216 lines
8.4 KiB
Markdown
216 lines
8.4 KiB
Markdown
§15
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Kepler's problem
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35
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Using (1), we find the energy in the form
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E
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(3)
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Hence
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Expressing the co-ordinates X2 sin o, y = l cos of the particle m2 in terms of
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by means of (2), we find that its path is an arc of an ellipse with horizontal semi-
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axis lm1/(m1+m2) and vertical semi-axis l. As M1 8 we return to the familiar simple pen-
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dulum, which moves in an arc of a circle.
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$15. Kepler's problem
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An important class of central fields is formed by those in which the poten-
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tial energy is inversely proportional to r, and the force accordingly inversely
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proportional to r2. They include the fields of Newtonian gravitational attrac-
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tion and of Coulomb electrostatic interaction; the latter may be either attrac-
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tive or repulsive.
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Let us first consider an attractive field, where
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U=-a/r
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(15.1)
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with a a positive constant. The "effective" potential energy
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(15.2)
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is
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of the form shown in Fig. 10. As r 0, Ueff tends to + 00, and as
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r
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8
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it tends to zero from negative values ; for r = M2/ma it has a minimum value
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Ueff, min = -mx2/2M2.
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(15.3)
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Ueff
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FIG. 10
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It is seen at once from Fig. 10 that the motion is finite for E <0 and infinite
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for E > 0.
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36
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Integration of the Equations of Motion
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§15
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The shape of the path is obtained from the general formula (14.7). Substi-
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tuting there U = - a/r and effecting the elementary integration, we have
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o =
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- constant.
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Taking the origin of such that the constant is zero, and putting
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P = M2/ma, e= [1 1+(2EM2/mo2)]
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(15.4)
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we can write the equation of the path as
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p/r = 1+e coso.
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(15.5)
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This is the equation of a conic section with one focus at the origin; 2p is called
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the latus rectum of the orbit and e the eccentricity. Our choice of the origin of
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is seen from (15.5) to be such that the point where = 0 is the point nearest
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to the origin (called the perihelion).
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In the equivalent problem of two particles interacting according to the law
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(15.1), the orbit of each particle is a conic section, with one focus at the centre
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of mass of the two particles.
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It is seen from (15.4) that, if E < 0, then the eccentricity e < 1, i.e. the
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orbit is an ellipse (Fig. 11) and the motion is finite, in accordance with what
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has been said earlier in this section. According to the formulae of analytical
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geometry, the major and minor semi-axes of the ellipse are
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a = p/(1-e2) = a2E, b =p/v(1-e2)Mv(2mE) =
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(15.6)
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y
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X
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2b
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ae
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2a
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FIG. 11
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The least possible value of the energy is (15.3), and then e = 0, i.e. the ellipse
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becomes a circle. It may be noted that the major axis of the ellipse depends
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only on the energy of the particle, and not on its angular momentum. The
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least and greatest distances from the centre of the field (the focus of the
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ellipse) are
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rmin = =p/(1+e)=a(1-e),
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max=p(1-e)=a(1+e). = = (15.7)
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These expressions, with a and e given by (15.6) and (15.4), can, of course,
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also be obtained directly as the roots of the equation Ueff(r) = E.
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§15
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Kepler's problem
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37
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The period T of revolution in an elliptical orbit is conveniently found by
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using the law of conservation of angular momentum in the form of the area
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integral (14.3). Integrating this equation with respect to time from zero to
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T, we have 2mf = TM, where f is the area of the orbit. For an ellipse
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f = nab, and by using the formulae (15.6) we find
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T = 2ma3/2-(m/a)
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= ma((m2E3).
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(15.8)
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The proportionality between the square of the period and the cube of the
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linear dimension of the orbit has already been demonstrated in §10. It may
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also be noted that the period depends only on the energy of the particle.
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For E > 0 the motion is infinite. If E > 0, the eccentricity e > 1, i.e. the
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the path is a hyperbola with the origin as internal focus (Fig. 12). The dis-
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tance of the perihelion from the focus is
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rmin ==pl(e+1)=a(e-1), = =
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(15.9)
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where a = p/(e2-1) = a/2E is the "semi-axis" of the hyperbola.
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y
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p
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ale-1)
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FIG. 12
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If E = 0, the eccentricity e = 1, and the particle moves in a parabola with
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perihelion distance rmin = 1p. This case occurs if the particle starts from rest
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at infinity.
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The co-ordinates of the particle as functions of time in the orbit may be
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found by means of the general formula (14.6). They may be represented in a
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convenient parametric form as follows.
