Consider an inward-sloping surface of revolution with a vertical axis (a “bowl” or “funnel”). Let an object slide frictionlessly on this surface. Let it be at a height where the surface has radius r, and inclination θ from the horizontal; what must the (horizontal, tangential) velocity v be so that the object remains at that height? What is the period T of the object’s “orbit”? What function r(z) gives a surface for which the square of the period is directly proportional to the radius cubed at all heights (so that it mimics Kepler’s third law)?
The object experiences two forces; gravity downward, and a normal force at an angle θ inward from the vertical. For a horizontal orbit, we need the net force to be purely horizontal. Letting N be the magnitude of the normal force, and m be the mass of the object, the balance of vertical force components means , and so
. The net force is then the horizontal component of the normal force:
This must be the centripetal force for the uniform circular motion:
.
The length of the object’s path is 2πr, so the period is
Thus, the period squared is
,
so for this to be proportional to r3 (with proportionality constant c), we have
,
where . Now, if we have function r(z), then the derivative r‘(z) gives the outward slope relative to the vertical, and so equals the tangent of the angle of the surface from the vertical, which is
; thus
, giving us the differential equation
;
this is a separable differential equation with solution
, (
,
); we have z<b. We can alternately describe this by the inverse function
; below is a graph of the resulting funnel shape.
[Note: this funnel shape gives a height, and thus potential energy, inversely proportional to the radius of the orbit; compare to the inverse proportionality to radius of gravitational potential energy in actual Keplerian orbits. Thus, this last step could also have been determined by considering the energy of the orbiting object.]
Tags: Centripetal Force, Friday Physics, physics
