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 space elevator consists of a heavy counterweight placed near geostationary orbit, a thread that connects it to the ground (assume this is massless), and elevators that run on the threads (also massless). The mass of the counterweight is $10^{7} \mathrm{~kg}$. Mass is continuously delivered to the counterweight at a rate of $0.001 \mathrm{~kg} / \mathrm{s}$. The elevators move upwards at a rate of $20 \mathrm{~m} / \mathrm{s}$. Assume there are many elevators, so their discreteness can be neglected. The elevators are massless. The counterweight orbits the Earth.


Find the minimum possible displacement radially of the counterweight. Specify the sign.

Solution:
As the orbit is geostationary, we can balance forces to write that
$$
\frac{G M_{E}}{R_{\mathrm{GS}}^{2}}=\omega^{2} R_{\mathrm{GS}} \Longrightarrow R_{\mathrm{GS}}=\left(\frac{G M_{E} T^{2}}{4 \pi}\right)^{1 / 3}
$$
The total mass of masses $=1790 \mathrm{~kg} \ll 10^{6} \mathrm{~kg}$ which means the displacements are small. The total gravity of masses is
$$
\int_{R_{E}}^{R_{E s}} \frac{G M_{E}}{x^{2}}(\lambda d x)=\operatorname{GME} \lambda\left(\frac{1}{R_{E}}-\frac{1}{R_{\cos }}\right)=2650 \mathrm{~N}
$$
where $\lambda=5 \cdot 10^{-5} \mathrm{~kg} / \mathrm{m}$. The total centrifugal of masses is
$$
\int_{R_{E}}^{R_{E}} \frac{4 \pi^{2}}{T^{2}} \times(\lambda d x)=\frac{2 \pi^{2}}{T^{2}} \lambda\left(R_{E s}^{2}-R_{E}^{2}\right)=230 \mathrm{~V}
$$
Hence, the total force is $2420 \mathrm{~N}$. The outwards force is just
$$
F_{\text {out }}=\omega^{2} a M-\frac{\mathrm{d}}{\mathrm{d} x}\left(\frac{G M_{E}}{x^{2}}\right) a M
$$
where $a$ is the horizontal displacement which is much less than $R_{E}, R_{\mathrm{GS}}$. Hence, this can be rewritten as
$$
F_{\text {out }}=\left(\frac{4 \pi^{2}}{T^{2}}-2 \frac{G M_{E}}{R_{\mathrm{GS}}^{2}}\right) a M=2420 \mathrm{~N}
$$
This implies that $a=15.21 \mathrm{~km}$.

Final answer: $15.21$ km