(a) Earth can be thought of as a sphere of radius 6400 km. Any object (or a person) is performing circular motion around the axis of earth due to earth's rotation (period 1 day). What is acceleration of object on the surface of the earth (at equator) towards its centre ? How does these accelerations compare with g = 9.8 m/s²?

(b) Earth also moves in circular orbit around sun once every year with on orbital radius of 11 1.5 10 * m . What is the acceleration of earth (or any object on the surface of the earth) towards the centre of the sun? How does this acceleration compare with g = 9.8 m/s2?

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Vishal Baghel | Contributor-Level 10

7 months ago

This is a Short Answer Type Question as classified in NCERT Exemplar

Explanation – a) radius of earth =6400km= 6.4 $* 10^{6} m$

Time period = 1 day = 24 $* 60 * 60$ = 86400s

Centripetal acceleration a= w²r= R(2 $π / T$ )²=4 $π$ ²R/T

= $\frac{4 * {(\frac{22}{7})}^{2} * 6.4 * 10^{6}}{({24 * 60 * 60)}^{2}}$ = 0.034m/s²

b) time = 1yr=365 $* 24 * 60 * 60$ days= 365=3.15 $* 10^{7} s$

centripetal acceleration = Rw²= $\frac{4 π^{2} R}{T^{2}}$

^{$= \frac{4 * {\frac{22}{7}}^{2} * 1.5 * 10^{11}}{({3.15 * 10^{7})}^{2}} = 5.97 * 10^{- 3} m / s$ 2}

$\frac{a_{c}}{g} = \frac{5.97 * 10^{- 3}}{9.8} = \frac{1}{1642}$

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