# Video: Determining the Orbital Period of an Object with an Elliptical Orbit

Ed Burdette

The asteroid Eros has an elliptical orbit about the Sun, with a perihelion distance of 1.13 AU and aphelion distance of 1.78 AU. What is the period of its orbit?

04:41

### Video Transcript

The asteroid Eros has an elliptical orbit about the Sun, with a perihelion distance of 1.13 AU and aphelion distance 1.78 AU. What is the period of its orbit?

Letβs start here by highlighting some of the vital information given to us. We were told that this asteroid orbits the Sun and that its perihelion distance, that is, the closest it ever approaches the Sun, is 1.13 astronomical units, where an astronomical unit is roughly equal to the distance between the Earth and the Sun.

This asteroid also has an aphelian distance or furthest distance from the Sun of 1.78 astronomical units. We want to know its period. And weβll call that capital π.

Letβs draw a diagram of the orbit of this asteroid. The asteroid Eros orbits the Sun in an elliptical orbit. In this diagram, weβve shown the asteroid at its point of farthest distance from the Sun, its aphelion distance, which weβve called π sub π.

When the asteroid is closest to the Sun, at its perihelion distance, we call that distance from the Sun π sub π.

To figure out what is the period, that is, how long does it take this asteroid to move in one complete revolution around the Sun, weβll rely on Keplerβs second law of planetary motion.

Keplerβs second law tells us that the square of the orbital period, capital π, is proportional. Keplerβs second law tells us that the square of the period π is equal to the cube of the semimajor axis of the elliptical orbit, which weβll call π.

On our diagram of the orbit, if we draw a coordinate plane with the center of the ellipse at the origin, then π, the semimajor axis, is equal to the distance from the origin to the furthest point of the ellipse.

In our problem, we havenβt been given π, but we have been given π sub π and π sub π. These are values that originate not at the origin of our coordinate plane but at the center of the Sun.

However, since the ellipse is symmetrical about the origin of the coordinate plane weβve drawn, we can write down a relationship between π, π sub π, and π sub π.

Notice that if we add together π sub π and π sub π, we get a distance which equals the entire horizontal extent of our orbit. This horizontal extent is equal to two times π, the semimajor axis. So therefore, we can write that π equals π sub π plus π sub π divided by two.

This relationship lets us refer back to Keplerβs second law. If we write the second law as it applies to our scenario, we see that we can substitute in for π our term π sub π plus π sub π divided by two.

When we make that substitution, we find that we now have an equation for the period that weβre working to solve for in terms of given information π sub π and π sub π.

Letβs take the square root of both sides of our equation. And when we do that, the square and the square root on the left-hand side cancel one another out. This leaves us with an equation that reads the period π equals π sub π plus π sub π divided by two, all raised to the three-halves power.

We can now substitute in the given values for π sub π and π sub π. We see that π is equal to 1.78 AU plus 1.13 AU divided by two raised to the three-halves power.

When we enter these values into our calculator, we find that the period π is equal to 1.75 years. Thatβs how long it will take this asteroid to make one complete orbit around the Sun.