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 earth`s rotation (period 1 day ). What is acceleration o object on the surface of th earth 9at equator ) towards its centre ? What is its latitude ` 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) Here, ` R =6400 km =6.4 xx 10^6 m, T= 24 xx 60 xx 60 s` At latitude ` theta (=0^2)`, the value of (R )remains the same, hence no change in the value of ` a _c (= 0.034 m//s^2)` Now, ` a_c/g = (0.034)/(9.8) = 1/(288) ` which is much smaller than `1` ` (b) R= 1.5 xx 10^(11) m, T= 365 xx 24 xx 60 s = 3.15 xx 10^7 s` `a_c = (R4 pi^2)/T^2 = ((1.5 xx 10^(11) ) xx 4 xx (3. 142)^(2)) /((3.15 xx 10^7)^2) =5.97 xx 10^3 m//s^2` ` a_c /g = (5.97 xx 10^(-3))/(9.8) = 1/(1642)`. |
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