Inverse Square Law as a special case of a Central Force

Both gravity, Fg = (-Gmm/r2)r, and electricity, Fe = (kqq/r2)r, are examples of a special type of central force that has an inverse square law.  Let’s apply the results of the previous section on general central force to this special case.

To account for both gravity and electricity, we will use the symbol K to represent either

(–Gm1m2) or (kq1q2).  Please note that K will be negative for gravity, while K can either be positive (two like charges) or negative (two opposite charges) for electricity.

1.  Potential Energy

Since we have a central force, we know that the force is conservative and hence we can define a potential energy:

V(r)  =  - r-standardòr F(r) dr   =   - r-standardòr (K/r2) dr  =   -[(-K/r) – (-K/rstandard)] .

The usual choice for the standard position is infinity, so V(r) = K/r .

2.  Effective Potential Energy

In the General Central Force, part 5, we defined an effective potential energy, Veff(r), as:

Veff(r)  =  V(r) + L2 /2mr2 ,

So in our case here of an inverse square central force, we have:

Veff(r) = K/r + L2/2mr2 .

so that the conservation of energy is essentially one-dimensional:

½ m r’2 + Veff(r)  =  E   =   constant .

3.  Special case:  stable equilibrium (circular motion)

In the General Central Force section, part 8, we noted that there will be a stable equilibrium position where dVeff/dr = 0 if Veff is a minimum at that location.  Let’s see if and when the inverse square law has such a stable equilibrium position:

dVeff(r)/dr  =  d[K/r + L2/2mr2]/dr  =  -K/r2 – 2L2/2mr3  =  0  (for equilibrium)

Since r, L, and m are all positive, we get two cases:

Case a)  If K>0, there is no positive value of r that satisfies the equation, so there is no stable equilibrium radius, and hence no circular motion.  This makes sense because K>0 implies a repulsive force.

Case b)  If K<0, then there is a stable value of r:  requilibirum = -L2/mK .  Note that even though there is a negative sign in this formula, the requillibrium is positive since K<0.  It makes sense that there would be a stable radius (circular motion), since K<0 implies an attractive force which balances the “centrifugal force”, or in better words, the attractive force causes the centripetal acceleration.

Let’s look at this circular motion from an energy standpoint.

E = ½ m r’2 + Veff(r)         but r’ = 0 for circular motion, so

E = Veff(requilibrium)  =  K/requilibrium + L2/2mrequilibrium2  .

Substituting in the requilibrium value from above  (req = -L2/mK), we get

E  =  K/(-L2/mK) + L2/{2m[-L2/mK]2}  =   - mK2/L2 + mK2/2L2  =  -mK2/2L2 .

Note that this energy is negative.  That means the object is “bound” by the central force.  Also note that as L gets bigger, requilibrium gets bigger and the Energy gets closer to zero.  Recall that L = mvqr  =  mr2q’, or q’ = L/mr2  .  Using requilibrium in this expression, we get:

q  =  L/{m[-L2/mK]2}   =   K2m/L3 .

Also note that vq = rq’, so the circular speed in the orbit would be:

vq  =  requilibrium q  =  (-L2/mK)*(K2m/L3)  =  -K/L ,

which says the orbital speed will be smaller as L gets larger (and requilibrium gets larger).

Note that L = r × p  so that L  =  rmvq  ,  or  vq  =  -K/L  =  -K/rmvq ,  or

vq  =  SQRT[-K/mr].  As the radius gets bigger, the circular speed gets smaller.  There is just one speed for each radius for circular motion.

Note that you only need to know one of the following to determine everything else about the circular motion: r, q’, E or L since we have the following relations among them:  requilibirum = -L2/mK , E  =  -mK2/2L2,  L = mvr = mr2q .

4.  Finding r(t) for an Inverse Square Central Force – straightforward approach

Using the procedure in part 6 under the previous General Central Force section, we have:

½ mr’2  +  Veff(r)  =  E                          (where Veff(r)  =  V(r) + L2 /2mr2 )

dr/dt  =  r’  =  [2(E-Veff(r))/m ]½

0òt dt  =  roòr {1 / [2(E-Veff(r))/m ]½ } dr .

For an inverse square law force, F = K/r2, we get (see part 2 above)

Veff(r) = K/r + L2/2mr2 .

0òt t dt  =  roòr {1 / [2(E-Veff(r))/m ]½ }dr ,        or

t    =   roòr {1 / [2(E-{ K/r + L2/2mr2} )/m ]½ }dr  .

Multiplying top and bottom of the integrand by r gives

t    =  (m/2)½ roòr {r / [Er2- Kr - L2/2m ]½ }dr  .

The above integral is of the general form:

ò {r / [cr2+br+a]½ }dr   =  [cr2+br+a]½ /c  -  (b/2c) ò {1 / [cr2+br+a]½ }dr

and

ò {1 / [cr2+br+a]½ }dr   =  (-1/[-c]½ ) arcsin{ (2cr+b) / [4ac-b2]½ ) }

as long as c < 0 and (4ac-b2) < 0 .

