General Central Force

1. Definition and Notations:

Here vectors will be displayed in bold blue, unit vectors in bold plum, and scalars (and components) in regular black.

Since force is a vector, F should be a vector, F.

If the force depends on position only, we can express force as a function of position as F(r), which in three dimensions means F(x,y,z) or F(r,,) .

A central force is one where the force is directed along the radial direction, r, and not in the other two directions,  or  . Its magnitude also depends only on the radial distance, r, and not on  or . This can be expressed as F = F(r) r . Be aware, however, that the radial direction, r, does depend on  and : F = F(r) r(,). r(x,y,z) = xx + yy + zz = r sin() cos() x + r sin() sin() y + r cos() z = rr where r = r(,) = sin()cos() x + sin() sin() y + cos() z .

2. Is a central force (always, sometimes, never) a conservative force? To answer this, recall Stokes Theorem:

 (F)  n dS = closed loop F  dl (where S is an area)

lf so that if F = 0 everywhere, then closed loop F  dl = 0, and so li F  dl is independent of path, and so a potential energy can be defined:

rf PE = - ri F  dl . Let’s now calculate F for a central force (using spherical coordinates): F = [r  /r +  (1/r)  / +  (1/{r sin()})  / ]  F(r) r = (F(r)/r) (r  r) + F(r) (r  r/r) + (1/r) (F(r)/) (  r) + (F(r)/r) (  (r/)) + (1/{r sin()}) (F(r)/) (  r) + (1/{r sin()}) F(r) (  r/) . There are six terms, so let’s look at these one by one.

The first term is zero since (r  r) = 0 (since any vector crossed itself is zero). The second term is zero since r/r = 0 (since r depends only on  and ). The third term is zero since F(r)/ = 0 (since F(r) does not depend on ). The fourth term is zero since r/ = , and (  ) = 0. The fifth term is zero since F(r)/ = 0 (since F(r) does not depend on ). The sixth term is zero since r/ = sin() , and (  ) = 0.

Therefore, F = 0 for F = F(r)r, so a central force is always a conservative force! We can then define a potential energy function:

r PE(r) = - r-standard F(r) dr .

3. Torque and Angular Momentum for a Central Force

Starting with the definition of torque as  ≡ r  F, for a central force, F = F(r)r, we have (recalling that r = rr):  = r  F(r)r = r F(r) r  r = 0. Central forces can create no torque.

Now we recall the relation between torque and angular momentum (L):  = dL/dt . Since torque under a central force is zero, the angular momentum must remain constant. Let’s now get an expression for angular momentum in polar coordinates. We can use 2-D polar (actually 3-D cylindrical) coordinates since a central force has no component of either motion or force perpendicular to the (r  v) plane, i.e., the z plane. By definition, L ≡ r  p = r  mv .

In cylindrical form, v = dr/dt = d(rr)/dt = (dr/dt)r + r(dr/dt) = r'r + r' (where r' = dr/dt, and ' = d/dt) . We will use this ' notation to only denote time derivatives, not any other derivatives.

Therefore, L = rr  m(r'r + r') = mrr'(r  r) + mr2'(r  ) = 0 + mr2' z = constant.

Note that if we know the initial conditions, ro and 'o, we can calculate the value of L. If we just consider the magnitude, since the direction will not change,

L = mr2' = constant. Eq. 1

This equation provides a relation between the radius, r and the angular speed, ' (which we sometimes have called ). We can write this as:

' = L/mr2 . Eq. 1a

4. Newton’s Second Law for a Central Force

 F = ma is a vector equation, which means it is one equation for each dimension. For a central force, F = F(r)r, in polar coordinates we have Fr = F(r) , and F = 0 . The acceleration expressed 2 2 2 2 in polar form is: ar = r''– r' (where r'' = d r/dt is the usual linear acceleration in the radial direction, and r' 2 2 2 =  r is the circular acceleration); and a = r'' + 2r'' (where r'' is r(d /dt ), the acceleration in the circular direction, and 2r'' is that strange Coriolis acceleration). Therefore, in component form, Newton’s Law becomes: r component: F(r) = m(r'' – r'2) , or F(r) + mr'2 = mr'' Eqs. 2 [where the term mr’2 = m2r can be considered the “centrifugal” force]

 component: F = 0, so 0 = m(r'' + 2r'') . Notice that both equations have both r and  so they are simultaneous differential equations. However, if we use the angular momentum equation, ' = L/mr2 in the radial equation, we get:

F(r) + mr(L/mr2)2 = mr'', or F(r) + L2/mr3 = mr'' Eq. 2a which is an expression that is equivalent to a one-dimensional problem since there is no mention of . If we can solve this for r(t), we can then use the  component equation to solve for '(t) and then q(t). We could also use ' = L/mr2 with r(t) and both L and m constant to solve for '(t) and then q(t).

