# Eros has an elliptical orbit about the sun, with a perihelion distance of 1.13 AU and aphelion...

## Question:

Eros has an elliptical orbit about the sun, with a perihelion distance of {eq}1.13 \ AU {/eq} and aphelion distance of {eq}1.78 \ AU {/eq}. What is the period of it's orbit?

## Kepler's Laws of Motion:

Any planet orbiting its star or any satellite orbiting a planet obeys Kepler's Laws of planetary motion. Those three laws can be stated as follows:

• All planets orbit the Sun along elliptical paths, with the Sun located at one of the foci of the ellipse;
• The area swept by the planet's radius-vector in a given time interval is the same along the orbit;
• The squares of periods of motion of two planets are in the same ratio as the cubes of major semi-axes of their respective orbits.

The major semi-axis of Eros is:

{eq}a = \dfrac {1.13 \ AU + 1.78 \ AU}{2} = 1.455 \ AU {/eq}

According to the Kepler's Third Law, the ratio of periods of the Earth and Eros, squared, equals the ratio of major semi-axes of their respective orbits.

{eq}\left (\dfrac {a}{1 \ AU} \right )^3 = \left (\dfrac {T_{Eros}}{1 \ year} \right )^2 {/eq}

Solving for the period, we obtain:

{eq}T_{Eros} = 1 \ year \times \left (\dfrac {a}{1 \ AU} \right )^{3/2} {/eq}

Calculating, we get:

{eq}T_{Eros} = 1 \ year \times \left (\dfrac {1.455 \ AU}{1 \ AU} \right )^{3/2} \approx \boxed {1.76 \ year} {/eq}