# Apparent weight / Rotating coordinate system

1. Apr 30, 2013

1. The problem statement, all variables and given/known data

Show that, owing to the rotation of the Earth on its axis, the apparent weight of an object
of mass m at latitude λ is :

m((g-$ω^{2}$R$cos^{2}$λ)$^{2}$-($ω^{2}$Rcosλsinλ)$^{2}$)$^{1/2}$

where ω is the angular velocity of the Earth and R its radius.
The first space travellers to reach the planet Isobarios land in the polar regions and discover,
to their surprise, that the apparent weight of objects, measured with a spring balance, does
not vary with latitude even though the planet is rotating. The planet is spherically symmetric
and the length of the “day” is 2hr46min40sec. Calculate its mean density.

2. Relevant equations

g*=-g R/R -ωx(ωxR)

3. The attempt at a solution

I've taken ω= ωcosλ$\hat{y}$ - ωsinλ $\hat{z}$

since we have gravity going in the z direction i took R= (R$\hat{x}$, R$\hat{y}$, 0)

i then crossed these two terms to find ωRsinλ$\hat{x}$ - ωRsinλ$\hat{y}$ -ωRcosλ$\hat{z}$

i then crossed it again with ω to find

2Rcos2λ + ω2Rsin2λ )$\hat{x}$ - ω2Rsin2λ $\hat{y}$ + ω2Rsinλcosλ$\hat{z}$

i tidied the first formula to get

ω2R$\hat{x}$ + ω2Rcosλ(cosλ$\hat{z}$-sinλ$\hat{y}$) - g$\hat{z}$

thats the first part of my problem, im not sure how to get rid of the ω2R$\hat{x}$ as without that i can just use trig and the answer will pop out.

on the 2nd part of the question were told lambda is independent of apparent gravity, im not exactly sure what that implies so ive left that part for now.

i know that g= GM/R2

taking M=V x ρ we then have g= 4πGRρ and we also know its angular speed since length of day was given ω=2π/T = 104s exactly

Im stuck on what to do next after those steps, and im sure the answer is staring me in the face but i just cant see it

thanks for the help

2. May 2, 2013

### voko

Why is this so? Why does the direction of angular velocity change with latatitude? Is this possible for a planet, presumably a solid body?