### 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?

Letβs 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 weβll call that capital π.

Letβs draw a diagram of the orbit of this asteroid. The asteroid Eros orbits the Sun in an elliptical orbit. In this diagram, weβve shown the asteroid at its point of farthest distance from
the Sun, its aphelion distance, which weβve 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, weβll rely on Keplerβs second law of planetary motion.

Keplerβs second law tells us that the square of the orbital period, capital π,
is proportional. Keplerβs 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 weβll 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 havenβt 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 weβve
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 Keplerβs 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 weβre working to solve
for in terms of given information π
sub π and π
sub π.

Letβs 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. Thatβs how long it will take this asteroid to make one complete orbit around
the Sun.