InterviewSolution
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InterviewSolution
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(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 `? |
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Answer» (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)` |
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