Elliptical Orbits: Periods & Speeds

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  • 0:03 What Is An Elliptical Orbit?
  • 1:26 Period & Speed Equations
  • 2:44 Example Calculations
  • 4:52 Lesson Summary
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Lesson Transcript
Instructor: David Wood

David has taught Honors Physics, AP Physics, IB Physics and general science courses. He has a Masters in Education, and a Bachelors in Physics.

After watching this lesson, you will be able to explain what an elliptical orbit is and calculate the period and speed of an object in an elliptical orbit, if given enough information. A short quiz will follow.

What is an Elliptical Orbit?

The orbits of the planets might look circular, but they're really not. They're actually oval shaped, squashed circles in various states of squashiness.

An ellipse is an oval. So an elliptical orbit is an oval-shaped orbit. Though to be more precise, it's anything from a circular orbit all the way to an orbit that isn't quite parabolic. An elliptical orbit is officially defined as an orbit with an eccentricity less than 1. Circular orbits have an eccentricity of 0, and parabolic orbits have an eccentricity of 1. So circular orbits technically ARE elliptical. But parabolic orbits are NOT elliptical.

Most of the orbits of the planets are actually pretty close to circular. Exceptions include Pluto and Mercury that have eccentricities of roughly 0.25. Still pretty close to zero, but definitely not circular.

But this is physics, so we want equations to describe these elliptical orbits. Johannes Kepler, a 17th-century German mathematician, was the first to take Newton's Laws and apply them to what would happen if orbits weren't perfectly circular. Through that work, he came up with what later became known as Kepler's Laws, a series of laws that describe the motion of orbiting bodies no matter the shape of the orbit. And these laws form the basis of our understanding of orbiting bodies to this day.

Period and Speed Equations

To figure out the time period of an orbit - the time it takes to complete one full orbit - we need to use the equation version of Kepler's Third Law. The Third Law equation tells us that the square of the time period, T (measured in seconds), divided by the maximum radius of the orbit, r (measured in meters), cubed, is equal to (4pi)^2, divided by the gravitational constant big G (which is always 6.67 x 10^-11), multiplied by the mass of the star, M (measured in kilograms).

T^2 / r^3 = (4pi)^2 / GM

Technically, to do this properly, M is really the mass of the star added to the mass of the orbiting object, but very often the orbiting object is so much smaller than the star that you might as well just put in the mass of the star.

The equation for the orbital velocity, v, of the orbiting object is given by this equation, which can also be derived from Kepler's Laws:

orbital velocity equation: v = sqrt (GM * ((2/r) - (1/a)))

It says that the orbital velocity, v (measured in meters per second), is equal to the square-root of the gravitational constant big G, multiplied by the mass of the star, M (measured in kilograms), measured by 2 over the distance between the orbiting bodies in meters, minus 1 over the maximum distance between the two bodies (otherwise known as the length of the semi-major axis).

So if you have all the missing variables, you can just plug numbers in and solve.

Example Calculation

Okay so now it's time to go through an example of how to use the equations. Let's say a small rock is orbiting a large rock at a maximum distance of 1 x 10^14 meters. If the mass of the large rock is 4 x 10^40 kilograms, what is the time period of the orbit? And what is the orbital velocity of the small rock when the two rocks are 5 x 10^13 meters apart?

First of all, as usual, we should write down what we have. We know that the maximum radius, r, is 1 x 10^14 meters, and we know that the mass of the large rock, M, is 4 x 10^40. That's all we actually need for the first calculation, where we're asked to find T, the time period, so T equals question-mark.

Plug these numbers into the Third Law equation, and solve for T, and we find that T = (4pi^2 * 1 x 10^14 meters^3) / ((6.67 x 10^-11) / (4 x 10^40)). Type all of that into a calculator, and we get a time period of 3.85 x 10^6 seconds.

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