A planet of mass m moves around the Sun in a circular orbit…

A planet of mass $m$ moves around the Sun in a circular orbit with velocity $v = 34.9\ \mathrm{km/s}$ (relative to the Sun). Find its orbital period, given the Sun's mass. Which planet is it? Also prove the relation $\left[2\pi\gamma M_S/v^2\right] = \mathrm{s}$.

Step 1: Idea

For a circular orbit the speed alone determines the radius: gravity must supply the centripetal force, and a given speed matches only one distance. Once we know the radius, the period is the circumference divided by the speed.

Step 2: Orbit radius

\[ m\frac{v^2}{R} = \gamma\frac{M_Sm}{R^2}, \]

\[ R = \frac{\gamma M_S}{v^2}. \]

\[ R = \frac{6.67 \cdot 10^{-11} \cdot 2 \cdot 10^{30}}{(3.49 \cdot 10^4)^2} = 1.10 \cdot 10^{11}\ \mathrm{m}. \]

Step 3: Period

\[ T = \frac{2\pi R}{v} = \frac{2\pi\gamma M_S}{v^3}, \]

\[ T = \frac{2\pi \cdot 1.10 \cdot 10^{11}}{3.49 \cdot 10^4}. \]

\[T \approx 1.97 \cdot 10^7\ \mathrm{s} \approx 228\ \mathrm{days}\]

Step 4: Which planet?

The radius is $1.10 \cdot 10^{11}/1.5 \cdot 10^{11} = 0.73$ of the Earth's orbit, and the year is about 225 days. The planet is faster than the Earth (29.8 km/s), so it is closer to the Sun; the only planet at 0.73 AU is Venus.

Step 5: The unit relation

Check the units of the expressions ($\gamma M_S$ has the unit $\mathrm{m^3/s^2}$): \[\left[\frac{2\pi\gamma M_S}{v^3}\right] = \frac{\mathrm{m^3\,s^{-2}}}{\mathrm{m^3\,s^{-3}}} = \mathrm{s},\]

\[ \left[\frac{2\pi\gamma M_S}{v^2}\right] = \frac{\mathrm{m^3\,s^{-2}}}{\mathrm{m^2\,s^{-2}}} = \mathrm{m}. \]

So it is $2\pi\gamma M_S/v^3$, the period, that is measured in seconds. The expression $2\pi\gamma M_S/v^2$ equals $2\pi R$, the length of the orbit, and is measured in metres.

Check

Venus: $R = 1.08 \cdot 10^{11}\ \mathrm{m}$, $T = 224.7\ \mathrm{d}$, $v = 35.0\ \mathrm{km/s}\ \checkmark$

Kepler's third law: $ (228/365)^{2/3} = 0.73\ \checkmark $