(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?

0 2 Views | Posted 4 months ago
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    Answered by

    alok kumar singh | Contributor-Level 10

    4 months ago

    This is a Short Answer Type Questions 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

    = 4 × 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= 4 π 2 R T 2

    = 4 × 22 7 2 × 1.5 × 10 11 ( 3.15 × 10 7 ) 2 = 5.97 × 10 - 3 m / s 2

    a c g = 5.97 × 10 - 3 9.8 = 1 1642

     

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