(a) Earth can be thought of as a sphere of radius ` 64 00 km`. Any object (or a person ) is performing circula motion around the axis os earth due to earths rotation (period 1 day ). What is acceleration o object on the surface of th earth 9at equator ) towards its centre ? What is its altitude ` theta` ? How does these accelerations compare with `g=9.8 m//s^2 `?
(b) Earth also moves in circular orbit around sum every year withon orbital radius of ` 1.5 xx 10 ^(11) m`. What is the acceleration of earth ( or any object on the surface of the earth ) towards the centre of the sum ? How dies thsi acceleration comparte with ` g=9.8 ms^2 `?

Explanation

(a) Radius of the earth (R)=`6400km=6.4xx10^(6) m`
Time period (T) = 1 day `=24xx60xx60 s=86400 s`
Centripetal acceleration `(a_(c))=omega^(2)=R=R((2pi)/(T))^(2)=(4 pi^(2)R)/(T)`
`=(4xx(22//7)^(2)xx6.4xx10^(6))/((24xx60xx60)^(2))`
`=(4xx484xx64xx10^(6))/(49xx(24xx3600)^(2))`
`=0.034m//s^(2)`
At equator, `" "` latitude `theta = 0^(@)`
`:." " (a_(c))/(g)=(0.034)/(9.8)=(1)/(288)`
(b) Orbital radius of the earth around the sun (R) `=1.5xx10^(11)m`
Time period =1 Yr =365 day
`=365xx 24xx60xx60 s = 3.15 xx10^(7) s`
Centripetal acceleration `(a_(c))=R omega^(2)=(4pi^(2)R)/(T^(2))`
`=(4xx(22//7)^(2)xx1.5xx10^(11))/((3.15xx10^(7))^(2))`
`=5.97xx10^(-3)m//s^(2)`
`:. " " (a_(c))/(g)=(5.97xx10^(-3))/(9.8)=(1)/(1641)`