Finding Freefall Acceleration on a planet

[B3NR4Y]

Homework Statement

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.

Homework Equations

$$\vec{g} = \vec{g_0} + (\Omega \times R_E ) \times \Omega$$

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 R_E 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.

 

[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?

 

[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 R_E = (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

 

[TSny]

Haha I knew it!

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

OK. That expression looks correct.

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

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.

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

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