## Illustration 12.5: Kepler's Second Law

Please wait for the animation to completely load.

A planet orbits a star under the influence of gravity (**distance is given in astronomical units [A.U.] and time is given in years; the total area swept out by the planet's orbit is given in A.U. ^{2}**). The animation begins from the point of aphelion, the point where the planet is farthest from the star. The planet's orbit is elliptical, and its trail is shown as it orbits the star. Kepler's second law states that the planets sweep out equal areas in their orbits in equal times. What does this mean for the planet's orbit? If the planet had a circular orbit, the planet would undergo uniform circular motion and Kepler's second law is just a statement of equal speed; it confirms the statement of uniform circular motion. For elliptical orbits, therefore, the planet's motion must not be uniform. Restart.

Starting at t = 0, run the animation for 3 years (not real time, animation time!). How much area has been swept out by the planet in this time interval? There is 28.43 A.U.^{2} swept out. What about from 3 to 6 years? Again 28.431 A.U.^{2} is swept out. Does it matter where you are in the orbit? No. Try it for yourself. While the planet is closer to the star, its speed increases; when the planet is farther away from the star, its speed decreases.

So what does Kepler's second law really tell us? The sweeping out of equal areas is equivalent to telling us that angular momentum is conserved! We know that if angular momentum is conserved (see Chapter 11), there is no net torque. Here the only force on the planet is gravity, and gravity cannot create a torque because the radius arm and the force are always in the same direction.

Illustration authored by Steve Mellema, Chuck Niederriter, and Mario Belloni.

Script authored by Steve Mellema and Chuck Niederriter.

Physlets were developed at Davidson College and converted from Java to JavaScript using the SwingJS system developed at St. Olaf College.

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