pls derive G=g-w2.R.cosx with every explanation plsssss!!!!!!

Question

vikas askiitian expert · Answer

let there is a particle at rest on the surface of earth in latitude @.... seudo force is acting on particle outwards &

real accleration due to gravity acts along the center of earth , then

F_c = seudo force = mW²r (r is distance of particle from center)

f_g = force due to gravity = mg

net force towards center is F = (f_g² + F_c²+2f_gF_ccos(180-@)^1/2 (by vector law )

putting these values , F = m[(g²+(rW²)² - 2grW²cos@)]^1/2 ...........1

now F = mg_eff .................2

from 1 & 2

g_eff = [g²+(RW²)² - 2grW²cos@]^1/2 ..............3

now , W = 2pi/T (T = 24hrs)

W <<<<<<<<1 so we can neglect (rw²)² , now

g_eff = [g² - 2grW²cos@]^1/2

r = Rcos@ (R is radius of earth)

g_eff = [ g²- 2gRW²cos²@]^1/2

= g[ 1 - 2RW²cos²@/g ]^1/2 , this is similar to g[1-RW²cos²@/g]

g_eff = g - W²Rcos²@ ...................................4

this is the required result

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