 + m3r3 - m2r1 - m3r1 + m1r1 - m1r1] = (Gm1/d3)[MR - Mr1] = -(GMm1/d3)r1.
The direction of F1 is towards the origin, towards the CM.
For the star to move in a circle of radius r1 we need F1 = mv12/r1, v1 = r1ω.
(GMm1/d3)r1 = mr1ω2, ω = (GM/d3)½.

Focusing on star 2 and star 3 we will get the same result for ω.  ω = ±(GM/d3)½k