Assignment 6, solutions
Problem 1:
The source of the first gravitational
wave event observed by the LIGO collaboration in 2015 has been interpreted as
the merger of two black holes in a binary system, each with a mass of roughly 35
solar masses (implying a radius for the event horizon of about 100 km for each,
if assumed spherical), where a solar mass is 1.989*10^{30
}kg. A full understanding requires general relativity, but assume
Newtonian mechanics and Newtonian gravity as a first approximation for the
orbital motion. At the peak amplitude of the detected gravitational wave, its
measured frequency indicated that the two black holes were revolving around the
center of mass about 75 times per second.
What was the approximate separation of the centers for the two black holes at
this point in the merger event?
Solution:
Concepts: Kepler's third law, relative motion | |
Reasoning: 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). | |
Details of the calculation: Here U(r) = Gm^{2}/r, F(r) = Gm^{2}/r^{2},^{ } r = distance between their centers. µv^{2}/r = Gm^{2}/r^{2}. v^{2} = Gm^{2}/(µr). µ = m/2. v^{2} = G 2m/r. (2πr/T)^{2} = G 2m/r. r = [T^{2} G2m/(4π^{2})]^{⅓}. With T = (1/75) s and m = 35*1.989*10^{30 }kg we have r = 3.47*10^{5} m = 347 km. |
Problem 2:
A particle of mass m is released a distance b from a fixed origin of force that attracts the particle according to the inverse square law F(x) = –k/x^{2}. Find the time required for the particle to reach the origin. Use this result to show that, if the Earth were suddenly stopped in its orbit, it would take approximately 65 days for it to collide with the Sun. Assume that the Sun is as a fixed point mass and Earth’s orbit is circular.
Solution:
Concepts: Kepler’s third law | |
Reasoning: For a central force F(r) = -k/r^{2}, the closed orbits are elliptical orbits. All elliptical orbits with the same semi-major axis have the same period. The elliptical orbit with zero angular momentum is a straight line. The motion of a particle in this orbit is in one dimension and with the appropriate orientation of the coordinate system we have F(x) = -k/x^{2}. | |
Details of the calculation: The period of the motion of a particle with semi-major axis r can easily be found by considering the circular orbit. k/r^{2} = mv^{2}/r, (2πr/T)^{2}r = k/m, T^{2} = 4π^{2}mr^{3}/k. For the particle in the problem r = b/2 and the time t_{0} required to reach the origin is T/2. t_{0} = ½πb(½mb/k)^{½}. If the Earth were suddenly stopped in its orbit, its semi-major axis would be halved. (T_{new}/365 days) = (R_{new}/R)^{3/2} = 2^{-3/2}. T_{new} = 129 days. t_{0} = T_{new}/2 = 64.5 days.
We can also brute-force integrate to find t_{0}. |
Problem 3:
Find the maximum time a comet (C) of mass m following a parabolic trajectory around the Sun (S) can spend within the orbit of the Earth (E). Assume that the Earth's orbit is circular and in the same plane as that of the comet.
Solution:
Concepts: The Kepler problem | |
Reasoning: Both the Earth and the comet move in the central potential V(r) = -β/r. Let the radius of the circular orbit of the Earth be a. We need to find the time a comet following a parabolic orbit can spend between r_{min} and a. | |
Details of the calculation: The total energy of a particle following a parabolic orbit is zero. E = ½m_{c}(dr/dt)^{2} + U_{eff}(r) = 0, U_{eff}(r) = U(r) + M^{2}/(2m_{c}r^{2}) = -m_{c}β/r + M^{2}/(2m_{c}r^{2}). At r = r_{min} we have dr/dt = 0 and m_{c}β/r_{min} = M^{2}/(2m_{c}r_{min}^{2}). M^{2} = 2m_{c}^{2}βr_{min}. We have found the relationship between M and r_{min}. dr/dt = ((2/m)(E - U_{eff}(r))^{½}. The time the comet spend between r_{min} and a is t = 2∫_{rmin}^{a} dr/[(2/m_{c})(m_{c}β/r - M^{2}/(2m_{c}r^{2}))]^{½}. t = 2∫_{rmin}^{a} dr/[(2/m_{c})(m_{c}β/r - m_{c}βr_{min}/r^{2})]^{½}. t = (2/β)^{½}∫_{rmin}^{a} rdr/(r - r_{min})^{½}. ∫_{rmin}^{a} rdr/(r - r_{min})^{½} = (2/3)a^{3/2}(1 + 2r_{min}/a)(1 - r_{min}/a)^{½}. Let x = r_{min}/a. t = C(1 + 2x)(1 - x)^{½}. Let us find an extremum for t. dt/dx = 2C(1 - x)^{½} - ½C(1 + 2x)(1 - x)^{-½} = 0. 2(1 - x) - ½(1 + 2x) = 0. 3x = 3/2, x = ½. Since for r_{min} > a the time t spend within Earth's orbit is zero, the extremum is a maximum. The comet following a parabolic orbit spends the maximum time between r_{min} and a when r_{min} = a/2. Then t_{max} = (1/β)^{½}(4/3)a^{3/2}. Since (a^{3}/β)^{½} = T/2π (Kepler's third law), we have t_{max} = 2T/(3π). t_{max} = (2*365 days)/(3π) = 77 days. |
Problem 4:
A particle of mass m moves under the action of a central force whose
potential energy function is U(r) = kr^{4},
k > 0.
