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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.

Constants:

    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?

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    $\begingroup$ You have made an error whilst using your calculator and failed to take the cube root? $\endgroup$ – Farcher Oct 24 '18 at 7:20
  • $\begingroup$ Yeah @Farcher, forgot to take the cubed root. $\endgroup$ – Adrian Oct 24 '18 at 12:33
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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:

((3.986004418E14m^3*86400^2/(2pi)^2)^(1/3)

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

((3.986004418E14m^3*86164.0905^2/(2pi)^2)^(1/3)

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.

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  • $\begingroup$ 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 Oct 24 '18 at 12:19
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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

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

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