if earth rotates with an angular velocity of twice the present value then find the value of g at equators.
Asked by thakursonali2000 | 7th Oct, 2015, 05:02: PM
Expert Answer:
At the equator:
g' = g - Reω2
Substituting Re = 6.37 × 106 m and 
, we get,
g-g' = Reω2 = (6.37 × 106 m) (7.27 × 10-5 s-1)2 = 0.034 m s-2. (I)
When an angular velocity of twice the present value, then value of g at euqator will be g''
g'' = g - Reω'2 = g - Re(2ω)2 = 9.8 - 4 Reω2 = 9.8 - 4 × 0.034 = 9.66 m s-2 (from (I))
At the equator:
g' = g - Reω2
Substituting Re = 6.37 × 106 m and , we get,
, we get,g-g' = Reω2 = (6.37 × 106 m) (7.27 × 10-5 s-1)2 = 0.034 m s-2. (I)
Answered by Priyanka Kumbhar | 8th Oct, 2015, 10:42: AM
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