Therefore, we get:                     (2/m)½ t   =

[Er2- Kr - L2/2m]½ /E  +  (K/2E)*{ (-1/(-E)½)*arcsin[(2Er-K)/[(-4EL2/2m)-K2]½ ] } .

This gives t(r), but it looks like it might be really hard to get a nice expression for r(t).

5.  Find r(q) using the tricky approach (using u = 1/r).

From part 7 of the previous General Central Force section, we have:

(-m/L2u2)*F(1/u)  +  -u    =   d2u/dq2  .

For the inverse square law force, we have F  =  K/r2  =  Ku2 ,  so we now get:

-mK/L2 - u   =   d2u/dq2  ,    or     d2u/dq2  + u  =  -mK/L2  =  constant..

The inhomogeneous solution for u is simply:       uinhomogenous  =   -mK/L2  .

The homogeneous solution is simply:                  uhomogeneous  =  A cos(q-qc) .

Note:  when q = qc , uhomogeneous  is as big as it gets, and so, with r = 1/u, r is as small as it gets.  This means the when q = qc , we have the position of closest approach.

Therefore, we get:                      u(q)  =  -mK/L2  +  A cos(q-qc),      or

1/r   =   -mK/L2  +  A cos(q-qc)   ,      where L = mvqr  =  mr2q’.

6.  Initial Conditions

1/r   =   -mK/L2  +  A cos(q-qc)   ,      where L = mvqr  =  mr2q’.

To determine the constants L, A and qc in the above expression for r(q), we need to know something about the orbit.

Usually we determine the constants A and qc from the initial conditions on r, r’, q, and q’.  Note that r’ = 0 only for circular motion.  What should r’ be equal to?

Note that r’ = dr/dt.  To find this, we can take the time derivative of our basic equation:

d[1/r]/dt   =  d[ -mK/L2  +  A cos(q-qc)]/dt        or

d[1/r]/dr*dr/dt   =  d[ -mK/L2  +  A cos(q-qc)]/dq*dq/dt      which becomes

[-1/r2]r’ = -[A sin(q-qc)]*q        or

r’  =  [A sin(q-qc)]*r2q         and with mr2q’ = L, we have

r’ =    (AL/m)* sin(q-qc) .

If we know the initial radius, ro, and the initial angular speed, qo, we can use the expression L = mr2q’ to find L.

If we know the initial r, r’ and q, we can use 1/r   =   -mK/L2  +  A cos(q-qc)   and r’ =    (AL/m)* sin(q-qc)  to find  A and qc .

If we can determine the orbit, and from the orbit determine the area enclosed by the orbit, S, we can then use the expression we developed in the General Central Force section, part 8:

S = (L/2m)T                 where T is the period of orbit and S is the area of the orbit,

to determine the period of the orbit.

7.  Overview of r(q) for inverse square law force

1/r   =   -mK/L2  +  A cos(q-qc)   .

Let’s first make a short-hand substitution:   define B = -mK/L2 , so we have

1/r   =   B  +  A cos(q-qc)   .                Let’s further define  ε = A / │B│.

a)  If K>0 (repulsive force), then  B < 0 .

The radius, r, can’t be negative, therefore we must have  A > │B│, which means ε > 1 .

The minimum radius, rmin occurs when cos(q-qc) is greatest.  This means rmin occurs for q=qc, and so

1/rmin  =  -mK/L2  +  A  .

The angle at which the radius goes to infinity occurs when B + A*cos(q-qc) = 0.  Note that there will be two angles for which this occurs.  Not all angles will result in positive values for r which means that the orbit is not closed.

1/r   =   B  +  A cos(q-qc)

b)  If K<0 (attractive force), then  B>0 and then A can take on any value.  We look at four distinct cases:

b-1)  A = 0;  which implies that ε = 0,  and further:  1/r  =  B.  But this is the case where r = constant, and so is the circular motion case:    rcircular  =  1/B  =  -L2/mK .
This is the same result we obtained in part 3 above.

b-2)  A < B;  which implies that ε < 1.  Since A < B, 1/r never reaches zero, and so there is a closed orbit.  There is a minimum radius and a maximum radius:

1/rmin  =   B + A           which occurs when (q-qc) = 0 ,  and

1/rmax  =  B – A            which occurs when (q-qc) = 180o.

This turns out to be an ellipse (see the next section).

b-3)  )  A = B;  which implies that ε = 1.  Since A=B, 1/r reaches zero (r approaches infinity) only when  q-qo  = 180o .  As in case b-2, there is a minimum radius when

1/rmin  =   B + A           which occurs when (q-qc) = 0o.

Also, r approaches infinity as (q-qc) approaches 180o.

This turns out to be a parabola (see the next section).

b-4)  )  A > B;  which implies that ε > 1.  Since A > B, r approaches infinity (1/r = 0) at two angles such that  B + A cos(q-qc) = 0.  This turns out to be like case a above (for attractive force), and this is a hyperbola (again, see the next section).