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pls derive G=g-w2.R.cosx with every explanation plsssss!!!!!!
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 Fc = seudo force = mW2r (r is distance of particle from center) fg = force due to gravity = mg net force towards center is F = (fg2 + Fc2+2fgFccos(180-@)1/2 (by vector law ) putting these values , F = m[(g2+(rW2)2 - 2grW2cos@)]1/2 ...........1 now F = mgeff .................2 from 1 & 2 geff = [g2+(RW2)2 - 2grW2cos@]1/2 ..............3 now , W = 2pi/T (T = 24hrs) W <<<<<<<<1 so we can neglect (rw2)2 , now geff = [g2 - 2grW2cos@]1/2 r = Rcos@ (R is radius of earth) geff = [ g2- 2gRW2cos2@]1/2 = g[ 1 - 2RW2cos2@/g ]1/2 , this is similar to g[1-RW2cos2@/g] geff = g - W2Rcos2@ ...................................4 this is the required result
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
Fc = seudo force = mW2r (r is distance of particle from center)
fg = force due to gravity = mg
net force towards center is F = (fg2 + Fc2+2fgFccos(180-@)1/2 (by vector law )
putting these values , F = m[(g2+(rW2)2 - 2grW2cos@)]1/2 ...........1
now F = mgeff .................2
from 1 & 2
geff = [g2+(RW2)2 - 2grW2cos@]1/2 ..............3
now , W = 2pi/T (T = 24hrs)
W <<<<<<<<1 so we can neglect (rw2)2 , now
geff = [g2 - 2grW2cos@]1/2
r = Rcos@ (R is radius of earth)
geff = [ g2- 2gRW2cos2@]1/2
= g[ 1 - 2RW2cos2@/g ]1/2 , this is similar to g[1-RW2cos2@/g]
geff = g - W2Rcos2@ ...................................4
this is the required result
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