Free-fall Acceleration with Centrifugal Force

[CoreyJKelly]

Homework Statement

From Taylor's "Classical Mechanics", problem 9.15:

On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = g_0 at the North Pole and g = \lambda g_0 at the equator (with 0 \leq \lambda \leq 1). Find g(\theta), the free-fall acceleration at colatitude \theta as a function of \theta.

Homework Equations

g = g_0 + (\Omega \times R) \times \Omega

The Attempt at a Solution

The north pole corresponds to a colatitude of 0 degrees, and the equator corresponds to 90 degrees. Using these values, and a trial-and-error approach, I arrived at the equation

g(\theta) = g_{0}(Sin(\theta) + \lambda Cos(\theta))

the book gives the following answer:

g(\theta) = g_{0}\sqrt{Sin^{2}(\theta) + \lambda^{2} Cos^{2}(\theta)}

And I'm not really sure where to start with this. The equation I've given is the only one given for free-fall acceleration, but I can't see any way to manipulate it into anything like this answer. Any suggestions would be appreciated.

 

[kamerling]

You can compute g as a function of theta with g = g_0 + (\Omega \times R) \times \Omega

this is a vector equation and g_0 doesn't point in the same direction as (\Omega \times R) \times \Omega
Once you've done that you can find \Lambda by looking at what happens at the equator.