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 and lambda =30° will be equal to what ?

Answer 1

Answer ⇒ g' = g - 3/2 × rw².

Explanation ⇒

Using the formula,

g' = g - rw²Cos²λ

Now, Given,

λ = 30°

Substituting this value in the given question,

 g' = g - rw²(Cos²30)

∴ g' = g - rw²3/2

∴ g' = g - 3/2 × rw².

Hence, the net acceleration due to the gravity at equator is given as,

g' = g - 3/2 × rw².

Hope it helps.

Answer 2

Answer:ge+w^2R/4

Explanation:formula is given by

ge-Rw^2cos^2lambda

Hence,  ge-Rw^2cos 30 degree^2 +w^2R

=ge-3/4w^2R+ w^2R

=ge + w^2R/4