Subfield: Modern Physics

$$

\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}

$$


$$

\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 $r$ is given by



$$

F=P Q \pi r^{2}

$$



where $P$ is the radiation pressure and $Q$ is a dimensionless quality factor that depends on the relative size of the particle $r$ and the wavelength of light $\lambda$. Throughout this problem assume that the sun emits a single wavelength $\lambda_{\max }$; 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 $L_{\odot}$, find an expression for the radiation pressure a distance $R$ from the sun.
Context answer:
\boxed{$P=\frac{L \odot}{4 \pi R^{2} c}$}


Context question:
ii. Assuming that the particle has a density $\rho$, derive an expression for the ratio $\frac{F_{\text {radiation }}}{F_{\text {gravity }}}$ in terms of $L_{\odot}$, mass of sun $M_{\odot}, \rho$, particle radius $r$, and quality factor $Q$.
Context answer:
\boxed{$\frac{F_{\text {radiation }}}{F_{\text {gravity }}}=\frac{3 L_{\odot}}{16 \pi G c M_{\odot} \rho} \frac{Q}{r}$}


Context question:
iii. The quality factor is given by one of the following



- If $r \ll \lambda, Q \sim(r / \lambda)^{2}$

- If $r \sim \lambda, Q \sim 1$.

- If $r \gg \lambda, Q=1$



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 $\lambda$ 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 $v$ in terms of $M_{\odot}$, distance from sun $R$, and any other fundamental constants.
Context answer:
\boxed{$v=\sqrt{\frac{G M_{\odot}}{R}}$}


Context question:
ii. Because the particle is moving, the radiation force is not directed directly away from the sun. Find the torque $\tau$ on the particle because of radiation pressure. You may assume that $v \ll c$.
Context answer:
\boxed{$\tau=-\frac{v}{c} \frac{L \odot}{4 R c} Q r^{2}$}


Question: iii. Since $\tau=d L / d t$, the angular momentum $L$ of the particle changes with time. As such, develop a differential equation to find $d R / d t$, the rate of change of the radial location of the particle. You may assume the orbit is always quasi circular.

Solution:
The angular momentum is



$$

L=m v R=\frac{4}{3} \pi \rho r^{3} \sqrt{\frac{G M_{\odot}}{R}} R=\frac{4}{3} \pi \rho r^{3} \sqrt{G M_{\odot} R}

$$



Differentiating both sides with respect to time,



$$

-\frac{v}{c} \frac{L_{\odot}}{4 R c} Q r^{2}=\frac{4}{3} \pi \rho r^{3} \sqrt{\frac{G M_{\odot}}{R}} \frac{1}{2} \frac{d R}{d t}

$$



which simplifies to



$$

-\frac{1}{c^{2}} \frac{L_{\odot}}{R} Q=\frac{8}{3} \pi \rho r \frac{d R}{d t}

$$

Final answer: $-\frac{1}{c^{2}} \frac{L_{\odot}}{R} Q=\frac{8}{3} \pi \rho r \frac{d R}{d t}$