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Let us first consider elliptical orbits. With a and e given by (15.6) and (15.4)
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we can write the integral (14.6) for the time as
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t
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=
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=
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38
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Integration of the Equations of Motion
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§15
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The obvious substitution r-a = - ae cos $ converts the integral to
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sioant
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If time is measured in such a way that the constant is zero, we have the
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following parametric dependence of r on t:
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r = a(1-e cos ), t =
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(15.10)
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the particle being at perihelion at t = 0. The Cartesian co-ordinates
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x = r cos o, y = r sin (the x and y axes being respectively parallel to the
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major and minor axes of the ellipse) can likewise be expressed in terms of
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the parameter $. From (15.5) and (15.10) we have
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ex = = =
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y is equal to W(r2-x2). Thus
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x = a(cos & - e),
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y = =av(1-e2) $.
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(15.11)
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A complete passage round the ellipse corresponds to an increase of $ from 0
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to 2nr.
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Entirely similar calculations for the hyperbolic orbits give
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r = a(e cosh & - 1), t = V(ma3/a)(esinh - - $),
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(15.12)
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x = a(e-cosh ) y = a1/(e2-1)sinh &
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where the parameter $ varies from - 00 to + 00.
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Let us now consider motion in a repulsive field, where
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U
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(a>0).
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(15.13)
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Here the effective potential energy is
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Utt
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and decreases monotonically from + 00 to zero as r varies from zero to
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infinity. The energy of the particle must be positive, and the motion is always
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infinite. The calculations are exactly similar to those for the attractive field.
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The path is a hyperbola (or, if E = 0, a parabola):
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pr r = =1-e coso, =
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(15.14)
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where P and e are again given by (15.4). The path passes the centre of the
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field in the manner shown in Fig. 13. The perihelion distance is
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rmin =p(e-1)=a(e+1). =
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(15.15)
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The time dependence is given by the parametric equations
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= =(ma3/a)(esinh+) =
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(15.16)
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x
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= a(cosh & e ,
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y = av((e2-1) sinh &
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§15
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Kepler's problem
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39
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To conclude this section, we shall show that there is an integral of the mo-
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tion which exists only in fields U = a/r (with either sign of a). It is easy to
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verify by direct calculation that the quantity
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vxM+ar/r
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(15.17)
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is constant. For its total time derivative is v
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since M = mr xv,
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Putting mv = ar/r3 from the equation of motion, we find that this expression
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vanishes.
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y
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0
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(I+e)
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FIG. 13
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The direction of the conserved vector (15.17) is along the major axis from
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the focus to the perihelion, and its magnitude is ae. This is most simply
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seen by considering its value at perihelion.
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It should be emphasised that the integral (15.17) of the motion, like M and
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E, is a one-valued function of the state (position and velocity) of the particle.
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We shall see in §50 that the existence of such a further one-valued integral
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is due to the degeneracy of the motion.
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PROBLEMS
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PROBLEM 1. Find the time dependence of the co-ordinates of a particle with energy E = 0
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moving in a parabola in a field U = -a/r.
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SOLUTION. In the integral
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we substitute r = M2(1+n2)/2ma = 1p(1+n2), obtaining the following parametric form of
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the required dependence:
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r=1p(1+n2),
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t=
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y=pn.
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40
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Integration of the Equations of Motion
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§15
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The parameter n varies from - 00 to +00.
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PROBLEM 2. Integrate the equations of motion for a particle in a central field
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U = - a/r2 (a > 0).
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SOLUTION. From formulae (14.6) and (14.7) we have, if and t are appropriately measured,
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(a) for E > andM 0 and
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(b) for E>0 0nd and M 2/2m a,
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(c) for E <0 and Ms1
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In all three cases
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In cases (b) and(c) the particle"falls"to the centre along a path which approaches the
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origin as
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00. The fall from a given value of r takes place in a finite time, namely
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PROBLEM 3. When a small correction SU(r) is added to the potential energy U = -a/r,
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the paths of finite motion are no longer closed, and at each revolution the perihelion is dis-
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placed through a small angle so. Find 80 when (a) SU = B/r2, (b) SU = y/r3.
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SOLUTION. When r varies from rmin to rmax and back, the angle varies by an amount
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(14.10), which we write as
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in order to avoid the occurrence of spurious divergences. We put U= - -a/r+8U, and
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expand the integrand in powers of SU; the zero-order term in the expansion gives 2nr, and
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the first-order term gives the required change so:
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(1)
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where we have changed from the integration over r to one over , along the path of the "un-
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perturbed" motion.
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In case (a), the integration in (1) is trivial: 80 = -2nBm/M2 = -2nB/ap, where 2p (15.4)
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is the latus rectum of the unperturbed ellipse. In case (b) r2SU = y/r and, with 1/r/given by
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(15.5), we have 80 = -6naym2/M4 = -6ny/ap2.
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