I've been trying to find the length of an unbendable and unbreakable tether of zero mass extending off the surface of the Earth with a mass at its end, where the centripetal force would equal the gravitational force. However I am getting a number that seems to be way too high.


    Period of the Earth $(T_e)$ is $86400$ s
    Radius of the Earth $(r_e)$ is $6371000$ m
    Mass of Earth $(m_e)$ is $5.97 \times 10^{24}$ kg
    Gravitational Constant $(G)$ is $6.67 \times 10^{-11}$ N·kg⁻²·m²

Centripetal acceleration:

    $F = {m_o v^2 \over r}$ (1)

Gravity between object $(m_o)$ and the Earth $(m_e)$:

    $F = {G m_o m_e \over r^2}$ (2)

Geosynchronous orbit speed:

    $v = {2 \pi r \over T_e}$ (3)

My calculations:

    ${m_o v^2 \over r} = {G m_o m_e \over{r^2}}$ Combine 1 & 2
    ${v^2 \over r} = {G m_e \over r^2}$ Cancel out $m_o$
    $r = {G m_e \over v^2}$ Isolate r (4)
    $r = {G m_e \over ({2 π r \over T_e})^2}$ Combine 3 & 4
    $r = {G m_e T_e^2 \over {4 π² r^2}}$ Simplify
    $r^3 = {G m_e T_e^2 \over {4 π^2}}$ Isolate $r$
    $r = \sqrt[\leftroot{-1}\uproot{2}\scriptstyle 3]{G m_e T_e^2 \over {4 π^2}}$

Given this equation, the value for $r$ is ${7.54}\times{10^{22}}$, but I thought that the value was much smaller than the distance to the moon, which is a mean of $384,000,000$ m. What am I doing wrong?

  • 2
    $\begingroup$ You have made an error whilst using your calculator and failed to take the cube root? $\endgroup$
    – Farcher
    Commented Oct 24, 2018 at 7:20
  • $\begingroup$ Yeah @Farcher, forgot to take the cubed root. $\endgroup$
    – Adrian
    Commented Oct 24, 2018 at 12:33

3 Answers 3


Your algebra is fine. I put a slightly modified version of your equation into the Google calculator, using the gravitational parameter of Earth and got 42241.0957 km.

Here's the "search" string I used:


To get the correct geosynchronous radius of 42164km, we need to use the sidereal period of rotation, not the mean solar day.


yields 42164.1696 km.

My guess is that you accidentally forgot to take the cube root, and also got some wrong powers of 10 from mistaking grams for kilograms &/or metres for kilometres.

  • $\begingroup$ Yeah, seems that I forgot to take the cubed root. Doh! Oh, and thanks for the sidereal day. Didn't even think about that. $\endgroup$
    – Adrian
    Commented Oct 24, 2018 at 12:19

I think any object in a circular orbit would have reached that equilibrium. Also you might be interested in space elevators.? https://m.youtube.com/watch?v=dc8_AuzeYKE


You can check the calculation here. I got $\approx 10^7$, which is near $\frac{r_{moon}}{{100}}$


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