# Estimate flattening of the Earth

1. Apr 16, 2015

### Yoonique

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

The flattening of the Earth is defined as
f ≡ 1 - (Rp - Re)
where Re is the radius of the equator and Rp is half of the distance between the North Pole and the South Pole. Estimate the flattening of the Earth.

2. Relevant equations
Fg = Gm1m2/r2

3. The attempt at a solution
apole = GME/Rp2
aequator = GME/RE2 = ω2RE

How do I approach this problem, I'm so clueless. How do I carry on from here?

Last edited: Apr 16, 2015
2. Apr 16, 2015

### haruspex

Maybe 1 - (Rp / Re)?
Treat the Earth as a rotating ball of water. Since the water at the surface does not run towards the poles, despite the spin, what can you say about the surface - i.e. what is it that is constant over the surface?

3. Apr 17, 2015

### Yoonique

Is a typo, suppose to be 1 - (Rp / Re). Ok, after reading about some article is it regarding about hydrostatic equilibrium? Because I haven't learn about it yet. So I think there must be a constant radial force all over the surface of the sphere for it to be in hydrostatic equilibrium? At the poles, the force is just GMEMW/Rp2. At the equator, the force is GMEMW/Re2 + ω2Re. For both of the forces to be balanced, Rp must decrease and Re must increase thus resulting in the flattening of the Earth? Am I right? I don't know anything about hydrostatic equilibrium so my concept would be wrong.

Last edited: Apr 17, 2015
4. Apr 17, 2015

### haruspex

No, it's not a constant (magnitude) force. If you were in a boat on the surface, anywhere, which way would gravity act? What does that tell you about potential energy levels?

5. Apr 17, 2015

### Yoonique

The gravity would act towards the center of the Earth. All points on the surface are equipotential when the sphere of water rotates so the surface water does not move around surface area of the sphere, it remains at 'rest'? So I assume this is the reason for hydrostatic equilibrium? The surface water at the poles must have the same potential as the surface water at the equator.

Last edited: Apr 17, 2015
6. Apr 18, 2015

### haruspex

Right, so can you figure out what the potential is at a given latitude and distance from earth's centre? (For the purposes of calculating the gravitational potential, pretend the Earth is spherical, and you are considering a point outside that sphere.)

7. Apr 18, 2015

### Yoonique

I'm not very sure how does the centripetal force contribute to the potential. I assume potential is potential energy per unit mass and potential energy is ∫-F(r) dr
Potential the poles: -GMe/Rp
Potential the equator: -1/Mw ∫F(R) dR= -1/Mw ∫[GMeMw/Re2 + Mwω2Re dR = GMe/Re - ω2Re2/2

Last edited: Apr 18, 2015
8. Apr 18, 2015

### haruspex

Sorry to do this to you but I'm going to change tack. I believe consideration of potentials will work, but I was finding it confusing and I wish to avoid confusing you.
Instead, consider the net force at some point on the surface at latitude theta. This will not point at the earth's centre. You can get an expression for the aproximate angular deviation from the radial line. You can use that angle to find out how rapidly radius of the surface changes as theta changes. Solving the ODE gives the profile.

9. Apr 18, 2015

### Yoonique

(GMe/r2)cosθ = ω2rcosθ

How do I continue from here?

Last edited: Apr 18, 2015
10. Apr 18, 2015

### haruspex

That doesn't look like what I had in mind.
Consider a point at distance r from Earth's centre, at latitude theta. There is a gravitational force towards Earth's centre. Working in terms of pseudo forces, there is a centrifugal force. The resultant of these two forces makes an angle, alpha, to the radial line. Find alpha as a function of r, theta, etc.
Next, the local horizontal (i.e. the surface tangent) must be at right angles to this. Consider a small change $d\theta$ to the latitude. Find dr as a function of that and alpha.
Combining these gives you a differential equation.

11. Apr 18, 2015

### Staff: Mentor

Is that necessary? The first approach looked fine and just two steps away from the solution if you plug in some parameters at the right places.
It is an approximation, of course, but it will get the first order right and the second order is much smaller.

12. Apr 18, 2015

### haruspex

Yes, as I posted, I believe my original approach works, but it looked like it was going to be hard explaining why the signs are as they are. Please take it further.

13. Apr 18, 2015

### Yoonique

The answer should be 0.003 but I got 0.002 using the potential method. So this method isn't accurate enough?

14. Apr 18, 2015

### Yoonique

I can't really visualize or understand this as I'm not used to using a rotating frame of reference. So centrifugal force is in the opposite direction of centripetal force right? So if I'm in the same rotating frame of reference, I'm seeing the point being push outwards? The resultant force is the apparent weight?
Because in an inertial frame of reference, the resultant of gravitational force and normal force provide the centripetal force. The apparent weight is the normal force. I'm little confused in the case of a rotating frame of reference.

Last edited: Apr 19, 2015
15. Apr 19, 2015

### haruspex

No.

16. Apr 19, 2015

### Yoonique

Is the resultant force the apparent gravitational force felt?

17. Apr 19, 2015

### haruspex

Yes.

18. Apr 19, 2015

### Yoonique

Do I need to approximate anything to find a relationship between α and θ? I can't find a direct relationship ;/

19. Apr 19, 2015

### haruspex

An approximation valid for small alpha would certainly be reasonable.

20. Apr 19, 2015

### Yoonique

Why does the potential method yield 1.7x10-4 while the given answer is 3.4x10-4?