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14 November 2020

What does "r" mean in equations for gravitational force and centripetal force?

On one hand, that's easy for me to answer: 

In the gravitational force equation F = GMm/r,2 the r represents the distance between the centers of the two objects.  I often use the variable d to emphasize this meaning.

In the centripetal force equation F = mv2/r, the r represents the radius of the circular motion.

These facts are also easy for students to read and recall.  It's not as easy for students to put into practice.  They see the letter r, hear "radius," and plug in any random distance they can pluck out of the problem stem.

And, this <pop> pull-a-radius-value-out-of-their-tuckus method is very often successful in a gravitation problem.  When an object is on the surface of a planet, the r value is in fact the planet's radius.  When a satellite undergoes a circular orbit around a central planet, the orbital radius r is in fact the same as the r distance between the satellite and the planet's center.

So why does it matter if students truly understand the difference between these two meanings of r?  In what possible physical situation in introductory physics does this difference even matter?  Here's one.

Two stars, each of equal mass M, maintain a constant distance x apart and rotate about a point midway between them at a rate of one revolution in every time t1.

(a) Why don’t the two stars crash into one another due to the gravitational force between them?

(b) Derive an expression for the mass of one star.  Use given variables and fundamental constants only.  You must annotate your calculation – if your response has no words, you will redo it from scratch in consultation.

This is a difficult question for students to conceptualize, especially because while we've done plenty of straightforward orbit problems, students have very often remembered comforting algorithms and not necessarily internalized physical meaning.  And I won't answer questions from students before they turn it in.  (They can discuss and argue with each other as much as they want!)  In class the day this problem is due, I don't start by "going over" how to solve it.  And on that model, I won't simply go into my solution here.  Instead, I'll show you the quiz with which I begin class:

For #1, r in this equation represents the distance between the centers of the two planets.  The problem says explicitly - the stars are always 8.0 x 1010 m apart.  (Most common misconception:  because in every problem they've previously done a satellite orbits around a central planet, they think that the "point midway between" the stars is the location of a planet of some sort - or that this "midpoint" is what exerts the force on the star.)

For #2, r in this equation represents the radius of the orbit.  The problem says explicitly that the stars "rotate about a point midway between them".  Since the stars are 8.0 x 1010 m apart, the point midway between them is half that distance from one of the stars, or 4.0 x 1010 m.  (It's not correct that one star is fixed with the other orbiting around the fixed star.  That's not what the problem says, nor is it how binary stars behave.  And half of 8.0 x 1010 m doesn't mean divide the exponent by two: half of 80 billion meters is certainly not 4 hundred thousand meters!)

For #3, students know to use this equation when they know the period of an object's circular motion.  It comes from the fact that the orbital speed is constant - for constant speed, speed is distance/time.  The relevant time here is the period, the time for one orbit.  The corresponding distance, then, is the circumference of the circular orbit.  From geometry class, that circumference is 2πr, where r is the radius of the circle.  We want the same distance as in number 2, the 4.0 x 1010 m radius of the orbit.

For #4, the distractors are practically word-for-word from past student responses.  As I discuss the quiz, I pick an incorrect answer to explain why it's incorrect:

For (A), first we discuss and agree that the gravitational forces on each star are indeed a Newton's Third Law force pair.  Then I go to a student's desk, hand the student a string, and pull.*  Is the force of me on Mr. Chamberlain equal to the force of Mr. Chamberlain on me?  So is that a N3L force pair?  Yes.  Are we orbiting around each other in a circle?  No.  So the logic of choice (A) is not logic at all.

*In the Before Times, I'd clasp hands with a student and pull lightly.

Next I ask students to close their eyes.  All who chose this incorrect answer, raise your hands high.  Now put your hands back down, and open your eyes.  The point is for students to acknowledge their misconceptions. It's okay that they made the mistake - after all, in my class everyone gets an A- until the AP exam, quizzes are given and graded but don't "count".  I don't want students to feel shame for being wrong.  But more importantly, I don't want them to sour-grapes style convince themselves that they knew that and that they didn't really make a mistake.  No.  Own the misconceptions, then don't make them any more!

For (B), first we discuss and agree that the gravitational forces on each star are indeed a Newton's Third Law force pair.  I again go to a student's desk, hand the student a string, and pull. The forces on each of us are equal. Are we orbiting each other?  No?  Then (B) is wrong.

For (D), I ask a student to point to the midpoint between us.  Only objects can exert forces... so what object at the "midpoint" can exert a force?  No object.  So (D) is wrong.

And finally, for the correct choice (C), I draw a picture of the two stars orbiting.  I draw the direction of one star's instantaneous velocity, which is tangent to the orbit.  I ask about the direction of the net force on that star, which is toward the center.  Everyone sees that the velocity is indeed perpendicular to the net force.  And we discuss how that's a restatement of one of our circular motion facts: when an object moves at constant speed in a circle, its acceleration is toward the circle's center.  The velocity will always be tangent to the circle, which by geometry is perpendicular to the direction toward the center.





1 comment:

  1. Thank you so much for writing this up. We're starting our circular motion unit next week and there's a day on my agenda where I've literally just put "the Jacobs Blog thing" :) This is perfect!

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