# Question #74d32

Apr 16, 2017

For a pendulum of length $l$, period is:

$T = 2 \pi \sqrt{\frac{l}{g}}$

Ie $T \propto \frac{1}{\sqrt{g}}$

So, for context, you are playing on the fact that gravity's magnitude is gonna be less on Mt Everest as it is very high up.

If we take logs of the original equation:

$\ln T = \ln 2 \pi + \frac{1}{2} \left(\ln l - \ln g\right)$

Taking the differentials, assuming that $l$ stays the same:

$\frac{\mathrm{dT}}{T} = - \frac{1}{2} \frac{\mathrm{dg}}{g} q \quad \triangle$

If we next look at what is meant by $g$, we see that it is a play on Newton's Law of Gravitation:

$F = m g = \frac{G M m}{r} ^ 2 \implies g = \frac{G M}{r} ^ 2$

We can do the same thing with logs:

$\ln g = \ln G M - 2 \ln r$

$\implies \frac{\mathrm{dg}}{g} = - 2 \frac{\mathrm{dr}}{r} q \quad \square$

Combining $\triangle$ and $\square$ means that:

$\frac{\mathrm{dT}}{T} = \frac{\mathrm{dr}}{r}$

Karachi is pretty much at sea level, Mt Everest is $8 , 848 \text{ m}$ above sea level. Ergo $\mathrm{dr} = 8 , 848$.

In terms of $r$, the earth has a radius of about: $6 , 371 \times {10}^{3} \text{ m}$.

So:

$\frac{\mathrm{dT}}{T} \approx + 0.14 \text{ %}$