5. Conservation of Energy

Since a central force is a conservative force, we can define a potential energy. If there are no other forces, then energy is conserved and we can write this in an equation:

KE + PE = E = constant

2 2 2 2 2 v = vrr + v , and v = vr + v = (r') + (r') so we get: ½ m v2 + PE(r) = ½ m r'2 + ½ m r2 '2 + PE(r) = E . If we use the angular momentum expression, ' = L/mr2, this becomes: ½ m r'2 + ½ m r2 (L/mr2)2 + PE(r) = E = ½ m r'2 + ½L2/mr2 + PE(r) . If we treat the term ½L2/mr2 as an additional potential energy which is equivalent to treating the centripetal acceleration as a centrifugal force (since it only depends on r), we can define an effective potential energy: 2 2 PEeff(r) = PE(r) + L /2mr . Eq. 3 We now have an equation that is essentially one-dimensional:

2 ½ m r' + PEeff(r) = E = constant . Eq. 4 As always, this is a first order differential equation whose solution is r(t). Then we can use ' = L/mr2 to solve for q'(t) and then q(t). 6. Straightforward solutions a) Use the Conservation of Energy (1-D) equation with the equivalent potential energy, and use the technique we used earlier in the course to first get r(t) and then to get (t):

2 2 2 ½ mr' + PEeff(r) = E where PEeff(r) = PE(r) + L /2mr

½ dr/dt = r' = [2(E-PEeff(r))/m ]

t r ½ 0 dt = ro {1 / [2(E-PEeff(r))/m ] } dr Assuming we can actually do the integration over r, we get an expression for t(r). Then we must find the inverse to get r(t). This is sometimes very difficult. If we can get r(t), we can then use the angular momentum expression: L = mr2' to get an expression for ':

'(t) = L/mr(t)2 . We can then use ' = d/dt to get:

 t t 2 0 d = 0 ' dt = 0 [L/mr(t) ] dt . If we can do the integration, we then have an expression for (t) and the problem is solved. b) Use Newton’s Second Law to try to get the trajectory r() [instead of r(t) and (t)]. While this will lose some information (time information), it has the advantage of giving us the orbits directly. We start with the equivalent 1-D relation (see part 4 above):

F(r) + L2/mr3 = mr'' . Eq. 2a Since we want r(), we still have to deal with dr/dt, but we note that (recalling that ' = L/mr2) : r' = dr/dt = (dr/d)*(d/dt) = (dr/d)*' = (dr/d)*(L/mr2) . r'' = d(r')/dt = (dr'/d)*(d/dt) = d[(dr/d)*(L/mr2)]/d * ' = (d2r/d2)*(L/mr2)*' + (dr/d)*(L/m)*(d[1/r2]/d)*' note: d[1/r2]/d = (d[1/r2]/dr)*(dr/d) = (-2/r3)*(dr/d) r'' = (d2r/d2)*(L/mr2)*' + (dr/d)*(L/m)*(-2/r3)*(dr/d)*' (again using ' = L/mr2 ) r'' = (d2r/d2)*(L2/m2r4) - (2L2/m2r5)*(dr/d)2 . Using this, we can now write Newton’s 2nd Law (r component equation, Eq. 2a) as:

F(r) + L2/mr3 = (d2r/d2)*(L2/mr4) - (2L2/mr5)*(dr/d)2 . Eq. 5

This doesn’t look all that easy to solve for r(q). 7. Tricky way to solve for r()

Sometimes things look nicer in one form than in another. We will try a trick that involves substitution: let’s define u = 1/r (or r = 1/u).

L = mr2' = m'/u2, or ' = Lu2/m r' = dr/dt = (dr/du)*(du/dt) = (dr/du)*(du/d)*(d/dt) = (d(1/u)/du)*(du/d)*' = (-1/u2)*(du/d)*' = (-1/u2)*(du/d)*(Lu2/m) = (-L/m)*(du/d). r'' = dr’/dt = (dr'/d)*(d/dt) = d[(-L/m)*(du/d)]/d * ' = (-L/m)*(d2u/d2)*(Lu2/m) = (-L2u2/m2)*d2u/d2 . Newton’s Second Law (r component equation) F(r) + L2/mr3 = mr’'' Eq. 2a thus becomes: F(1/u) + L2u3/m = m*(-L2u2/m2)*d2u/d2 , or (-m/L2u2)*F(1/u) + (-m/L2u2)*(L2u3/m) = d2u/d2 , or (-m/L2u2)*F(1/u) + -u = d2u/d2 . Eq. 6

This equation to solve for u(q) looks better than the straightforward way to solve for r(q) above (Eq. 5) because the r'' term gives only one term instead of two terms when we use this particular substitution. This is because the r' term does not have any u’s in it – a result of our particular choice for substitution. 8. Special Cases

8a) One of Kepler’s Laws of planetary motion Consider a small area of the orbit, dS, that is swept out in a small unit of time, dt:

vdt = r'dt dS

r

The distance covered by vr dt does not cover an area since it’s direction is parallel to r. The distance covered by v dt does cover an area of a triangle, so dS = ½r*(v dt) = ½r*(r'dt) = ½r2'dt. We now use the relation ' = L/mr2 to get

dS/dt = ½r2(L/mr2) = L/2m = constant. Eq. 7

This is one of Kepler’s laws for planetary motion: a planet sweeps out equal areas in equal time intervals. Note that this law applies for ANY central force, not just for the inverse square gravity force.

