Question

R is the radius of earth and w is its angular velocity and ge is the acceleration of gravity at
equator. The effective value of acceleration due to gravity at the latitude lamda = 30° will be equal to what?​

Answer 1

it is given that,

radius of the earth = R

angular velocity = ω

acceleration due to gravity at latitude is given as $g=g_0-R\omega^2cos^2\lambda$

where, $g_0$ is the acceleration due to gravity at the earth's surface, λ is latitude.

here, λ = 30°

so, g = $g_0$ - Rω²cos²30°

or, g - $g_0$ = -Rω²(√3/2)² = -3/4ω²R

hence, affective acceleration due to gravity is -3/4ω²R

[note : The negative sign indicates that the decrease in gravitational acceleration is due to the rotation of earth. ]

Answer 2

Answer:

g' = gE + ¼w²R

Explanation:

*formula, you know,

g'= g° - w²Rcos²y

[ y = angle of the latitude/ lambda,

g°= g of earth ]

now, we have to answer in terms of gE , so,

at equator , y=0°

or , gE = g°-w²R

or, g°=gE+ w²R ________ 1)

secondly, at y=30°

g' = g° - w²Rcos²30°_________2)

now, it's done,

if you substitute the value of g° from 1) in 2), you'll get the answer, mentioned above.