- #1
Fightfish
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Homework Statement
Estimate the flattening of the earth, given by 1 - \frac{R_{P}}{R_{E}}, where R_{P} is half the distance between the north pole and south pole and R_{E} the radius of the equator.
The attempt at a solution
Well, I did manage to obtain a solution of sorts, but it involved a lot of approximations along the way, which I am not sure if they are all justifiable.
We start off with the base assumption that the flattening of the Earth is very small, such that there is relatively little deviation from a spherical shape.
Then, Newton's law of gravity is approximately true for this oblate spheroid. Modeling the Earth as a frictionless incompressible fluid, in order for a point at the pole surface to be in hydrostatic equilibrium with one at the equatorial surface, we have:
Now, if the Earth were a perfect sphere, M = \frac{4}{3}\pi R_{S}^{3} \rho. Since the flattening of the Earth is very small, we expect \frac{R_{E}^{3}}{R_{S}^{3}} \approx 1. Then,
As we expect \frac{R_{E}}{R_{P}} to be very close to 1, this implies that \frac{3 \omega^{2}}{4G\pi\rho} \ll 1, and hence
Thus, the flattening of the earth
Does this look reasonable?
Estimate the flattening of the earth, given by 1 - \frac{R_{P}}{R_{E}}, where R_{P} is half the distance between the north pole and south pole and R_{E} the radius of the equator.
The attempt at a solution
Well, I did manage to obtain a solution of sorts, but it involved a lot of approximations along the way, which I am not sure if they are all justifiable.
We start off with the base assumption that the flattening of the Earth is very small, such that there is relatively little deviation from a spherical shape.
Then, Newton's law of gravity is approximately true for this oblate spheroid. Modeling the Earth as a frictionless incompressible fluid, in order for a point at the pole surface to be in hydrostatic equilibrium with one at the equatorial surface, we have:
\frac{GM}{R_{P}^{2}} = \frac{GM}{R_{E}^{2}} + R_{E} \omega^{2}
\frac{R_{E}^{2}}{R_{P}^{2}} = 1 + \frac{R_{E}^{3} \omega^{2}}{GM}
\frac{R_{E}^{2}}{R_{P}^{2}} = 1 + \frac{R_{E}^{3} \omega^{2}}{GM}
Now, if the Earth were a perfect sphere, M = \frac{4}{3}\pi R_{S}^{3} \rho. Since the flattening of the Earth is very small, we expect \frac{R_{E}^{3}}{R_{S}^{3}} \approx 1. Then,
\frac{R_{P}}{R_{E}} = (1 + \frac{3 \omega^{2}}{4G\pi\rho})^{-\frac{1}{2}}
As we expect \frac{R_{E}}{R_{P}} to be very close to 1, this implies that \frac{3 \omega^{2}}{4G\pi\rho} \ll 1, and hence
\frac{R_{P}}{R_{E}} \approx 1 - \frac{3 \omega^{2}}{8G\pi\rho}
Thus, the flattening of the earth
1 - \frac{R_{P}}{R_{E}} \approx \frac{3 \omega^{2}}{8G\pi\rho} = \frac{3 \pi}{2GT^{2}\rho}
Does this look reasonable?
