Constant of gravity in earth fixed coordinate system

I have this problem:

If the constant of gravity is measured to be $g_0$ in an earth fixed coordinate system, what is the difference $g-g_0$ where $g$ is the real constant of gravity as measured when the earth stands still.

The earth fixed coordinate system is rotating, so that the absolute acceleration $g$ is given by

$g=g_0+2\vec\omega\times\frac{d}{dt}\vec r+\vec\omega\times (\vec \omega\times \vec r)$

This is the Coriolis term and the Centrifugal term, the other ones disappear since earth rotates in a steady pace and does not itself accelerate.

I know the answer is supposed to be

$g-g_0\approx g_0\frac{x^2-2x}{2}\cos^2(\phi)$

Where $x=\frac{R \omega^2}{g_0}$. How can I come to this conclusion? My attempts so far have been to try to evaluate the expression given earlier. But I cannot for my life figure out how that yields the supposed answer.

(Note: I've edited the "correct" answer, multiplying what was written originally with g_0 after some of the answers were written)

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Let's assume that we are considering the effective acceleration due to gravity felt by an object at rest relative the the Earth's surface at lattitude $\phi$. Since the object is at rest, $\dot{\vec r} = 0$, and the Coriolis term vanishes.

With the $z$-axis along the axis of rotation, we can write $$\vec\omega = \omega \hat z$$ On the other hand, the position vector is given by $\vec r = R\hat r$. It follows that the centrifugal force is $$\vec F_\mathrm{cf} =-\vec\omega\times(\vec\omega\times \vec r) = -\omega^2R \,\hat z\times(\hat z \times \hat r) =-\omega^2 R\,((\hat z \cdot\hat r)\hat z - \hat r)$$ Since it is the radial component of the centrifugal term contributes to the component of the effective acceleration due to gravity perpendicular to the surface of the Earth, we take the dot product of this expression with $\hat r$ to obtain this component; $$\vec F_\mathrm{cf}\cdot \hat r = \omega^2 R(1-\cos^2\theta) = \omega^2r\sin^2\theta$$ where $\theta$ is the spherical polar angle measured from the positive $z$ axis. Since the polar angle $\theta$ and lattitude $\phi$ are related by $\theta = \pi/2-\phi$, this result can also be written as $$\vec F_\mathrm{cf}\cdot \hat r = \omega^2 R\cos^2\phi$$ The perpendicular component of the effective acceleration due to gravity is therefore $$g_0 = g-\omega^2 R\cos^2\phi$$ I'm a bit baffled by the expression you say one is supposed to get. For one thing, the dimensions don't seem to work out since $x$ is dimensionless, but is being equated to something with dimensions of acceleration. However, if one were to take the result I just derived, then notice that it implies $$\frac{g-g_0}{g} = \frac{\omega^2 R \cos^2\phi}{g_0+\omega^2R\cos^2\phi} = \frac{x \cos^2\phi}{1+x\cos^2\phi} \approx (x-x^2\cos^2\phi)\cos^2\phi$$ where in the last equality we have assumed that $x$ is small, namely that the spin of the earth is slow. This is as close as I can seem to get to the desired expression.

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I added $g_0$ in front of the expression, which should give it the right units. I am very sorry about writing that down wrong in the first place. However, excellent answer. Thank you very much. –  Althalos May 11 '13 at 10:26

The Coriolis term is neglected since it does not add to the uniform acceleration. We can consider only the case of starting motion.

The last term is calculated straight out, and then added to the first by the rule of vector sum. To ease that:
1. Consider Cartesian coordinates aligned with the Earth axis of rotation. That will give you a simple way to get the modulo and direction of $\vec{\omega}\times(\vec{\omega}\times\vec{r})$.
2. Recall the modulo and direction of the $\vec{g}_0$ vector.

Also, the dimensions are not quite right with your supposed answer.

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