Golden mean in this equation Years ago I had started with this equation: $2^{1/3}=(R/2)(\sqrt{1+8/R^3}-1)$ And arrived at the result $2^{1/3}=\phi R$ Where phi, the golden ratio, is (sqrt(5) +1)/2. But at the moment can't retrace the steps that led to this result. I'm hoping someone can help me. I took equation 5 from P. K. Aravind's pdf and adapted it to the more general case of a vertical orbital tether. I was looking at a vertical tether whose top throws payloads into parabolic trajectories. ![enter image description here](

Replacing $R$ by its value,

$$2^{1/3}=(2^{1/3}/2\phi)(\sqrt{1+8\phi^3/2}-1),$$

$$2\phi=\sqrt{1+4\phi^3}-1,$$

$$(2\phi+1)^2=1+4\phi^3,$$

$$\phi^2+\phi=\phi^3,$$ which is right. You can backtrack.

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If you want to solve from scratch, let $s:=1/R$ and

$$2^{4/3}s+1=\sqrt{1+8s^3},$$

$$8s^3-2^{8/3}s^2-2^{7/3}s=0.$$

Simplify by $4s$ and solve the quadratic equation.