# At which colatitude $\theta$ is the error between $\vec g_{eff}$ and $\vec g$ maximum? [closed]

Suppose a plumb bob hangs without swinging, then the string defines the effective direction of gravity. Suppose you are holding the bob on the surface of the earth at colatitude $$\theta$$, where $$\theta$$ measures the spherical angle as measured from the north pole, then the error $$\epsilon$$ between $$\vec g_{eff}$$ and $$\vec g$$ is given by

$$\epsilon \approx \frac{R \Omega^2 sin(\theta) cos(\theta)}{g-R \Omega^2 sin^2(\theta)}$$

We can find the maximum error, by differentiating $$\epsilon$$ with respect to $$\theta$$. I was curious just if we can answer this question conceptually.

• It seems the denominator with $R\Omega$ has not correct units ? Oct 20, 2022 at 4:23
• @ Cretin2 good catch, I have corrected it. Thanks! Oct 20, 2022 at 4:25
• The easiest way is to graph the function, no? And you only need a domain of 0 to 2pi. Oct 20, 2022 at 4:54
• If you think that amount of calculation is tedious, just wait until you learn E&M or Quantum. Oct 20, 2022 at 11:45
• @ Michael Seifert, experienced that in E&M already haha. I was just curious if there is a way to conceptually answer this. Oct 20, 2022 at 15:20

I will define $$\alpha \equiv R\Omega^2/g$$ so that we can write $$\epsilon = \frac{\alpha\sin(\theta)\cos(\theta)}{1-\alpha\sin^2(\theta)}.$$ We can differentiate this using the quotient rule to find that $$\frac{d\epsilon}{d\theta} = \frac{\alpha(1-\alpha\sin^2(\theta))(\cos^2(\theta)-\sin^2(\theta)) + 2\alpha^2\sin^2(\theta)\cos^2(\theta)}{(1-\alpha\sin^2(\theta))^2};$$ this requires a little work but isn't too tedious. Setting $$d\epsilon/d\theta = 0$$ to find the maximum $$\theta$$, we get (after some algebra) $$(1-\alpha)\tan^2(\theta_{\rm max}) = 1 \implies \theta_{\rm max}=\arctan\left(\sqrt{\frac{1}{1-\alpha}}\right).$$ Note that this implies the constraint $$\alpha < 1$$, or $$R\Omega^2 < g$$.