The Mathematics of Circular Motion,and Multiple Star Systems

Circular Motion

Before considering the more involved example of a multiple star system, consider a moon moving in a simple circular orbit around a planet, which can be considered stationary. The planet's gravitational field results in the moon having an acceleration towards it; the moon is constantly falling towards the planet. However, the moon also has its own velocity, which is tangential to the circle around which it is moving, and so at right angles to the acceleration of gravity exerted on it by the planet. If the motion is treated in a series of large steps, as in the diagram below, can be seen why the resulting motion is an orbit.

At each step, the orbiting moon is falling toward the planet under gravity. It is also moving, with its own velocity, tangential to the orbit. The result is that it falls not toward the planet, but into the next step of its motion. These steps can be reduced in size until they are infinitely small, which produces a circular orbit.

For motion in a circle, some of the simplest pieces of information are the linear speed and the angular speed. The linear speed is the velocity tangential to the orbit that has already been mentioned. Let P be the period time taken for one complete orbit. Let the radius of the orbit be r.

While the linear speed is the rate of change of distance around the orbit, the angular speed is the rate of change of angle. The angular speed is designated the Greek letter omega ( w ). The angle moved through is theta ( q ). For angular speed, it is normal to work in radians, a unit of angle such that there are 2 p radians to 360 degrees. In the orbital period P, the orbiting body therefore moves through 360 degrees = 2 p radians.

These two equations for v and w both contain T, so:

Rearranging, this gives v = r w This equation relates r and w , and so is important to circular motion.

The gravitational force that was mentioned earlier on is what sustains the circular motion, and is called the centripetal force. The centripetal acceleration is the acceleration that results from the centripetal force. A little calculus will obtain formulae for the centripetal acceleration, and through this for the centripetal force:

Double Stars

These systems differ from conventional circular motion because of the fact that both bodies are moving in an orbit around a common centre of mass. The diagram below shows the motion of the members of a double star system in circular orbits. We will remain with circular orbits, since elliptical orbits result in excessive levels of calculus for the purposes of this article.

Star A has mass M A , and star B has mass M B. Note that the black point at the centre is in fact the centre of the circular orbits of both bodies.

The two members of a system orbit such that they are always opposite each other, so that at any point the line between joining the two bodies will pass through the point around which they orbit. Using Newton's Law of gravitation, we can calculate the force of attraction between the two bodies:

The centripetal force is known, so the centripetal acceleration can be found:

Now we have the centripetal accelerations for each of the systems, and system is now reduced to applying the mathematics of circular motion to each body in turn.

The Mathematics in the Model

We have produced an animated model of circular double star orbits that is necessary for the exercise. In order to run this model, it is necessary to have a copy of the freeware program Modellus. Further details about this download are available here.

The mathematics involved in the model is explained in full here.

Kepler's third law is an equation relating the period and radius of orbit for a planet orbiting around a star. This was later adapted for double stars, and a result of this is the following equation that relates the masses and the orbital radii of the two members of a double star system:

Applying the same techniques, but substituting to remove r A instead, the following equation is obtained for r B :

These two equations are the first two lines of the model, and allow the two radii of orbit to be calculated from the masses and the separation, which are the only inputs to the model.

The most difficult part of the model is now over; the rest is simpler. The equation on the third line may look imposing but is easy to derive:

This formula allows the period time to be calculated from the initial input conditions of separation, M A , and M B. Note that these are the same inputs as are required for the previous formulae for the radii. Note also that the period for both stars must be the same if they are to remain opposite each other at all times.

Since the stars have the same periods, they must also have the same angular velocities.

The remainder of the model relies upon the concept of parametric equations. Consider the centre of mass - the centre of both the circles. Let this be the origin of a graph. Parametric co-ordinates express the x and y co-ordinates in terms of theta: the angle made with the x-axis by the line joining the point to the origin. For a circle, the parametric equations are x = rcos q and y = rsin q , where r is the radius of the circle

This result can be derived using trigonometry.

In the model, theta_A and theta_B are the angles made with the x-axis by stars A and B respectively. They are calculated for a time t by:

Therefore, as in the model: theta_A = omega x t

The stars must remain opposite each other at all times, so theta_A and theta_B must differ by 180 degrees, which is p radians.

This leads to the: theta_B = theta_A + p

The model's last four lines simply use theta_A and theta_B to calculate the x and y co-ordinates at any time, using the previously mentioned parametric co-ordinates of a circle.