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If I understand correctly, according to Kepler, a planet orbiting a star should cover a constant surface area of its ellipsoid or circular orbit per time unit, regardless of the distance of that planet to the star.

I have tried to prove this by using a scenario in which a planet is orbiting in a perfect circle around a star, so that the distance is the radius ##r##. I would expect that changing the distance ##r## would still make the planet cover a constant area of the circle per time unit through changing its orbital velocity.

The time period ##P## in which this planet orbits a full circle around the star is defined by:

$$\sqrt{\frac{4\pi^2\cdot r^3}{G(M+m)}}=P$$

Dividing the circumference of the circle ##2\pi r## by that formula would give the orbital velocity per second. Divding that orbital velocity again by the circumference of the circle would give the fraction of the circumference that the planet covers per second. Multiplying that fraction by by the surface area of the circle ##\pi r^2## would give the surface area that the planet orbits per second.

Rewriting all this in the formula and simplifying would eventually give me:

$$\sqrt{\frac{r \cdot G(M+m)}{4}} = A$$

In which ##A## is the covered surface area per second. This formula still shows that ##A## is dependent on the distance ##r##. From what I understand, this shouldn’t be the case since a planet revolving around a star should cover a constant area regardless of distance.

I have a feeling that this "constant area per unit time" rule only applies within the orbit of the planet itself, not when changing its distance. Within a circular orbit, ##A## would indeed be constant.

Furthermore, if the orbit of a planet ##1## is an ellips with a semi-major axis ##a_1##, then the planet only sweeps equal areas per time unit within that particular ellips with that semi-major axis ##a_1##. An equally massive planet ##2## orbiting in an ellips but has a different semi-major axis ##a_2## wouldnotsweep the same area per time unit as planet ##1##, but it will sweep equal areas within its own elliptical orbit with semi-major axis ##a_2##.

Is this the cause of my misunderstanding?

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# I Some confusion about Kepler’s formula

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