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: A spacecraft is orbiting in a very low circular orbit at a velocity $v_{0}$ over the equator of a perfectly spherical moon with uniform density. Relative to a stationary frame, the spacecraft completes a revolution of the moon every 90 minutes, while the moon revolves in the same direction once every 24 hours. The pilot of the spacecraft would like to land on the moon using the following process:So the final answer is 

1. Start by firing the engine directly against the direction of motion.So the final answer is 

2. Orient the engine over time such that the vertical velocity of the craft remains 0 , while the horizontal speed continues to decrease.

3. Once the velocity of the craft relative to the ground is also 0 , turn off the engine.

Assume that the engine of the craft can be oriented instantly in any direction, and the craft has a TWR (thrust-to-weight ratio, where weight refers to the weight at the moon's surface) of 2, which remains constant throughout the burn. If the craft starts at $v_{0}=500 \mathrm{~m} / \mathrm{s}$, compute the delta-v expended to land, minus the initial velocity, i.e. $\Delta v-v_{0}$.

Solution:
The trick in this question is to work in dimensionless units. Let $v$ be the ratio of the craft's velocity to orbital velocity. Then, if the craft has horizontal velocity $v$, the acceleration downwards is the following:

$$
a_{V}=\frac{v_{0}^{2}}{r}-\frac{\left(v_{0} v\right)^{2}}{r}=\frac{v_{0}^{2}}{r}\left(1-v^{2}\right)=g_{m}\left(1-v^{2}\right)
$$

As the TWR is 2 , the total acceleration the engine provides is $a_{0}=2 g_{m}$, where $g_{m}$ is the surface gravity of the moon. As this total acceleration is the sum of horizontal and vertical components, and the vertical component cancels out the downwards acceleration:

$$
a_{H}=\sqrt{\left(2 g_{m}\right)^{2}-\left(g_{m}\left(1-v^{2}\right)\right)^{2}}=g_{m} \sqrt{2^{2}-\left(1-v^{2}\right)^{2}}
$$

And $a_{H}$ is related to $v$ (which is dimensionless!) by the following relation:

$$
\begin{gathered}
\frac{\mathrm{d}\left(v_{0} v\right)}{\mathrm{d} t}=-a_{H} \\
\frac{\mathrm{d} v}{\mathrm{~d} t}=-\frac{g_{m}}{v_{0}} \sqrt{2^{2}-\left(1-v^{2}\right)^{2}} \\
\frac{\mathrm{d} v}{\sqrt{2^{2}-\left(1-v^{2}\right)^{2}}}=-\frac{g_{m}}{v_{0}} \mathrm{~d} t
\end{gathered}
$$

At the start, $v=1$. However, at landing, velocity of the craft is 0 relative to the surface, not a stationary frame! Therefore, we use the orbital periods data to determine the final $v$ to be $\frac{1.5}{24}=\frac{1}{16}$. Then integrating:

$$
\int_{v=1}^{v=1 / 16} \frac{\mathrm{d} v}{\sqrt{2^{2}-\left(1-v^{2}\right)^{2}}}=-\frac{g_{m}}{v_{0}} t
$$

And as delta-v is related to time by $\Delta v=a_{0} t$ :

$$
\Delta v=a_{0} t=a_{0} \frac{v_{0}}{g_{m}} \int_{1 / 16}^{1} \frac{\mathrm{d} v}{\sqrt{2^{2}-\left(1-v^{2}\right)^{2}}}=a_{0} \frac{v_{0}}{a_{0}} 2(0.503)=500 \cdot 1.006=503.06 \frac{\mathrm{m}}{\mathrm{s}}
$$

And subtracting:

$$
\Delta v-v_{0}=3.06 \frac{\mathrm{m}}{\mathrm{s}}
$$

Final answer: $3.06$ m/s