(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) `=6400 km `
`=6.4 xx10^(6)m`
Time period (T) `=1 "day" =24xx60xx60 s=88400s`
Centripetal acceleration `(a_(c))`
`=R omega ^(2)`
`=R((2pi)/(T))^(2)=(4pi^(2)R)/(T^(2))`
or `a_(c)(4xx(3.14)^(2)xx6.4 xx10^(6))/((88400)^(2))`
or `a_(c) =0.034 ms^(-2)`
At lattitude `theta =0^(@)` , R remain the same , so `a_(c)` remains the same . Now `(a_(c))/(g) =(0.034)/(9.8) =(1)/(288)`
which is very small as compared to g
(b) `T=365 xx24 xx60xx60`
`3.14xx10^(7s`
`ac=omega^(2)R=(4pi ^(2)R)/(T^(2))`
`=(4xx(3.14)^(2)xx1.5xx10^(11))/((3.15xx10^(7))^(2))`
`5.97 xx10^(-3) ms ^(-2)`
`therefore (a_(c))/(g)=(5.97xx10^(3))/(9.8)=(1)/(1642)`