We can go even further to get:  dS = (L/2m) dt, or if we integrate over the complete orbit, we get: S = (L/2m)T where T is the period of orbit and S is the area of the orbit. Eq. 8

8b) Oscillations near equilibrium If we use the effective potential (see part 5 above): 2 2 PEeff(r) = PE(r) + L /2mr , Eq. 3 we now have an equation that is essentially one-dimensional:

2 ½ m r' + PEeff(r) = E = constant . Eq. 4

If there is a stable equilibrium position (where dPEeff/dr = 0), and PEeff is a minimum at this value 2 2 of r (d PE/dr │r-eq > 0), then we should be able to get circular motion (where r'=0) when r=req , since the central force (if it is attractive) can cause the centripetal acceleration - or stated alternately, where the central force inward balances the centrifugal force outward. If we provide a little more energy than necessary for this stable circular motion, we should get oscillations (in the value of r) about this stable value of req . (See the first part of the course, under Potential Wells and Oscillations.) The frequency, osc, in which r oscillates about req is given by

½ 2 2 osc = [k/m] , where k = (d PEeff/dr )│r-equilibrium . Eq. 9

Note: the above angular frequency, osc, is the frequency that r oscillates around the circular radius, requilibrium. This osc is not the same thing as ' = circ , which is the angular frequency that the motion has going around the center of the force.

Example: Given a spherically symmetric harmonic oscillator with PE = ½ kr2. (To visualize this, picture a mass on a spring in outer space – no gravity – where the spring is free to rotate about one end with the mass on the other end.) a) What is the frequency (q' = ωcirc) for circular motion at a radius of req ? b) What is the frequency (osc) for small oscillations of r about req ? c) Describe the orbits for small oscillations of r about req .

a-1) For circular motion, r = req = constant, so r' = 0 and r'' = 0, so the r component equation of nd 2 Newton’s 2 Law: F(r) = m(r''–r' ) Eq. 2 becomes, with ' = circ :

2 Fr = mar = m(-circ r) and 2 Fr = -dPE/dr = -d[½kr ]/dr = -kr . Putting these two expressions together gives: 2 ½ - kr = -mcirc r , or circ = [k/m] . 2 2 2 a-2) For a stable orbit, we want dPEeff /dr = 0 where PEeff = ½kr + L /2mr Eq. 3 so we get the condition: 2 3 4 2 dPEeff /dr = kr – L /mr = 0, or req = L /mk . 2 We have L = mvreq = m(circreq)req = mcircreq , or 2 ½ ½ circ = L/mreq = L/m(L/[mk] ) = [k/m] .

Note that circ does not depend on the radius! As the radius gets bigger, circ stays the same and the v = circ*r gets bigger.

½ b) For small oscillations about the bottom of the potential well, osc = [k’/m] 2 2 2 2 2 2 2 2 4 where k’ = (d PEeff/dr )│r-eq = (d [½ kr + L /2mr ] /dr )│req = k + 3L /mreq , so ½ 2 4 ½ 2 2 4 ½ 2 2 ½ osc = [k’/m] = [ {k + 3L /mreq }/m ] = [(k/m) + 3L /m req ] = [circ + 3corc ] , or

osc = 2circ . c) For one cycle around the circle, the radius grows bigger and gets smaller and grows bigger again twice. The origin is still right in the center. This is an oval, not an ellipse.

x x

Homework Problem #21: For an inverse square law attractive force with effective 2 2 potential energy, PEeff(r) = -K/r + L /2mr (where K>0), a) find the frequency of small radial oscillations, osc , about the equilibrium (circular) radius of orbit; b) show that this frequency of small radial oscillations, osc , is equal to the frequency of the circular orbit, circ . (This makes the circular orbit into an elliptical one instead of an oval one).

8c) Interesting special case

What happens if the force is a little different than an attractive inverse square law, say F = (-K/r2.1) r instead of F = (-K/r2) r (where K>0)? If you follow the procedure above, you’ll find 2.1 1.1 that osc = 0.95 circ for the case of F(r) = -K/r where PE(r) = -K/r instead of osc = circ for the case of F(r) = -K/r2 where PE = -K/r as in homework problem 21. You can see a plot of this orbit if you run the Gravity program on the computer homework, Volume 1 program #9. This program uses a numerical method to compute the orbit. Thus, by watching the planets orbit, we can test whether or not the force is truly an inverse square law force.