# What is wrong with my calculations of Venus' orbital period?

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.

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

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.