# Free-fall Acceleration with Centrifugal Force

1. Feb 23, 2008

### CoreyJKelly

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

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 $$\theta$$.

2. Relevant equations

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

3. 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.

2. Feb 24, 2008

### 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.