If there were no rotation, celestial bodies would appear as perfect (though not absolutely perfect) spheres due solely to the gravitational interaction between matter. However, the vast majority of celestial bodies possess rotation, which causes their equatorial radii to be longer than their polar radii. BoJone has roughly considered the shapes presented by celestial bodies under the combined action of gravity and inertial centrifugal force. By comparing these results with some celestial bodies in the solar system, it was found that they match to a certain extent. I am sharing this here for the readers’ reference.
Taking Earth as an example, if there were no rotation, Earth should be a sphere. Considering this in a 2D plane, the equation for Earth would be x^2+y^2=R^2, where R is the Earth’s radius, specifically the polar radius. Due to rotation, the equator is “stretched” accordingly; the closer a point is to the equator, the more it is “pulled.” Assume the Earth rotates uniformly with an angular velocity \omega. When there is no rotation, the horizontal component of the gravitational force acting on a point (x,y) on the Earth’s surface (assuming unit mass) is \frac{GM}{R^2}\cdot \frac{\sqrt{R^2-y^2}}{R} (we use an expression containing y instead of x because the x-axis changes after rotation, while y remains constant). Here, we only need to consider the horizontal component because the centripetal force is in the horizontal direction, with a magnitude of \omega^2 \sqrt{R^2-y^2}. The centripetal force is provided by the horizontal component of gravity, which is equivalent to reducing the gravity. Thus, the horizontal component of gravity after rotation becomes \frac{GM}{R^2}\cdot \frac{\sqrt{R^2-y^2}}{R}-\omega^2 \sqrt{R^2-y^2}.
Next, we make a relatively subjective assumption: the “stretching” ratio is inversely proportional to the horizontal component of gravity (we will later find that this assumption is quite accurate). Thus, we find that the original circular x-coordinate is stretched by a factor of: \frac{\frac{GM}{R^2}\cdot \frac{\sqrt{R^2-y^2}}{R}}{\frac{GM}{R^2}\cdot \frac{\sqrt{R^2-y^2}}{R}-\omega^2 \sqrt{R^2-y^2}} This result happens to be a constant, equal to \frac{1}{1-\frac{\omega^2 R^3}{GM}}. That is, the shape after rotation becomes an ellipse, with the equation: \left(1-\frac{\omega^2 R^3}{GM}\right)^2 x^2+y^2=R^2 Using an approximation, \frac{1}{1-\frac{\omega^2 R^3}{GM}}\approx 1+\frac{\omega^2 R^3}{GM} (given |\frac{\omega^2 R^3}{GM}| \ll 1), meaning the length in the horizontal direction is stretched by \frac{\omega^2 R^3}{GM}.
Let’s test this with some examples. For Earth, checking Wikipedia gives an equatorial radius of 6,378 km and a polar radius of 6,356 km, a difference of about 22 km. Calculating with the above formula: R\cdot \frac{\omega^2 R^3}{GM}=R\cdot \frac{(\omega R)^2}{v_1^2}=21.76\text{ km} Where v_1=7.9\text{ km/s} is the first cosmic velocity, R=6356\text{ km}, and \omega=\frac{2\pi}{86400\text{ s}}. As we can see, this is a very good approximation. To prove this is not “pure coincidence,” let’s test it with data from Jupiter.
Jupiter is the largest planet in the solar system, with an equatorial radius of 71,492 km and a polar radius of 66,854 km, a difference of 4,638 km (almost the size of an Earth, haha), and it rotates once every 9.9 hours. Calculating with the formula: R\cdot \frac{\omega^2 R^3}{GM}=R\cdot \frac{(\omega R)^2}{v_1^2}\approx 5100\text{ km} Although not extremely precise, BoJone believes the agreement is quite good, considering how crude this model is! Finally, we have obtained an oblateness factor \frac{\omega^2 R^3}{GM}! Everyone is welcome to make further corrections!
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