Answer to Question #151705 in Molecular Physics | Thermodynamics for Tana

Question

Imagine a space elevator with m = 30 kg/m extending from the surface (6400 km) to the geosynchronous height of 40,000 km. Take g to be changing with the height. If a stray satellite hits the top of the elevator find the time that the wave will take to reach the surface. The answer should be a few hours or so.

Answer 1

$m = 30kg/m\\ R = 6400 km\\ h = 40,000km\\$

$\begin{aligned} g_n &= g(1+\frac{h}{R_e})^{-2}\\ &= 9.8(1+\frac{40000}{6400})^{-2}\\ &= 0.186 \end{aligned}$

$\begin{aligned} T &= \dfrac{2π}{R_e}\sqrt{\dfrac{(R_e+h)^3)}{g}}\\ &= \dfrac{2π}{6400×10^3}\sqrt{\dfrac{((40000+6400)×10^3)^3)}{g}}\\ &= 99 × 10^3 sec \end{aligned}$

$\begin{aligned} v &= \dfrac{2πr}{T}\\ \\ &= \dfrac{2π(R_e + h)}{T}\\ \\ &= \dfrac{2π((40000+ 6400)×10^3)}{99×10^3}\\ \\ &= 2.9×10^3ms^{-1} \end{aligned}$

$\begin{aligned} t &= \dfrac{s}{v_0}\\ \\ &= \frac{40000×10^3}{2.9×10^3} = 13,793 sec \end{aligned}$

converting 13793 seconds to hour, we divide the number of seconds by 3600

$\dfrac{13793}{3600} = 3.83 \ hours =$

$3\ hours \ \ 50\ minutes$

Therefore, it will take the wave 3 hours 50 minutes to reach the surface.