Subfield: Mechanics

- 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: Adithya is launching a package from New York City $\left(40^{\circ} 43^{\prime} \mathrm{N}\right.$ and $\left.73^{\circ} 56^{\prime} \mathrm{W}\right)$ to Guam $\left(13^{\circ} 27^{\prime} \mathrm{N}\right.$ and $\left.144^{\circ} 48^{\prime} \mathrm{E}\right)$. Find the minimal launch velocity $v_{0}$ from New York City to Guam. Ignore the rotation of the earth, effects due to the atmosphere, and the gravitational force from the sun. Additionally, assume the Earth is a perfect sphere with radius $R_{\oplus}=6.37 \times 10^{6} \mathrm{~m}$ and mass $M_{\oplus}=5.97 \times 10^{24} \mathrm{~kg}$.

Solution:
We first want to find the angular distance between New York City and Guam. Let this be $\theta$. Let New York City be point $A$ and Guam be point $B$. Consider the north pole $P$ and the spherical triangle $P A B$. By the spherical law of cosines,
$$
\cos \theta=\cos \left(90^{\circ}-\phi_{A}\right) \cos \left(90^{\circ}-\phi_{B}\right)+\sin \left(90^{\circ}-\phi_{A}\right) \sin \left(90^{\circ}-\phi_{B}\right) \cos \left(\ell_{B}-\ell_{A}\right) .
\tag{23}
$$



Equation 23 simplifies to

$$
\cos \theta=\sin \phi_{A} \sin \phi_{B}+\cos \phi_{A} \cos \phi_{B} \cos \left(\ell_{B}-\ell_{A}\right)
\tag{24}
$$

from which we find $\theta=115.05^{\circ}$.

Now that we have determined the angular distance, we will proceed with the orbital mechanics problem. By the vis-viva equation, the speed at the launch point is

$$
v_{0}=\sqrt{G M_{\oplus}\left(\frac{2}{R_{\oplus}}-\frac{1}{a}\right)}
\tag{25}
$$

where $a$ is the semimajor axis for the orbit. It is clear that in order to minimize $v_{0}$, we must minimize $a$. The orbit is an ellipse with focii $F_{1}$ and $F_{2}$, where $F_{1}$ is the center of the earth. By the definition of an ellipse,

$$
A F_{1}+A F_{2}=2 a
\tag{26}
$$

Since $A F_{1}=R_{\oplus}$, is suffices to minimize $A F_{2}$. By symmetry, line $F_{1} F_{2}$ is the perpendicular bisector of $A B$. Since $F_{2}$ is on a fixed line, to minimize $F_{2}$, we place it at the foot of the perpendicular from $A$ to this line. Then, we obtain $A F_{2}=R_{\oplus} \sin \left(\frac{\theta}{2}\right)$. Thus,

$$
a=\frac{1+\sin \left(\frac{\theta}{2}\right)}{2} R_{\oplus}
$$

and from Equation 25 we find

$$
v_{0}=\sqrt{\frac{G M_{\oplus}}{R_{\oplus}} \frac{2 \sin \frac{\theta}{2}}{1+\sin \frac{\theta}{2}}} .
$$

Final answer: $7564$ m/s