Physics

Physics · modern-physics

Problem

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. 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}

Browse the problem library