Solveeit Logo
Login

Question

Question: R is the radius of the earth and \(\omega\) is its angular velocity and \(g _ { p }\) is the value...

R is the radius of the earth and ω\omega is its angular velocity and gpg _ { p } is the value of g at the poles. The effective value of g at the latitude λ=60∘\lambda = 60 ^ { \circ } will be equal to

A

gp−14Rω2g _ { p } - \frac { 1 } { 4 } R \omega ^ { 2 }

B

gp−34Rω2g _ { p } - \frac { 3 } { 4 } R \omega ^ { 2 }

C

gp−Rω2g _ { p } - R \omega ^ { 2 }

D

gp+14Rω2g _ { p } + \frac { 1 } { 4 } R \omega ^ { 2 }

Answer

gp−14Rω2g _ { p } - \frac { 1 } { 4 } R \omega ^ { 2 }

Explanation

Solution

g=gp−Rω2cos⁡2λg = g _ { p } - R \omega ^ { 2 } \cos ^ { 2 } \lambda=gp−ω2Rcos⁡260∘g _ { p } - \omega ^ { 2 } R \cos ^ { 2 } 60 ^ { \circ } =gp−14Rω2g _ { p } - \frac { 1 } { 4 } R \omega ^ { 2 }