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?
