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 R= 6.37 × 106 m and begin mathsize 14px style straight omega equals fraction numerator 2 straight pi over denominator straight T end fraction equals fraction numerator 2 cross times 3.14 over denominator 24 cross times 60 cross times 60 end fraction equals 7.27 cross times 10 to the power of negative 5 end exponent space straight s to the power of negative 1 end exponent end style, we get,
g-g' = Reω= (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 Reω= 9.8 - 4 × 0.034 = 9.66 m s-2  (from (I))

Answered by Priyanka Kumbhar | 8th Oct, 2015, 10:42: AM

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