Dear Greg,

I have been reviewing with my students and want to pose this question to them but am having a bit of difficulty with the solution:

The dwarf planet Pluto goes around the sun in an elliptical orbit. Consider Pluto only. Is its angular momentum about the sun conserved? Justify your answer.

I know that in an elliptical orbit the distance between the sun and Pluto would change and therefore angular momentum would not be conserved. However what is throwing me off is considering Pluto only. Does this mean you do not take into consideration the elliptical path?

Angular momentum is conserved when the system experiences no net torque. Pluto alone is the system. What forces act on Pluto? Just the (gravitational) force of the sun. Does that force provide a torque about the sun? No - the force of the sun on Pluto is always directed toward the sun, so there is no lever arm for that force about the sun. (Another way to say it: the line of the force of the sun on Pluto goes through the axis of rotation, which is the sun itself.) So angular momentum is conserved.

How does that square with the elliptical orbit? Angular momentum for the point-object Pluto is

*mvr*, and that can’t change. So when the distance from the sun*r*is small, the speed*v*is large, and vice versa. That is in fact the case for all planets. The earth moves faster around the sun in (northern hemisphere) winter, when we’re about 3% closer to the sun, than in summer when we’re a wee bit farther away.
GCJ

I think the confusion comes from incompletely diagramming it. Draw the ellipse, with the sun at one focus, and the foci on a horizontal line. Put Pluto at a point that is at, say, "noon" if this were an elliptical clock. Draw a vector showing the tangential velocity of Pluto, and a vector representing the gravitational force exerted on Pluto by the sun. Note that the angle is not 90 degrees. This means there is a tangential component to the force being applied to Pluto. Of course, your answer is still correct; the gravitational force is parallel to the lever arm (which is what our drawing has forgotten), which is what matters here. But I've found that's where the confusion comes from; people take that angle I mentioned to be the "theta" of Torque = F l sin(theta), which basically is confusing this situation with any number of circular-motion situations in which a tangential component of a force would indeed be applying torque to the system.

ReplyDelete