Calculate the ratios of the following accelerations: the acceleration $w_{1}$ due to the gravitational force on the Earth's surface, the acceleration $w_{2}$ due to the centrifugal force of inertia on the Earth's equator, and the acceleration $w_{3}$ caused by the Sun to the bodies on the Earth.

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Solution

$w_{1} = \frac{γ M}{R^{2}} = \frac{6.67 \times 10^{- 11} \times 5.96 \times 10^{24}}{{(6.37 \times 10^{6})}^{2}} = 9.8 m / s^{2}$
$w_{2} = w^{2} R = {(\frac{2 π}{T})}^{2} R = {(\frac{2 \times 22}{24 \times 3600 \times 7})}^{2} 6.37 \times 10^{6} = 0.034 m / s^{2}$
and $w_{3} = \frac{γ M_{s}}{R_{m e a n}^{2}} = \frac{6.67 \times 10^{- 11} \times 1.97 \times 10^{30}}{{(149.50 \times 10^{6} \times 10^{3})}^{2}} = 5.9 \times 10^{- 3} m / s^{2}$
Then $w_{1} : w_{2} : w_{3} = 1 : 0.0034 : 0.0006$

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Calculate the ratios of the following accelerations: the acceleration w1 due to the gravitational force on the Earth's surface, the acceleration w2 due to the centrifugal force of inertia on the Earth's equator, and the acceleration w3 caused by the Sun to the bodies on the Earth.

w1=γMR2=6.67×10−11×5.96×1024(6.37×106)2=9.8m/s2w2=w2R=(2πT)2R=(2×2224×3600×7)26.37×106=0.034m/s2and w3=γMsR2mean=6.67×10−11×1.97×1030(149.50×106×103)2=5.9×10−3m/s2Then w1:w2:w3=1:0.0034:0.0006

Calculate the ratios of the following accelerations: the acceleration $w_{1}$ due to the gravitational force on the Earth's surface, the acceleration $w_{2}$ due to the centrifugal force of inertia on the Earth's equator, and the acceleration $w_{3}$ caused by the Sun to the bodies on the Earth.