# Finding Freefall Acceleration on a planet

1. Oct 27, 2015

### B3NR4Y

1. The problem statement, all variables and given/known data
On a certain planet, which is perfectly spherically symmetric, the free-fall acceleration has magnitude g = g0 at the North Pole and g = λ g0 at the equator (lambda is between zero and 1). Find g as a function of theta, the colatitude angle.

2. Relevant equations
$$\vec{g} = \vec{g_0} + (\Omega \times R_E ) \times \Omega$$

3. The attempt at a solution
The colatitude angle at the North Pole is 0 degrees, and at the equator it is 90 degrees.

I subtracted the vector g0 from both sides, to only have to deal with the magnitude of the cross product term. The magnitude of that term is Ω2 RE sin2 θ, which is equal to zero for theta equals 0, and for theta = 90 it equals lambda minus one. I'm not sure where to go from here to get what the book gives.

2. Oct 27, 2015

### TSny

What does the book give? (I bet you knew someone was going to ask this )

How did you get the sinθ to be squared?

3. Oct 27, 2015

### B3NR4Y

Haha I knew it!

$$g = g_0 \sqrt{cos^2 (\theta) + \lambda^2 sin^2 (\theta) }$$

I've done some manipulation using the fact that Ω2 RE = (λ-1)g0

And using the magnitude of g and manipulating it as much as possible, but that doesn't seem to be taking me anywhere

4. Oct 27, 2015

### TSny

OK. That expression looks correct.

I believe you have a sign error here. This expression would make $\lambda > 1$, which would imply that the acceleration is greater at the equator compared to the pole.

What is your expression for the magnitude of g? You first need to do the vector addition $\vec{g}_0 + (\vec{\Omega} \times \vec{R}_E) \times \vec{ \Omega}$.