(a) For what energy and angular momentum will the orbit be a circle of radius a
about the origin? What is the period of this circular motion?
(b) If the particle is slightly disturbed from this circular motion, what will
be the period of small radial oscillations about r = a?
Solution:
Concepts: A particle moving in a central potential U = U(r), F = f(r)(r/r), f(r) = -∂U/∂r. In a central potential the energy E and the angular momentum M are conserved. E = ½m(dr/dt)^{2} + M^{2}/(2mr^{2}) + U(r) = ½m(dr/dt)^{2} + U_{eff}(r), with U_{eff}(r) = M^{2}/(2mr^{2}) + U(r). md^{2}r/dt^{2} = -∂U_{eff}(r)/∂r. | |
Reasoning: A potential energy function U(r) is given. The particle moves under the action of a central force. | |
Details of the calculation: (a) For a circular orbit with radius a we need r = a, dr/dt = 0, d^{2}r/dt^{2} = 0. For a circular orbit at with radius a we therefore need ∂U_{eff}(r)/∂r|_{a} = 0. ∂U_{eff}(r)/∂r = 4kr^{3} - M^{2}/(mr^{3}). 4ka^{3} = M^{2}/(ma^{3}), a^{6} = M^{2}/(4mk), M = 2a^{3}(km)^{½}. E = U_{eff}(a) = ka^{4 }+ M^{2}/(2ma^{2}) = ka^{4 }+ 4mka^{6}/(2ma^{2}) = 3ka^{4}. The period for the circular motion is τ = 2π/ω, ω = dΦ/dt = M/ma^{2} = 2a(k/m)^{½}. τ = (π/a)(m/k)^{½}. (b) For a stable orbit we need a restoring force. Let r = a + ρ, r >> ρ. We need md^{2}ρ/dt^{2} = -αρ. md^{2}ρ/dt^{2} = -∂U_{eff}(ρ)/∂ρ = -(∂U_{eff}/∂ρ)|_{ρ=0} - (∂^{2}U_{eff}/∂ρ^{2})|_{ρ=0}ρ (series expansion). We need ∂^{2}U_{eff}/∂ρ^{2} > 0 at ρ = 0, or ∂^{2}U_{eff}/∂r^{2} > 0 at r = a. ∂U_{eff}(r)/∂r = 4kr^{3} - M^{2}/(mr^{3}). ∂^{2}U_{eff}(r)/∂r^{2} = 12kr^{2} + 3M^{2}/(mr^{4}). ∂^{2}Ueff/∂ρ^{2}|_{r=a} = 12ka^{2} + 3M^{2}/(ma^{4}) = 12ka^{2} + 3kma^{6}/(ma^{4}) = 24 ka^{2} > 0. The orbit with radius a is stable, α = 24 ka^{2}. The period of small radial oscillation about a is τ = 2π/ω_{r}, ω_{r} = (α/m)^{½} = (24 ka^{2}/m)^{½}. τ = (π/a)(m/(6k))^{½}. |
Problem 5:
A steel disk A with radius R moves with speed v = 10 m/s when it collides with a second identical disk B at rest. The collision is elastic and has an impact parameter “b”. After the collision, the speed of disk A is equal to 5 m/s. What is the value of the impact parameter “b”? Neglect friction.Solution:
Concepts: Conservation of energy and momentum | |
Reasoning: In the elastic collision kinetic energy and momentum are conserved. | |
Details of the calculation: Let disk A initially move in the x direction. The in the CM frame the initial velocity of disk A is 5 m/s i and the initial velocity of disk B is -5 m/s i. After the collision disk A has velocity components v_{xA} = -5 m/s cosθ, v_{yA} = 5 m/s sinθ, and disk B has velocity components v_{xB} = 5 m/s cosθ, v_{yB} = -5 m/s sinθ. In the lab frame v_{A} = 5 m/s (1 - cosθ) i + 5 m/s sinθ j, and v_{B} = 5 m/s (1 + cosθ) i - 5 m/s sinθ j. v_{A}^{2} = 25 m^{2}/s^{2} = 25 m^{2}/s^{2} [(1 - cosθ)^{2} + sin^{2}θ]. 2 - 2cosθ = 1, θ = 60^{o}. Impact parameter b: b = 2R sin30^{o} = R. |