A planet orbits a star under the influence of gravity (**t****he
distance is given in Astronomical Units (AU) and the time is given in years; the total area swept out by the planet's orbit is given in
AU ^{2}**).
Restart. The animation begins from the
point of aphelion, the point where the planet is furthest
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 undergoes
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, the planet's motion must not be uniform therefore.

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 28.431 AU^{2} swept out. What about
from 3 to 6 years? Again 28.431 AU^{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 further away from the star its speed decreases.

So what does Kepler's second law really tell us? The sweeping out equal areas is equivalent to telling us that angular momentum must be conserved! We know that if angular momentum is conserved (see Chapter 10), there is no net torque. Here the only force on the planet is gravity and gravity cannot create a torque: the radius arm and the force are in the same direction, always.

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

Script authored by Steve Mellema and Chuck Niederriter

© 2003 by Prentice-Hall, Inc. A Pearson
Company