Assignment 3

Goldstein 3.10 A planet of mass M is in an orbit of eccentricity e = 1 where

so that if we construct |E0|/v20 , we can eliminate the l and k. Thus we have

KEc |E0|2mM

21 + e1 e

Approximating e = 1 where is a small number, we get

KEc 12M

m|E0|

So we have two large numbers multiplying the initial energy of the planet. This is the kinetic energy whichthe comet must have if it is to change the orbit of the planet.

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Goldstein 3.13 (a) Show that if a particle describes a circular orbit under the influence of an attractivecentral force directed toward a point on the circle, then the force varies as the inverse-fifth power of thedistance.

(b) Show that for the orbit described the total energy of the particle is zero.

(c) Find the period of the motion.

(d) Find x, y, and v as a function of angle around the circle and show that all three quantities are infiniteas the particle goes through the center of force.

(a) To set up coordinates in this problem, let us take the circular orbit to be centered on a two dimensionalx-y coordinate system at (x, y) = (a, 0) where a is the radius of the orbit. Note that this places the y-axistangent to the orbit. The particle in its orbit will have a force, ~F , always directed toward the origin inthese coordinates. We will call this direction vector ~r. At an arbitrary point on the orbit, (x, y), the particlewill have coordinates with respect to the center of the orbit of (a cos, a sin) where is the angle whichparametrizes the location of the particle on the circle. We can, of course, also parametrize the point on thecircle using usual polar coordinates, i.e. (x, y) = (r cos , r sin ). In this way, we can relate the coordinates

x = a+ a cos = r cos y = a sin = r sin

and express the constraint that the particle remain on the circle as

a2 =(r cos a)2 + r2 sin

This simplifies tor = 2a cos

This is the orbit equation.Having found the orbit equation, we can now work backwards and find the potential that must yield

this orbit. From the energy equation, we know

V (r) = E T = E (12mr2 +

l2

2mr2)

where it is still true that the angular momentum is conserved and l = mr2. We can get r from taking aderivative of the orbit equation:

r = 2a sin Substituting in, we find

V (r) = E 12m 4a2 sin2

( lmr2

)2 l

2

2mr2

= E 2a2l2

mr4(1 cos2 ) l2

2mr2

= E 2a2l2

mr4

(1 r

2

4a2) l

2

2mr2

= E 2a2l2

mr4

and the potential has an inverse quartic form. The resulting force will be an inverse fifth power of r. Theextra constant, E, is just that, a constant and corresponds to the zero of potential. We set it to zero andthereby complete part (b).

(c) To get the period of the motion, we can recall Keplers third law about equal areas being swept out inequal times. Our current problem, of course, is not the original Kepler problem with an inverse square force.

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However, this law follows from conservation of angular momentum and not from the particular power lawfor the central force. As a result, it is still true that the angular momentum is given by

l = mr2

which is still related to the areal velocity, dA/dt, by

dA

dt=

12r2

In particular, we have that dA/dt is a constant and the time, , it takes to sweep out the entire circularorbit will come from 2mA = l with A being given by pia2. Thus, we have

=2pima2

l

(d) The components of velocity in our coordinates are given by taking the time derivatives of our coordinatesabove and using the form of the orbit equation:

x = r cos r sin = 4a sin cos = 2a sin 2 l

mr2

y = r sin + r cos

= 2a cos 2l

mr2

which both blow up as the origin of coordinates (and the center of force) is approached. Note that the totalveloctiy also blows up:

|v| =x2 + y2 =

2almr2

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Goldstein 3.19 A particle moves in a force field described by the Yukawa potential

V (r) = krer/a

where k and a are positive.

(a) Write the equations of motion and reduce them to the equivalent one-dimensional problem. Use theeffective potential to discuss the qualitative nature of the orbits for different values of the energy and theangular momentum.

(b) Show that if the orbit is nearly circular, the apsides will advance approximately by pi/a per revolution,where is the radius of the circular orbit.

(a) The Lagrangian for this system is

L =12m(r2 + r22

)+k

rer/a.

continues to be a cyclic coordinate, hence l = mr2 is a conserved quantity, namely the angular momentum.The other equation of motion is

0 = mr l2

mr3+k

r

(1r+

1a

)er/a

Multiplying this by r and integrating, we get a conserved energy

E =12mr2 +

l2

2mr2 k

rer/a

and we can define an effective potential, namely

Veff(r) l2

2mr2+ V (r)

Considering the effective potential and its form, we can extract considerable information about theorbits possible in this central force field. First of all, define the dimensionless parameter b = l2/(2mka) andthe dimensionless variable x = r/a. We now write the effective potential in these dimensionless quantites as

Veff =k

a

[ bx2 e

x

x

]We know that for critical points of this potential, we can get circular orbits. In particular, we find

V eff =k

a

1x3

[x(x+ 1) ex 2b

]The critical points of the effective potential will be found for radial values where

x(x+ 1) ex = 2b

Notice that the left hand side rises from 0 to a maximum and then falls off towards 0 as x. As a result,if b is small enough, there will be two solutions to this equation, call them x1 and x2 such that x1 < x2. Itis clear from a graph of the function that the slope of the left hand side is positive at x1 and negative atx2 (the x3 prefactor in V eff will not affect the consequences of this analysis). Hence the effective potentialwill have a local minimum at x1 and a local maximum at x2. If b gets larger, these two critical points mergeinto an inflection point in the effective potential and for yet larger b values, there are no solutions to thisequation and the effective potential has no local critical points and hence will be purely repulsive for all

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incoming particles of any energy and cannot result in any bound orbits. Recall that b will be a measure ofthe angular momentum of the incoming particles.

