Subfield: Modern Physics

- Proton mass, $m_{p}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Neutron mass, $m_{n}=1.67 \cdot 10^{-27} \mathrm{~kg}$
- Electron mass, $m_{e}=9.11 \cdot 10^{-31} \mathrm{~kg}$
- Avogadro's constant, $N_{0}=6.02 \cdot 10^{23} \mathrm{~mol}^{-1}$
- Universal gas constant, $R=8.31 \mathrm{~J} /(\mathrm{mol} \cdot \mathrm{K})$
- Boltzmann's constant, $k_{B}=1.38 \cdot 10^{-23} \mathrm{~J} / \mathrm{K}$
- Electron charge magnitude, $e=1.60 \cdot 10^{-19} \mathrm{C}$
- 1 electron volt, $1 \mathrm{eV}=1.60 \cdot 10^{-19} \mathrm{~J}$
- Speed of light, $c=3.00 \cdot 10^{8} \mathrm{~m} / \mathrm{s}$
- Universal Gravitational constant,

$$
G=6.67 \cdot 10^{-11}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{kg}^{2}
$$

- Solar Mass

$$
M_{\odot}=1.988 \cdot 10^{30} \mathrm{~kg}
$$

- Acceleration due to gravity, $g=9.8 \mathrm{~m} / \mathrm{s}^{2}$
- 1 unified atomic mass unit,

$$
1 \mathrm{u}=1.66 \cdot 10^{-27} \mathrm{~kg}=931 \mathrm{MeV} / \mathrm{c}^{2}
$$

- Planck's constant,

$$
h=6.63 \cdot 10^{-34} \mathrm{~J} \cdot \mathrm{s}=4.41 \cdot 10^{-15} \mathrm{eV} \cdot \mathrm{s}
$$

- Permittivity of free space,

$$
\epsilon_{0}=8.85 \cdot 10^{-12} \mathrm{C}^{2} /\left(\mathrm{N} \cdot \mathrm{m}^{2}\right)
$$

- Coulomb's law constant,

$$
k=\frac{1}{4 \pi \epsilon_{0}}=8.99 \cdot 10^{9}\left(\mathrm{~N} \cdot \mathrm{m}^{2}\right) / \mathrm{C}^{2}
$$

- Permeability of free space,

$$
\mu_{0}=4 \pi \cdot 10^{-7} \mathrm{~T} \cdot \mathrm{m} / \mathrm{A}
$$

- Magnetic constant,

$$
\frac{\mu_{0}}{4 \pi}=1 \cdot 10^{-7}(\mathrm{~T} \cdot \mathrm{m}) / \mathrm{A}
$$

- 1 atmospheric pressure,

$$
1 \mathrm{~atm}=1.01 \cdot 10^{5} \mathrm{~N} / \mathrm{m}^{2}=1.01 \cdot 10^{5} \mathrm{~Pa}
$$

- Wien's displacement constant, $b=2.9$. $10^{-3} \mathrm{~m} \cdot \mathrm{K}$
- Stefan-Boltzmann constant,

$$
\sigma=5.67 \cdot 10^{-8} \mathrm{~W} / \mathrm{m}^{2} / \mathrm{K}^{4}
$$

Question: While exploring outer space, Darth Vader comes upon a purely reflective spherical planet with radius $R_{p}=$ $40,000 \mathrm{~m}$ and mass $M_{p}=8.128 \times 10^{24} \mathrm{~kg}$. Around the planet is a strange moon of orbital radius $R_{s}=$ $6,400,000 \mathrm{~m}\left(R_{s} \gg R_{p}\right)$ and mass $M_{s}=9.346 \times 10^{19} \mathrm{~kg}\left(M_{s} \ll M_{p}\right)$. The moon can be modelled as a blackbody and absorbs light perfectly. Darth Vader is in the same plane that the planet orbits in. Startled, Darth Vader shoots a laser with constant intensity and power $P_{0}=2 \times 10^{32} \mathrm{~W}$ at the reflective planet and hits the planet a distance of $\frac{R_{p}}{2}$ away from the line from him to the center of the planet. Upon hitting the reflective planet, the light from the laser is plane polarized. The angle of the planet's polarizer is always the same as the angle of reflection. After reflectance, the laser lands a direct hit on the insulator planet. Darth Vader locks the laser in on the planet until it moves right in front of him, when he turns the laser off. Determine the energy absorbed by the satellite. Assume the reflective planet remains stationary and that the reflective planet is a perfect polarizer of light.

<img_4328>

Solution:
The original angle of reflectance can be found with some optical geometry to be $\theta_{0}=\sin ^{-1}\left(\frac{1}{2}\right)=$ $\frac{\pi}{6}$. By Malus' Law, when the laser hits the planet with an angle $\theta$, the final power after reflection is $P_{0} \cos ^{2}(\theta)$. Also, note that the satellite has to be at an angle $2 \theta$ due to the law of reflection. Therefore, by Kepler's Third law, the time for the satellite to reach $\theta=0$ is
$$
t=2 \theta \sqrt{\frac{r^{3}}{G M_{p}}} \Longrightarrow d t=2 \sqrt{\frac{r^{3}}{G M_{p}}} d \theta
$$

Thus, the total power absorbed by the satellite is

$$
\int_{0}^{\theta_{0}} P_{0} \cos ^{2}(\theta) d t=2 P_{0} \sqrt{\frac{r^{3}}{G M_{p}}} \int_{0}^{\theta_{0}} \cos ^{2}(\theta) d \theta=P_{0} \sqrt{\frac{r^{3}}{G M_{p}}}\left(\theta_{0}+\sin \left(\theta_{0}\right) \cos \left(\theta_{0}\right)\right)=1.33 \cdot 10^{35} \mathrm{~J} .
$$

Final answer: $1.33 \cdot 10^{35}$ J