iii. Since , the angular momentum of the particle changes with time. · OlympiadBench #1493

Physics · modern-physics

Problem

iii. Since τ=dL/dt\tau=d L / d t, the angular momentum LL of the particle changes with time. As such, develop a differential equation to find dR/dtd R / d t, the rate of change of the radial location of the particle. You may assume the orbit is always quasi circular. Given constants: g=9.8 N/kgG=6.67×10−11 N⋅m2/kg2k=1/4πϵ0=8.99×109 N⋅m2/C2km=μ0/4π=10−7 T⋅m/Ac=3.00×108 m/skB=1.38×10−23 J/KNA=6.02×1023( mol)−1R=NAkB=8.31 J/(mol⋅K)σ=5.67×10−8 J/(s⋅m2⋅K4)e=1.602×10−19C1eV=1.602×10−19 Jh=6.63×10−34 J⋅s=4.14×me=9.109×10−31 kg=0.511MeV/c2(1+x)n≈1+nx for ∣x∣≪1sin⁡θ≈θ−16θ3 for ∣θ∣≪1cos⁡θ≈1−12θ2 for ∣θ∣≪1\begin{array}{ll} g=9.8 \mathrm{~N} / \mathrm{kg} & G=6.67 \times 10^{-11} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{kg}^{2} \\ k=1 / 4 \pi \epsilon_{0}=8.99 \times 10^{9} \mathrm{~N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} & k_{\mathrm{m}}=\mu_{0} / 4 \pi=10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A} \\ c=3.00 \times 10^{8} \mathrm{~m} / \mathrm{s} & k_{\mathrm{B}}=1.38 \times 10^{-23} \mathrm{~J} / \mathrm{K} \\ N_{\mathrm{A}}=6.02 \times 10^{23}(\mathrm{~mol})^{-1} & R=N_{\mathrm{A}} k_{\mathrm{B}}=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K}) \\ \sigma=5.67 \times 10^{-8} \mathrm{~J} /\left(\mathrm{s} \cdot \mathrm{m}^{2} \cdot \mathrm{K}^{4}\right) & e=1.602 \times 10^{-19} \mathrm{C} \\ 1 \mathrm{eV}=1.602 \times 10^{-19} \mathrm{~J} & h=6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.14 \times \\ m_{e}=9.109 \times 10^{-31} \mathrm{~kg}=0.511 \mathrm{MeV} / \mathrm{c}^{2} & (1+x)^{n} \approx 1+n x \text { for }|x| \ll 1 \\ \sin \theta \approx \theta-\frac{1}{6} \theta^{3} \text { for }|\theta| \ll 1 & \cos \theta \approx 1-\frac{1}{2} \theta^{2} \text { for }|\theta| \ll 1 \end{array} M⊙=1.989×1030 kgL⊙=3.828×1026 WRearth =1.5×1011 mλmax⁡=500 nm\begin{array}{lrr} M_{\odot} & = & 1.989 \times 10^{30} \mathrm{~kg} \\ L_{\odot} & = & 3.828 \times 10^{26} \mathrm{~W} \\ R_{\text {earth }} & = & 1.5 \times 10^{11} \mathrm{~m} \\ \lambda_{\max } & = & 500 \mathrm{~nm} \end{array} Extra Supplementary Reading Materials: Radiation pressure from the sun is responsible for cleaning out the inner solar system of small particles. a. The force of radiation on a spherical particle of radius rr is given by F=PQπr2F=P Q \pi r^{2} where PP is the radiation pressure and QQ is a dimensionless quality factor that depends on the relative size of the particle rr and the wavelength of light λ\lambda. Throughout this problem assume that the sun emits a single wavelength λmax⁡\lambda_{\max }; unless told otherwise, leave your answers in terms of symbolic variables. Context question: i. Given that the total power radiated from the sun is given by L⊙L_{\odot}, find an expression for the radiation pressure a distance RR from the sun. Context answer: P=L⊙4πR2cP=\frac{L \odot}{4 \pi R^{2} c} Context question: ii. Assuming that the particle has a density ρ\rho, derive an expression for the ratio Fradiation Fgravity \frac{F_{\text {radiation }}}{F_{\text {gravity }}} in terms of L⊙L_{\odot}, mass of sun M⊙,ρM_{\odot}, \rho, particle radius rr, and quality factor QQ. Context answer: Fradiation Fgravity =3L⊙16πGcM⊙ρQr\frac{F_{\text {radiation }}}{F_{\text {gravity }}}=\frac{3 L_{\odot}}{16 \pi G c M_{\odot} \rho} \frac{Q}{r} Context question: iii. The quality factor is given by one of the following - If r≪λ,Q∼(r/λ)2r \ll \lambda, Q \sim(r / \lambda)^{2} - If r∼λ,Q∼1r \sim \lambda, Q \sim 1. - If r≫λ,Q=1r \gg \lambda, Q=1 Considering the three possible particle sizes, which is most likely to be blown away by the solar radiation pressure? Context answer: particles of size near λ\lambda are most likely to be blown away, and even then, only if the density is small enough. Extra Supplementary Reading Materials: b. The Poynting-Robertson effect acts as another mechanism for cleaning out the solar system. Context question: i. Assume that a particle is in a circular orbit around the sun. Find the speed of the particle vv in terms of M⊙M_{\odot}, distance from sun RR, and any other fundamental constants. Context answer: v=GM⊙Rv=\sqrt{\frac{G M_{\odot}}{R}} Context question: ii. Because the particle is moving, the radiation force is not directed directly away from the sun. Find the torque τ\tau on the particle because of radiation pressure. You may assume that v≪cv \ll c. Context answer: τ=−vcL⊙4RcQr2\tau=-\frac{v}{c} \frac{L \odot}{4 R c} Q r^{2}

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