Gravitation Question 330
Question: R is the radius of the earth and co is its angular velocity and $ [{g_{p}}] $ is the value of g at the poles. The effective value of g at the latitude $ [\lambda = 60{}^\circ ] $ will be equal to
Options:
A) $ [g_{p}-\frac{1}{4}R{{\omega }^{2}}] $
B) $ [g_{p}-\frac{3}{4}R{{\omega }^{2}}] $
C) $ [g_{p}-R{{\omega }^{2}}] $
D) $ [g_{p}+\frac{1}{4}R{{\omega }^{2}}] $
Show Answer
Answer:
Correct Answer: A
Solution:
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$ [g=g_{p}-R{{\omega }^{2}}{{\cos }^{2}}\lambda ] $
$ [g=g_{p}-{{\omega }^{2}}R{{\cos }^{2}}{{60}^{o}}=g_{p}-\frac{1}{4}R{{\omega }^{2}}] $
BETA
