Question

If the earth were a perfect sphere of
radius 6.37 x 106 m, rotating about its axis with a period of 1 day (= 8.64 x 104 s), how much would the acceleration due to gravity (g) differ from the poles to the equator?​

Answer 1

Answer:

$\displaystyle{g_{pole}-g_{equ.}=3.365\times10^{-2} \ m/sec}$

Explanation:

We know acceleration due to gravity at latitude λ is given as :

$\displaystyle{g^{'}=g\left[1-\dfrac{R \ \omega^{2}}{g}\cos^2\lambda\right]}$

We know at the pole  λ = 90 then g will be

$\displaystyle{g^{'}=g\left[1-\dfrac{R \ \omega^{2}}{g}\cos^290\right]}\\\\\\ \displaystyle{g^{'}=g\left[1-\dfrac{R \ \omega^{2}}{g}\times0\right]}\\\\\\\displaystyle{g^{'}=g \ .. \ ..(i)}$

We also know at equation  λ = 0 then g will be

$\displaystyle{g^{'}=g\left[1-\dfrac{R \ \omega^{2}}{g}\cos^20\right]}\\\\\\\displaystyle{g^{'}=g\left[1-\dfrac{R \ \omega^{2}}{g}\times1\right]}\\\\\\\displaystyle{g^{'}=g-R \ \omega^{2} \ .. \ ..(ii)}$

Subtracting ( i )  and  ( ii )

$\displaystyle{g_{pole}-g_{equ.}=g-[g-R \ \omega^{2}]}\\\\\displaystyle{g_{pole}-g_{equ.}=R \ \omega^{2}}\\\\\displaystyle{Given \ R=6.37\times10^6 \ m}\\\\\displaystyle{We \ know}\\\\\displaystyle{\omega=\dfrac{2\pi}{T}}$

$\displaystyle{\omega=\dfrac{2\pi}{8.64\times10^4} \ rad/sec \ [Given \ T]}$

Putting values here we get

$\displaystyle{g_{pole}-g_{equ.}=6.37\times10^6\times\left[ \dfrac{2\pi}{8.64\times10^4}\right]^2}\\\\\\\displaystyle{g_{pole}-g_{equ.}=\dfrac{251.22}{74.64}\times10^{-2}}\\\\\\\displaystyle{g_{pole}-g_{equ.}=3.365\times10^{-2} \ m/sec}$

Thus we get answer .