To categorize the possible orbits, we first assume that b, or the angular momentum is small enough thatthere will be a local minimum and maximum in the effective potential. In this case, there will be both boundorbits and unbound orbits. The bound orbits will result for particles with energies Veff(x1) E < Veff(x2)and with turning points at radii such that xtp < x2. There will be a circular orbit at E = Veff(x1). Theunbound orbits come in essentially two types. For energies E > Veff(x2), particles will come in, experience asingle turning point at a radius x < x1 due to the repulsive barrier, deflect and move away from the center.There are other unbound orbits for which the energy of the particle lies below the local maximum of theeffective potential: E < Veff(x2). These are orbits that come in from infinity, reach a turning at a radiuslarger than the location of the local maximum: xtp > x2 and then move off to infinity again.

If the angular momentum is exactly such that the effective potential has an inflection point, there willbe the sole (unstable) bound orbit while the others will be unbound orbits with a single turning point.

If the angular momentum is large enough there will be no bound orbits and all particles will be unboundwith a single turning point.

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Goldstein 3.21 Show that the motion of a particle in the potential field

V (r) = kr+

h

r2

is the same as that of the motion under the Kepler potential alone when expressed in terms of a coordinatesystem rotating or precessing around the center of force.

For negative total energy, show that if the additional potential term is very small compared to theKepler potential, then the angular speed of precession of the elliptical orbit is

=2pimhl2

.

The perihelion of Mercury is observed to precess (after correction for known planetary perturbations) atthe rate of about 40 of arc per century. Show that this precession could be accounted for classically if thedimensionless quantity

=h

ka

(which is a measure of the perturbing inverse-square potential relative to the gravitational potential) wereas small as 7 108. (The eccentricity of Mercurys orbit is 0.206 , and its period is 0.24 year.)

The new part of the potential will contribute a term to the force which will have the same r-dependanceas the centripetal term: l2/mr3. Indeed, the r equation of motion becomes

0 = mr l2

mr3+

k

r2 2h

r3

while the equation of motion (conservation of angular momentum) is unchanged, i.e. l = mr2. Wecan rewrite this in the same form as the original equation for the Kepler problem with a shifted angularmomentum: l =

l2 + 2mh, or

0 = mr l2

mr3 kr2

If we now want to use l in this form, we also need to go back to the equation and realize a rescaling inthat equation:

l =l

lmr2 = mr2

where we have rescaled by a constant amount

=

1 +

2mhl2

In terms of r and , the equations take the same form as the original Kepler problem and will have thesame solutions, but now referenced to r(t) and (t) and using l for the angular momentum. In particular,the orbit equation is

1r= C

[1 + e cos

( 0

)]where

C =mk

l2

and

e =2El2

mk2

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If we want to think in terms of the original variables in the problem, r and , we write

1r= C

[1 + e cos

(1 + 2mh/l2 ( 0)

)]and we can view the rescaling of as a change in the period of the orbit given by

P =2pi

1 + 2mh/l2

This suggests the view that an orbiting particle (with e < 1) will come back to the same r value withoutnecessarily closing the orbit as will have traversed an amount less than 2pi. Hence, the orbit itself willprecess. Assuming that the ratio mh/l2 is small, the difference with the Keplerian value will be

= 2pi 2pi1 + 2mh/l2

2pi mhl2

The angular velocity of the precession will be given by / where is the time it takes to traverse theorbit. We can take it to be the unperturbed, Keplerian value. So the angular velocity, is

=2pimhl2

Notice that this is the amount that the orbit fails to catch up to its original (Keplerian) value, i.e. theprecession is backwards.

Note that the angular velocity can be expressed in terms of the orbital parameters as

=2pimhl2

=2pimh

mka(1 e2) =2pi

(1 e2)

and the effect is most pronounced for highly elliptical orbits, e.g. 1 e

Goldstein 3.22 The additional term in the potential behaving as r2 in Exercise 21 looks very much likethe centrifugal barrier term in the equivalent one-dimensional potential. Why is it then that the additionalforce term causes a precession of the orbit, while an addition to the barrier, through a change in l, does not?

The easiest way to see this is that the additional term in the potential in the last problem affects onlythe equation of motion for r. The equation of motion for (conservation of angular momentum) is leftunchanged. In consequence of this, as we saw, there must be a rescaling of the angular coordinate as well asthe angular momentum in order to interpret the equations with respect to the original problem. This led,then, to the interpretation that our orbit was precessing.

In the case where we simply increase (or decrease) the angular momentum itself, while there is a change(rescaling if you will) in the angular momentum, this change appears in both the r and equations withoutcorrespondingly requiring a rescaling of .

In equations, we have the following when we add h/r2 to the potential. The effective potential becomes

Veff =l2

2mr2 k

r+

h

r2

=l2

2mr2 k

r

where l2 = l2+2mh is the new angular momentum. Note, however, that the conserved angular momentumis technically unchanged:

l = mr2.

To incorporate the new angular momentum here, we must rescale

l = mr2( ll)

On the other hand, consider increasing the angular momentum as l = l + l0 where l0 is that angularmomentum needed to give us l (for comparisons sake). In this case, the effective potential becomes

Veff =(l + l0)2

2mr2 k

r

=l2

2mr2 k

r

but the definition of the conserved angular momentum must also shift:

l = l + l0 = mr2

In effect, the addition of angular momentum does not change the nature of the orbits in any way. Theequations can be thought of as invariant under the shift in angular momentum. The types of orbits remainthe same. However, adding an additional term to the potential, even of the same form as the centrifugalbarrier, causes a material change in the equations (in this case, in the evolution of ). We interpret thischange through a rescaling of the angle and hence a precession in the orbit.

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