# How is orbital period calculated if perihelion and aphelion are known? For example, the orbit of a spacecraft about the sun has a perihelion distance of 0.5 AU and an aphelion of 3.5 AU., what is its orbital period?

Nov 21, 2015

Using Kepler's 3rd law, we get an orbital period of about $2.83$ years.

#### Explanation:

The orbital period can be calculated using Kepler's 3rd law, which states that the square of the orbital period is proportional to the cube of the average distance from the sun. In other words;

${T}^{2} \propto {R}^{3}$

Where $T$ is the orbital period and $R$ is the body's average distance from the sun, or the semi-major axis. Furthermore, this ratio is the same for any body that orbits the sun. We can therefore rewrite this expression as a ratio in terms of some constant, $C$.

${T}^{2} / {R}^{3} = C$

We know that the orbital period for the Earth is $1 \text{ year}$, and the Earth's semi-major axis is defined as $1 \text{ AU}$. Using this information we can solve for $C$.

$C = {T}_{\text{Earth"^2/R_"Earth"^3 = (1 " year")^2/(1 " AU")^3 = 1 "year"^2/"AU}}^{3}$

Remember that $C$ is the same for all bodies orbiting the Sun. We can calculate the semi-major axis for the spacecraft by taking the average of its perihelion and aphelion.

R_"spacecraft" = ("perihelion" + "aphelion")/2

R_"spacecraft" = (.5 " AU" + 3.5 " AU")/2

${R}_{\text{spacecraft" = 2 " AU}}$

Now we can calculate the orbital period.

${T}^{2} = C {R}^{3}$

${T}^{2} = {\left(1 \text{year"^2/"AU"^3)(2 " AU}\right)}^{3}$

${T}^{2} = 8 {\text{ years}}^{2}$

$T = 2 \sqrt{2} \text{ years}$

So the spacecraft has an orbital period of about $2.83$ years.

*Note: Kepler's 3rd law works for things orbiting bodies other than the sun, but the constant, $C$ will be different.