I am trying to use Kepler's second law to find the duration of Venus's orbit. I am assuming circular orbits (using Earth and Venus, so low eccentricity). Here is my process:

Assuming that the radius of Earth's orbit is 150 million km, then the swept area in one day is $\frac{1}{365.25}\times\pi\times 150^2 \approx 194 \text{ million km}^2$.

Venus must sweep the same area in the same time. Assuming a orbital radius of 108 million km for Venus, and using $A = \frac{\theta}{360}\pi r^2$, we can find the central angle for the swept sector, that is, the angle traveled in one Earth day:

$194 = \frac{\theta}{360}\pi \times108^2 \implies \theta = 1.90 ^{\circ}$ per Earth day.

Hence the orbital period should be $\frac{360}{1.90}\approx 189$ Earth days.

Of course, the orbital period of Venus is $224.7$ Earth days. The difference between 189 and 224.7 appears to be well beyond the error introduced by my assumption of circular orbits.

What am I doing wrong?

I know this is perhaps a circuitous way of doing this calculation. My goal is to write a mathematics exercise that uses the area of sectors in a meaningful way.

  • $\begingroup$ +1 for showing all work and asking a very clear question! $\endgroup$
    – uhoh
    Oct 25, 2018 at 10:05

1 Answer 1


Kepler's laws state that a planet sweeps equal areas in equal times as it moves in its elliptical orbit. It doesn't state that different planets will sweep the same area.

The "equal areas" law can be derived from the "conservation of angular momentum". In fact dA/dt = L/(2m) (where A is the area, L is the angular momentum and m is the (reduced) mass).

Different planets will sweep out different areas. To calculate the period you used Kepler's third Law: $T^2 = k a^3$ (T= orbital period,a = semi-major axis). If, for convenience you take a in AU and T in Earth Years, then the constant $k=1$.

For Venus, a = 0.72. so $T=\sqrt{0.72^3}=0.61$ or about 223 days.

Hyperphysics has a section on Kepler's Laws


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