Question

If $R$ is the radius of earth, $\omega$ is its angular velocity and $g_p$ is the value of acceleration due to gravity at the poles, then effective value of acceleration due to gravity at the latitude $\lambda ={60}^o$ will be equal to

• 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. $g_p+\frac{1}{4}R{\omega }^2$

Answer 1

Answer: (A)

$g_p-\frac{1}{4}R{\omega }^2$

$g=g_p-R{\omega }^{2}co{s}^2\lambda g=g_p-{\omega }^{2}Rco{s}^2{60}^{\circ }g=g_p-\frac{1}{4}R{\omega }^2$