Question

At what height (in km) over the Earth's pole the free-fall acceleration decreases by one percent?

Answer 1

Let the height be ${}^{\prime }{h}^{\prime }$

Given,acceleration is one percent less, which is $0.99g$

$\Rightarrow \frac{99}{100}g=\frac{GM}{{(R+h)}^2}$

As, $g=\frac{GM}{R^2}$

$\frac{99}{100}={(1+\frac{h}{R})}^{-2}$

Using binomial rule, as $h<<R$

$\frac{99}{100}=(1-\frac{2h}{R})$

$\Rightarrow h=\frac{R}{200}$

Radius of earth is $6400km$

Therefore, $h=2400/200=32km$