Space Elevator

Space Elevator theory and calculations

A space elevator is a proposed type of planet-to-space transportation system. The main component would be a cable (also called a tether) anchored to the surface and extending into space. The design would permit vehicles to travel along the cable from a planetary surface, such as the Earth’s, directly into space or orbit, without the use of large rockets. An Earth-based space elevator would consist of a cable with one end attached to the surface near the equator and the other end in space beyond geostationary orbit (35,786 km altitude). The competing forces of gravity, which is stronger at the lower end, and the outward/upward centrifugal force, which is stronger at the upper end, would result in the cable being held up, under tension, and stationary over a single position on Earth. With the tether deployed, climbers could repeatedly climb the tether to space by mechanical means, releasing their cargo to orbit. Climbers could also descend the tether to return cargo to the surface from orbit.


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F_g : gravitational force
F_c : centripetal force
R : 6.371\times 10^6m Earth radius
M : 5.972\times 10^{24} kg Earth mass
\omega : \frac{2\pi}{86400} rad/s Earth’s angular velocity
G : 6.674\times 10^{-11} m^3kg^{-1}s^{-2} gravitational constant

R_c : cable radius
\rho : cable density
m : cable mass
l : cable length (calculated below)
\sigma : cable stress (calculated below)

Gravitational force

    \[ F_g=G\frac{Mm}{r^2} \]

    \[F_g=\int_R^{x+R}G\frac{Mm}{r^2} dr \]

    \[F_g=GMm\int_R^{x+R}\frac{1}{r^2} dr \]

Centripetal force

    \[ F_c=\frac{v^2}{r}m \]

    \[ v=\omega r \]

    \[ F_c=\omega^2 r m \]

    \[ F_c=\int_R^{R+x}\omega^2mr dr \]

    \[ F_c=\omega^2m\int_R^{R+x} r dr \]

Solve Equation


    \[GMm\int_R^{R+x}\frac{1}{r^2} dr=\omega^2m\int_R^{R+x} r dr \]

    \[ \frac{GM}{\omega^2} \frac{-1}{r}\Big|_R^{R+x}=\frac{r^2}{2}\Big|_R^{R+x} \]

    \[ \frac{-2GM}{\omega^2} r^{-1}\Big|_R^{R+x}={r^2}\Big|_R^{R+x} \]

    \[ \frac{2GM}{\omega^2}(\frac{1}{R}-\frac{1}{R+x})=(R+x)^2-R^2 \]

    \[ \frac{2GM}{R\omega^2}(\frac{x}{R+x})=x(x+2R) \]

    \[ \frac{2GM}{R\omega^2}=(x+2R)(x+R) \]

    \[ 0=x^2+3Rx+2R^2-\frac{2GM}{R\omega^2} \]

    \[ x=\frac{-3R+\sqrt{9R^2-(8R^2-\frac{8GM}{R\omega^2})}}{2} \]

    \[ x=\frac{-3R+\sqrt{R^2+\frac{8GM}{R\omega^2}}}{2} \]

    \[ l\equiv x=1.4429\times 10^8 m \]


The stress, \sigma (i.e., the tension per unit of cross sectional area) is constant along the length of the cable. Calculate the stress at the end of the cable:

    \[ \sigma=\frac{F}{S}=\frac{m v^2}{r}\frac{1}{2\pi R_c}=\frac{m\omega^2 r}{2\pi R_c} \]

    \[\rho=\frac{m}{2 \pi R_c r} \]

    \[\sigma= \omega^2 r^2 \rho \]

    \[ \sigma=1.101\times 10^8 \rho \; N/m^2 \]

14 thoughts on “Space Elevator”

    1. Thank you for the advice. I agree with you, I am still developing the post because it is much more deeper than I thought at first.
      I will demosrtate that the space elevator could be built using carbon nanotubes 😉





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